FM Jan 2020: What do we know about Fluid Mechanics and How Well Did I do? by Carrol Ng

Name_ID

carrol_ng82 — 2020-01-09 08:55:53 UTC

(Example)
Homework:
answer here.

Reflection:
to write something here.

Haziqah_18000890

2020-01-11 12:37:28 UTC

Homework:
1- PO1
Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.
2-
Newtonians fluids
= The viscosity of the fluid does not change when force applied to it ( Ex: alcohol )
Non-newtonians fluids
= The viscosity of the fluid change depends on force or pressure applied to it ( Ex: glue )

Reflection:
I have learnt the types of fluid and the properties including the formula and also there are two types of viscosity which are dynamic and kinematic.
:)

Syirin_18000899

2020-01-11 13:27:54 UTC

Homework:
> PO1: Apply knowledge of mathemathics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

>Newtonian Fluids: It doesnt matter about how big shear stress is applied on the fluids, the viscosity remains constant (if the temperature is constant). (eg:milk)

>Non Newtonian Fluids: Opposite to newtonian fluids, when the shear stress is applied on the fluids, the viscosity of the fluid changes. When a big shear stress is applied, it will become harder to get through it, when a small shear stress is applied, it will become easier to get through it. (eg: corn starch mixed with water)

Reflection:
>I have learned and understand about system and control volume [eg: open system (boiling soup in a saucepan without putting a lid], closed system (putting a lid on a saucepan while boiling a soup)
I can also differentiate the 2 types of fluids (ideal & real) and the 2 types of viscosity (dynamic and kinematic). From all the understanding, i can solve question using the suitable formula given.

Dayang_17009173

dayang_17009173 — 2020-01-11 14:45:29 UTC

Homework:
~PO1: Apply knowledge of mathemathics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

~Newtonian Fluids:
Its viscosity remains constant, no matter the amount of shear applied for a constant temperature. These fluids have a linear relationship between viscosity and shear stress.
(Ex: Alcohol)

~Non-Newtonian Fluids:
Its viscosity can change when under force to either more liquid or more solid.
(Ex: Ketchup)

~Reflection:
I have learnt and understand about the properties of fluids. I also learnt that there's 3 method of solving fluid dynamic problems which is; Analytical, Computational and Experimental.
Other than that, I think the class is fun bcs of the game we played and its good cs we didnt feel sleepy :3

Diana_18000915

2020-01-11 15:21:48 UTC

Homework :

PO1 : Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

Newtonian fluid : Substance that acts to force linearly. For example water, sea water

Non-Newtonian fluid : Substance that behaves like a solid or liquid depending on how much forces applied on it. For example ketchup, oobleck (mixture of cornstarch and water)

Reflection : I learnt the properties of fluid (liquid and gases), aerodynamic forces of fluid and fluid in engineering industries related to our daily life (open systems and closed system) and applying new equations/formulas.

Eriza_18001088

2020-01-12 00:08:19 UTC

Homework 1:
PO1 - Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

Homework 2 :
Newtonian fluid - The viscosity remains constant, regardless of how much shear is applied. Ex : Gasoline

Non-Newtonian Fluid - When shear is applied the fluid's viscosity changes. Ex : Glue

Reflection:
I have learned about the definition of fluid mechanics and able to differentiate between solid, liquid and gas phases. I also learned about open system (exchange energy and matter) and closed system (exchange only energy). Towards the end of the class, i am able to use the formulas to solve the exercises.

Chieng_17009369

2020-01-12 02:40:59 UTC

Homework 1 :
PO 1- Apply knowledge of mathematics, nature science, engineering fundamental and engineering specialisation to the solution of complex engineering problems.

Homework 2 :
Newtonian fluid- the viscosity remain constant and will not change if any force applied on it.(eg: water).

Non Newtonian fluid- the viscosity of fluid depends on the force applied on it.(eg: ketchup)

Reflection :
I have learnt the properties fluid (liquid and gas) such as viscosity (dynamic and kinematic).From what I have learnt, I can solve the question better.

Safea_18000912

safea_18000912 — 2020-01-12 03:35:22 UTC

Homework:
1. PO 1 emphasizes on applying knowledge of mathematics, natural science, engineering fundamentals and engineering specialization to the solution of complex engineering problems.

2. Difference between Homework Newtonians fluids and Non-newtonians fluids

>> Newtonians fluids: The viscosity of the fluid does not change when force applied to it (example: water, benzene)

>> Non-newtonians fluids: The viscosity of the fluid change depends on force or pressure applied to it (example: ketchup, gels)

Reflection: I've 4 new formulas in the first week of class. I've also mastered the basic concept of fluid mechanics. Active learning has really helped me understand this course better. Overall, I had fun during class and I hope my excitement towards this course will never fade. (jk)

Calvin_18000952

2020-01-12 04:05:35 UTC

Homework:
1. PO 1 - Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

2.
Newtonian Fluid - Fluid with constant viscosity. The shear rate is directly proportional to the shear stress.

Non-Newtonian Fluid - Fluid that does not follow Newton's law of viscosity. The viscosity of the fluid is dependent on shear rate or shear rate history.

Reflection:
I have learnt about the definition of fluid mechanics and able to understand why it is important in Civil Engineering. I have also learnt about the properties of fluid such as viscosity and also able to solve problems using formulas.

Siang_18001003

2020-01-12 04:20:44 UTC

Homework:
1. PO1: - Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialization to the solution of complex engineering problems.

2.
> Newtonian Fluids:
Its viscosity remains constant, no matter the amount of shear applied for a constant temperature. These fluids have a linear relationship between viscosity and shear stress.

> Non-Newtonian Fluids:
Fluid that doesn't follow the Newton's Law of viscosity. Its viscosity can change when under force to either more liquid or more solid.

> Reflection:
I have learnt the importance of studying fluid mechanics and its application in Civil Engineering. By using the formula given, we can solve the problems related to fluid mechanics.

Hakimi_18000885

2020-01-12 04:23:20 UTC

Homework :
1) PO 1 : apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialization to the solution of complex engineering problems

2) Newtonian fluids - it is a fluid that obey newton's law and show its linear relationship between shear stress and rate of angular deformation. Easy to say, if we apply a large amount of force into newtonian fluid, the viscosity does not change at all. Honey is an example of newtonian fluid.

3) Non-newtonian fluids - liquid that does not exhibit linear relationship between shear stress and rate of angular deformation. In other words, if we attempt to force an object into non newtonian fluid, the fluid become instantly viscous and it will behave like a solid. Corn starch is a great example of non newtonian fluids.
3) Reflection - i have now understand about fluid properties but still have no idea about how it will help me as a future civil engineer.

Wanah_18000797

2020-01-12 05:26:44 UTC

Homework:
1) PO1: Apply knowledge of mathemathics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

2) Newtonian Fluid- It's obey the Newton’s law of viscosity, such fluids exhibit linear relationship between shear stress and rate of angular deformation..

Non-Newtonian Fluid- Its do not follow the linear relation between shear stress and rate of angular deformation.

Reflection: I gain more knowledge about the fluids properties and have learnt new formula that can help me to solve problems that connected to it.

Shakyr_23964

2020-01-12 06:23:35 UTC

1. PO1: - Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialization to the solution of complex engineering problems.

2. Newtonian fluid:-•The fluids which obey Newton’s law of viscosity are called as Newtonian fluids such fluids exhibit linear relationship between shear stress and rate of angular deformation. E.g. - water , air etc.

3. Non Newtonian fluid:-
Fluids which do not follow the linear relation between shear stress and rate of angular deformation are termed as Non-Newtonian fluids.

4. Reflection
On the first week of lecture I have gained beneficial knowledge that related to fluid mechanics. After a brief discussion that we had with Dr Carol about the effect of temperature of liquid towards the viscosity, I can conclude that the reason for insulation of pipe in the drilling at offshore to ensure the crude oil can be transported easily.

Alya_18000917

2020-01-12 07:07:02 UTC

ketchup
1. PO1 : Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

2. Newtonian Fluids
-> viscosity remain constants, no matter the amount of shear applied (constant temperature)
-> example : water

Non-Newtonian
-> opposite of Newtonian Fluids
-> vixcosity of the fluid changes when shear is applied
-> example : ketchup

3. Reflection
I have come up with some ideas on how fluid mechanics relate with civil engineering, and also I have also learned some fluid properties.

KAI YING_17009034

2020-01-12 07:41:41 UTC

Homework :
1.
PO1 : Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

2.
---> Newtonian Fluids :
The viscosity of fluid will not affected, no matter how fast they are forced to flow through a pipe or channel. Their viscosity is independent of the rate of shear.
Example : Alcohol, water

---> Non-Newtonian Fluids :
The viscosity of the fluid changes to either more liquid or more solid when shear is applied. Their viscosity is dependent on the shear rate.
Example:
Quicksand (Viscosity increases when shear is applied),
Ketchup (Viscosity decreases when shear is applied)

Reflection:
I've learned the properties of fluid (Dynamic viscosity & Kinematic viscosity), Open & Closed System with examples, and some formulas that are needed to solve some questions. I've also understand how Fluid Mechanics is related to Civil Engineering and why we need to study it.

LEE CHEE YANG_17009078

2020-01-12 08:26:17 UTC

PO1:
HOMEWORK: 1
To acquire and apply engineering fundamentals to complex civil engineering fundamentals to complex civil engineering problems (Engineering knowledge).

HOMEWORK: 2

Newtonian Fluid- The fluid viscosity remains constant, no matter the amount of shear applied

NON-NEWTONIAN FLUIDS

When shear is applied , the viscosity of the fluid changes

REFLECTION:

I have learnt fluid properties , the type of fluid , viscosity and solving the problem with the formula .

Pei En_17010224

2020-01-12 08:29:29 UTC

PO1: Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

Newtonian fluid: It is defined as one with constant viscosity, with zero shear rate at zero shear stress , that is, the shear rate is directly proportional to the shear stress.

Non Newtonian fluid: It is defined as fluid whose viscosity is variable based on applied stress or force. The most common everyday example of a non newtonian fluid is cornstarch dissolved in water.

Reflection: I have learnt the properties of fluid through the lecture in week 1. I have learnt about system and control volume such as open system and closed system. I also can understand the type of viscosity such as dynamic and kinematics. I can solve the examples in slide using the formula that we learnt.

Syahmi_18001009

mohdsyahmisukri — 2020-01-12 08:43:56 UTC

Homework 1:
1- PO1
Apply knowledge of mathematics,natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

Homework 2:
2- Newtonians fluids
Fluid that does follow Newton's law of viscocity which remains it's constant viscocity when shear is applied for a constant temperature. For example, water,alcohol and glycerol.

3- Non-newtonians fluids
Fluid that are the opposite of Newtonians fluid which the viscocity of fluid can change when the shear is applied. For example,custard,honey and paint.

Reflection :
For the first week, I have learnt that fluid mechanics is about something that related to gas and water. I also know 4 formulas that related to fluid properties and understand well on how to use it. Fluid properties also have 3 component which system,boundary,and surroundings. System has two types - open system and closed system.

Hellen_19000903

2020-01-12 10:10:54 UTC

Homework 1
The PO 1 of BEng. Civil Engineering is : Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialization to the solution of complex engineering problems.

Homework 2
Newtonian fluids are fluids whose viscosity is constant over time and their stress - strain rate graph is linear..for example water.

Non-Newtonian fluids are fluids whose viscosity varies with time in response to the force applied. The stress - strain rate graph is either decreasing or increasing curve. For example yogurt.

Reflection
From the lessons in week 1, I learnt
i. What fluid mechanics is and it's difference from the other mechanics courses (Mechanics of Solids).
ii. The difference between an open system and closed system using examples from our day-to-day activities.
iii. The first four formulas needed in Fluid Mechanics which are crucial in Civil Engineering.

KAI CHUEN_19000374

2020-01-12 10:25:51 UTC

1) PO1

Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

2) Newtonian fluid:

A Newtonian fluid is a fluid that would have a constant viscosity no matter the amount of force or shear applied to the fluid, example: water, oil, etc

Non-Newtonian fluid:

Non-Newtonian fluid is the opposite of a Newtonian fluid whereas the viscosity of the fluid would change depending on the amount of shear applied to it. There are two types of non-Newtonian fluid, these is called dilatant and pseudoplastic. In a dilatant fluid, the viscosity of the fluid increases when shear is applied, example: corn flour and water (oobleck). While in a pseudoplastic fluid, the viscosity of the fluid decreases when shear is applied, example: ketchup.

3) reflection

Within these few days, ive learned about the properties of fluid: dynamic and kinematic viscosity. Also I have also learned about open and closed systems and differences of these 2 systems. Lastly, ive also learned different equation and the applications of those equations

Yong-18000873

2020-01-12 12:37:09 UTC

Homework:

1- PO1

-Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

2) Newtonian fluid are the fluid that its viscosity remains constant when shear is applied.Ex: Water
3) Non-Newtonian fluid are the fluid that its viscosity increases or decreases when shear is applied.Ex: Quicksand

4) For week 1 , i had been recalled some of the physics that were in foundation, I also learnt the properties of fluis and how to calculate specific weight, shear stress that involve dynamic viscosity and kinamatic viscosity.

5) What do I expect from this subject is I could learn the behaviour and the properties of fluid which I could apply in my future career.

Zulfaqar_17010089

2020-01-12 12:43:03 UTC

Homework

1. PO1
To apply the knowledge of mathematics,natural science,engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

2. Newtonians Fluids
The fluid's viscosity remains constant, no matter the amount of shear applied with a constant temperature.These kinds of fluids mainly have a linear relationship between viscosity and shear stress.Ex, water,oil and alcohol.

3. Non-Newtonians Fluids
The fluids have the opposite of the newtonians fluids which are when shear stress is applied to the fluid, the viscosity of the fluid changes.

Reflection
In the class i have understood the definition of the subject and learned the properties and phase of the fluids.I also learned how to solve problems relating to what we have learned in the class.

Lau_18000889

2020-01-12 12:47:37 UTC

Homework
1. PO1
- Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

2. NEWTONIAN FLUIDS
- Newtonian fluids are fluids that has a constant viscosity with the stresses that imposed on the fluid.

3. NON-NEWTONIAN FLUIDS
- Non Newtonian fluids are fluids that do not follow the Newton's Law of viscosity which the viscosity of the fluids will change after stresses is applied.

4. REFLECTION
- Through the 1st week of classes I have learnt some basic knowledge of Fluids Mechanics such as the fluids properties , calculation for specific gravity , specific volume and specific weight. The active learning section in class was fun and making the learning progress to become less stress. I am looking forward for more fun activity that can be carry out in the class so it will be fun while learning.

Aneeq_18001094

2020-01-12 13:33:18 UTC

Homework:
PO1 : Apply knowledge of mathematics,natural science,engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

Newtonian fluids: fluid which the viscous stresses arises from its flow, at every point, are linearly correlated to the local strain rate.

Non-newtonian fluid: fluid that does not follow newtons law of viscosity.

Reflection: I've learned that fluid is either gas or liquid. I've also learned new kind of formulas in the first class.

amira_18001056

amirasaleh — 2020-01-12 13:51:11 UTC

Homework :
1. PO1 : Apply knowledge of mathemtics, natural science, engineering fundamentals and engineering specialization to the solution of complex engineering problems.

2. a) Newtonian fluid : fluid's viscosity remains constant, despite the amount of shear applied for a constant temperature. For instance, alcohol and gasoline.
b) Non-newtonian fluid : fluid that disobey the Newton's law of viscosity. When shear is applied to non-Newtonian fluids, the viscosity of the fluid changes. For example, cornstarch and ketchup.

Reflection :
I've been enlightened about the fundamentals of Fluid Mechanics ever since Dr Carol started teaching in the first lecture. I now understand that fluid behaves in different ways and gasses are part of fluid too. In the second lecture, I too got the chance to know about methods of solving fluid dynamics problem, density and viscosity.

Tuan_18000941

2020-01-12 14:12:07 UTC

1) PO1
to apply our knowledge of mathematics, natural science, engineering fundamental and engineering specialization to the solution of complex engineering problems

2) Newtonian
-The fluids which obey Newton’s law of viscosity are called as Newtonian fluids such fluids exhibit linear relationship between shear stress and rate of angular deformation. E.g. - water , air . Then,it's a fluid in which the viscous stresses arising from its flow, at every point, are linearly correlated to the local strain rate witch is the rate of change of its deformation over time.

3) Non-Newtonian
- fluid whose viscosity is variable based on applied stress or force. The most common everyday example of a non-Newtonian fluid is cornstarch dissolved in water. Behavior of Newtonian fluids like water can be described exclusively by temperature and pressure. Fluids which do not follow the linear relation between shear stress and rate of angular deformation are termed as Non-Newtonian fluids.

-Reflection
I've learned properties of fluid mechanics and how to apply the density , viscosity and kinetic viscosity. Besides that, i've learned why civil engineering need to take mechanics of fluid and studied whats meaning by for all lecture 2.

Guan You_18001063

2020-01-12 14:16:02 UTC

1. PO1 :
Apply knowledge of mathemtics, natural science, engineering fundamentals and engineering specialization to the solution of complex engineering problems.

2. Newtonian fluid
Is a fluid in which the viscous stresses arising from its flow, at every point, are linearly correlated to the local strain rate—the rate of change of its deformation over time.

3. Non-Newtonian fluid
Is a fluid whose viscosity is variable based on applied stress or force.it does not follow the Newton's law of viscocity.

Reflection:
So far i love how Dr corral teaching style ,is quite fun .From your class i know how important is Fluid mechanic for a civil engineer.I also learnt the properties of fluid and the formulas in solving question.

2020-01-12 14:38:45 UTC

Hadzmin_17008211

Homework:
1. PO1 :
Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.
2.
i. Newtonian fluid : The viscosity of the fluid remain constant when different force applied to it in a constant temperature such as water and alcohol.
ii. Non-newtonian fluid : The viscosity of the fluid change depends on force is applied on them. Two behaviour of them can be divided into four group which is dilatant, pseudoplastic, rheopectic and thixotropic.

Reflection:
From this week, I have done an early research on fluid mechanics and i quite comfortable with the way the lecturer delivering her lecture. I'm hoping for an exciting week ahead.

Faris_18001086

2020-01-12 14:52:15 UTC

Homework 1 :
PO1
Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

Homework 2-
Newtonians fluids
A Newtonian fluid's viscosity remains constant, no matter the amount of shear applied for a constant temperature.. These fluids have a linear relationship between viscosity and shear stress.For example , water , alcohol, gasoline , mineral oil.

Non-Newtonian fluids
When shear is applied to non-Newtonian fluids, the viscosity of the fluid changes in respect to the amount of shear or stress applied to the fluid.There are four ways to describe the fluid which are dilatant ,pseudoplastic , rheopectic, thixotropic . Example for this fluids are ketchup , glue , cream.

Reflection :
I had so much fun last week . I love the way u taught us . Chill and rilex but got the point , knowledge and input . So far what I remember I learnt about the importance of fluid mechanics to civil engineers . Then I learnt about open and closed system which I can relate it with my surrounding . Next I learnt the methods for solving fluid dynamics problems.

Fatih_18001089

2020-01-12 15:19:17 UTC

Homework 1:
PO1
Apply knowledge of mathematics, natural science, engineering fundamentals and engineering specialisation to the solution of complex engineering problems.

Homework 2:
i) Newtonian Fluids
- follows the newton's law of viscosity which states that the relationship between shear stress and rate of angular deformation is directly proportional.
- the viscosity of the Newtonian fluids remain constant regardless how big the applied force is.
eg: Water

ii) Non-Newtonian Fluids
- These fluids does not obey the newton's law of viscosity
- Non-Newtonian fluids result in different viscosity when different amount of force is applied,
-When large force is applied to it,
the viscosity increases which makes the fluid harder to deform.
-When small force is applied, the viscosity decreases which makes the fluid easier to deform.
eg: corn starch

Reflection: I'll try my best to catch up as i missed the last week's lecture. So far, this course seems very interesting. I've read the lecture notes and i'm looking forward to have a better understanding with your explanation.

Beh Jie Lin_18001093

2020-01-12 15:30:44 UTC

Homework 1 :
Apply knowledge of mathematics, natural science,engineering fundamental and engineering specialisation to the solution of complex engineering problems.

Homework 2:
Newtonian fluid - the viscosity remains constant and will not change if any force applied on it. (Eg water)

Non newtonian fluid - liquid that does not exhibit linear relationship between shear stress and rate of angular deformation. (Eg corn starch)

Reflection:
throughout the two lecture in first week, I’ve learnt the learning activity that proposed by lecturer, it’s a special way that i never seen before, it’s a good try and i think that this quick learning will be always smooth and good throughout these 12weeks, and I’ve learnt the definition of fluid and the proprieties of fluid. And lastly, what’s fluid mechanics truly related to civil engineer that will helps me in future.

2020-01-13 03:44:44 UTC

Ong shi sheng_24853

Closure of week 1

2020-01-13 06:43:49 UTC

Posting after this tab will not be considered.

Eriza_18001088

2020-01-18 03:01:06 UTC

Reflection:
I've learned about

Definition of pressure
Types of pressure (Absolute,Gage, & Vacuum)
Characteristics of pressure (act perpendicular, equal values if fluid at rest, in closed vessel all parts have same pressure)
Relationship between Pressure-Density-Height (P=ρgh)
-Pressure with depth (P=Patm + ρgh)
Pascal's law (P=F/A)
Measurement of pressure (Barometer for Patm, & Manometer for pressure differences) and other pressure measurement devices.
Force is a vector that must have 3 items (magnitude, location of act, direction)
Horizontal and vertical component

Reference:
Munson, Young, Okiishi, Huebsch (2009) Fundamentals of Fluid Mechanics. Available at: http://civilcafe.weebly.com/uploads/2/8/9/8/28985467/fluid_mechanics.pdf

Syahmi_18001009

mohdsyahmisukri — 2020-01-19 06:58:18 UTC

Reflection:

For the second week, I have learnt about pressure
( P=F/A @ P = density x gh) and how to find the pressure in different level. Besides, i also know that barometer pressure is use to measure the atmospheric pressure. I also able to calculate the pressure in manometer. Hydrostatic pressure have 3 items which are magnitude, location of act and direction. For hydrostatic pressure on submerged curved surfaces, there are two components-horizontal force and vertical force.By using this formula, I am able to calculate the resultant force and know where the force will act on submerged curved surfaces.

Reference:
Batchelor, G. K. (2000).An Introduction to Fluid Dynamics. United Kingdom:Cambridge University Press. https://doi.org/10.1017/CBO9780511800955

Syirin_18000899

2020-01-19 07:31:02 UTC

Reflection:
i can summarize the definition of fluid statics: study about fluids at rest. The fluids can be liquid or gas. Hydrostatic pressure- when the fluid is liquid. the pressure exerted by a fluid when the force acting on it. Force is a vector that must have 3 items: magnitude, location of act and the direction. I also learned some formula to find force. There's different formula applied between hydrostatic pressure force on plane and curved surfaces. For curved surfaces, there's vertical and horizontal component. From these components, we can find resultant force.

Reference:
Donald F. Elger, Barbara A. Lebret, Clayton T. Crowe & John A. Roberson (2016). Engineering Fluid Mechanics 11th Edition.
http://dl140.zlibcdn.com/dtoken/189578d55397d0982bcf044c0ba3e2cf/[Donald_F._Elger,_Barbara_A._LeBret,_Clayton_T._Cr(z-lib.org).pdf

Haziqah_18000890

2020-01-19 08:03:09 UTC

Reflection:

I have learnt:

> The definition of pressure, P=F/A

> 3 types of pressure

-absolute-gage-vaccum

> Characteristic

- act perpendicular to the surface
-acting at any point are equal
- pascal law

> Barometer
(measured atmospheric)

> Manometer
(measured pressure difference)

> Force (horizontal & vertical)

- magnitude
- location
- direction

hc = centre of gravity   hp = pressure acting

Reference:

Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

Alya_18000917

2020-01-19 08:06:10 UTC

Reflection:
--> pressure,P = F/A
= p(rho)gh
--> type of pressure
1. absolute (actual pressure)
2. gage (difference between the absolute pessure and local atmospheric pressure)
3. vacuum (below atmospheric)
--> Pascal's Law
--> Pa = Pb
--> hydrostatic (fluid is a liquid)
--> aerostatic (fluid is a gas)

Reference :
T. Al-Shemmeri (2012) Engineering Fluid Mechanics. https://eprints.staffs.ac.uk/222/1/engineering-fluid-mechanics%5B1%5D.pdf

Hellen_19000903

2020-01-19 08:11:43 UTC

Reflection:
I have learnt :
i) The definition of pressure, the Si unit to be used and its formula.
ii) The types of pressure (absolute pressure, gage pressure and vacuum pressure) and the formulas that show their relationship.
iii)Pascal's law
iv) The instruments to measure pressure and how to compute the pressure using the barometer and manometer.
v) The definition and formula of hydro static pressure

Reference: Yunus A. Cengel & John Cimbala (2013). Fluid Mechanics Fundamentals and Applications. New York: McGraw-Hill. https://b-ok.cc/book/2517687/e082f9

Amira_18001056

amirasaleh — 2020-01-19 08:49:28 UTC

In week 2,I have learnt about :

- Pressure.

Pressure is a normal force exerted by a fluid per unit area.

- Types of pressure

There are 3 types of pressure which are the Vaccum, Gage and Absolute. Absolute pressure is the actual pressure at a given position. Gage pressure is the difference in pressure between the absolute pressure and the local atmospheric pressure. Vacuum pressure is the pressure below atmospheric pressure.

- Characteristics Of Pressure

1. Pressure acts perpendicularly.

2. Pressure acts equally regardless of direction.

3. Pascal’s Principle.

- Barometer and Manometer

Barometer > To measure atmospheric pressure

Manometer > To measure difference in pressure.

- Force

Vertical and Horizontal force

Reference :
R.C. Hibbeler. (2015), Fluid Mechanics. Pearson. https://b-ok.cc/ireader/2774689

Kai Ying_179009034

2020-01-19 09:07:39 UTC

Reflection:
In week 2, I have learnt about :

> PRESSURE
-- Pressure = Force / Area

-- 3 Types of pressure (Absolute Pressure, Gage Pressure, Vacuum Pressure) and the relationships & formulas between them.

P_gage = P_abs - P_atm
P_vac = P_atm - P_abs

-- Pressure = ρgh (density*gravitational force*height)

-- Barometer is used to measure the atmospheric pressure.

-- Manometer is used to measure pressure difference.

-- P₁= P_atm + ρgh (in manometer)

Reference:
Dr. R.K Bansal (2010), A Text Book of Fluid Mechanics and Hydraulics Machines (Revised Ninth Edition). New Delhi.
http://www.musaliarcollege.com/e-Books/ME/Fluid%20Mechanics%20&%20Hydraulic%20Machines.pdf

Siang_18001003

2020-01-19 13:02:58 UTC

Reflection for week 2

What have I learnt :

Pressure
- Pressure is the force acting on a surface area. Pressure at rest increases with depth as a result of added weight.

1Pa= Nm^2

P= F/A

Types of pressure
-Absolute pressure
-Gage pressure
-Vacuum pressure

-Pascal’s law: The pressure applied to a confined fluid increases the pressure throughout by the same amount.

Devices used to measure pressure:
-Barometer
-Manometer
-Deadweight tester

Reference:
Yunus Cengel &John Cimbala (2014) Fluid Mechanics: Fundamentals and Applications,

4^th Edition.
https://www.mheducation.com/highered/product/fluid-mechanics-fundamentals-applications-cengel-cimbala/M9781259696534.html

Calvin_18000952

2020-01-19 13:08:40 UTC

I've learnt about:
-Pressure:
Pressure means how much something is pushing on something else. It is expressed as force per unit area: P=F/A.

-Types of Pressure:
a. Absolute pressure
b. Gage pressure
c. Vacuum pressure

-Pascal Law
Any pressure applied to a fluid inside a closed system will transmit that pressure equally in all directions throughout the fluid.

-Vertical & Horizontal Forces in Fluid.

References:
Philip J.Pritchard, John W.Mitchell (2011). Fox and McDonald's Introduction to Fluid Mechanics, 9th Edition.
http://ftp.demec.ufpr.br/disciplinas/TM240/Marchi/Bibliografia/Pritchard-Fox-McDonalds_2011_8ed_Fluid-Mechanics.pdf

Wanah_18000797

2020-01-19 13:32:49 UTC

Reflection for week 2
I have learn :
1) Type of pressure
- Absolute, Gage, Vacuum
2) Characteristics of Pressure
- perpendicular, equal, same
3) Measurement of Pressure
-barometer, manometer
4) Hydrostatic Pressure Force
-magnitude, location of act, direction
5) how to calculate force on submerged plane surface and submerged curved surfaces

References:
Dr. R.K Bansal (2010), A Text Book of Fluid Mechanics and Hydraulics Machines (Revised Ninth Edition). New Delhi.
https://www.academia.edu/10119500/A_Text_Book_of_Fluid_Mechanics_and_Hydraulic_Machines_-_Dr._R._K._Bansal

Diana_18000915

2020-01-19 13:36:00 UTC

Reflection :

Pressure
- a normal force exercted by a fluid per unit area, P=F/A

- divided into three, which are absolute pressure, gage pressure and vacuum pressure.

- characteristics : perpendicular to the wall, must equal at any point/direction, transmitted to all part at same pressure

Equipments to measure pressure are barometer, manometer.

Two types of fluid statics which are Hydrostatics and Aerostatics. This week focused on hydrostatic force on curved surfaces. In this lecture, questions ask to determine the center of pressure, hp, center of gravity, hc, horizontal pressure force, FH , vertical pressure force, FV and other new formulas to find out resultant force and angle of inclination.

Reference :

Donald F. Elger, Barbara C. Williams, Clayton T. Crowe Washington State, John A. Roberson. Engineering Fluids Mechanics Tenth Edition. https://www.academia.edu/37581876/Engineering_Fluid_Mechanics_10th_2012_Wiley_4790_.pdf?auto=download

Reflection<3

dayang_17009173 — 2020-01-19 14:14:53 UTC

1)I have learnt:
-The definition of pressure,P=F/A (F=force, A=area)
2)Type of pressure:
-absolute pressure
-gage pressure
-vacuum pressure
3)Characteristic of pressure:
-act perpendicular to the wall
-value are equal at any point
-pascal law(P=F/A)
4)barometer
-measure atmospheric pressure
-Patm=ρgh
5)manometer
-pressure differences
6)3 components:
-magnitude, act of location and direction
7)Horizontal&vertical components

References:
Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

KAI CHUEN_19000374

2020-01-19 14:15:32 UTC

Reflection:

I have learned:

3 types of pressure, which are gage, absolute and vacuum pressure
I have also learned that pressure= force over area
I have learned that the more we go below a fluid, the higher pressure it is. To calculate the pressure, we use the formula Patm+Rgh
Manometer is used to measure pressure difference.
Pressure acts perpendicularly to the surface

Reference:

Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

Beh Jie Lin_18001093

2020-01-19 15:18:35 UTC

Reflection:
From this lecture, first i have learnt type of pressure,characteristics of pressure, measurement of pressure, etc.
Also, I learnt how to calculate the pressure in different situations with examples given by dr, throughout the examples, i learnt how to find the height with given formula which is p=h x rho x g ,in order to find the pressure.

Reference:
Donald F.Elger,Babara A.Lebret, Clayton T.Crowe& John A. Roberson(2016) Engineering fluid mechanics 11th Edition
http://dl140.zlibcdn.com/dtoken/189578d55397d0982bcf044c0ba3e2cf/%5BDonald_F._Elger,_Barbara_A._LeBret,_Clayton_T._Cr(z-lib.org).pdf

Fatih_18001089

2020-01-19 15:23:10 UTC

I've learned,
Fluid Static (Part 1)

1) There are 3 types of pressure;
- Pabs -> actual pressure
- Pgauge= Pabs - Patm
- Pvac= Patm - Pabs

2) Characteristics of pressure;
- acts perpendicular to the wall
contact to fluid
- value of pressure at any point
in a fluid at rest are equal
- F1/A1 = F2/A2

3) P= pgh

4) Measurements of pressure
Barometer
- measure Patm
Manometer
- Pressure difference

Hydrostatic Pressure force :
- no shear stresses trying to deform it
- deal with normal stress

Force:
1) Magnitude
2) Location of act
3) Direction

Plane surface :
1) Magnitude, F= pghc * A
2) Location, acting at center of pressure, hp
3) act perpendicular to wetted surface

Curved Surface:
Horizontal
1) Magnitude, F = pghc * A
2) Location - horizontally to the curved surface

Vertical
1) Magnitude, F = pgV
2) Location - vertically to curved surface, that is Xc
3) Resultant force.

Refference : Hibbler,R.C.(2015) Fluid Mechanics Pearson https://b-ok.cc/ireader/2774689

Pei En_17010224

yuen_17010224 — 2020-01-19 15:25:17 UTC

Reflection
I have learnt the definition of pressure that is p=f/A and p=pgh. There are three types of pressure that are gage, vacuum and absolute pressure. Barometer is used to measure the atmospheric pressure while manometer is used to measure the pressure difference. There are two types of force which are vertical and horizontal and they related to location, magnitude and direction.
Reference:
Bansal, R. K., & Brar, J. S. (2013). Textbook of machines in S I Units. Bangalore: Laxmi Publications.
https://books.google.com.my/books/about/A_Textbook_of_Fluid_Mechanics.html?id=FzQz6A6SnyoC&redir_esc=y

Safea_18000912

safea_18000912 — 2020-01-19 15:27:11 UTC

Reflection.
Ive learnt about pressure. the general formula for it is force over area. the unit used to quantify pressure is in pascal or newton per meter squared. besides that, i also learnt that pressure can be measured with another formula which multiplies the density with gravity and height.

there are 3 types of pressure, the absolute pressure, the gage pressure and the vacuum pressure. all these can be realated by two different equations: pgage =pabs - patm or pvac = patm - pabs.

there are also 3 characteristics of pressure. 1. presure of a fluid acts perpendicular to the wall in contact with the fluid. 2. the values of pressure acting at any point in a fluid at rest are equal regardless of the direction the force is acting on and 3, it obeys the pascal law of p1=p2.

pressure density and height can be related by the following formula: p=density x gravity x height

there is also a variation of pressure in the depth pf a liquid. when depth increases, the pressure increases. in gas, however, the variation of pressure with height can be negligible.

there are also several tools to measure pressure. a barometer is used to measure atmospheric pressure. it is al so often refered as barometric pressure. a manometer is used to measure small and moderate pressure differences. other pressure measurement devices includes: a bourdon tube, a pressure tranducer, a strain gage pressure tranducer, a prezoelectric tranducer and a dead weight tranducer.

fluid statics deal with the problems assicated with fluid when a fluid is at rest.

1. horizontal component
Fh = density x gravity x height of centroid x area
Hp = moment of inertia on the centrois over the area multiplied by the height of centroid plus the height of centroid

2. verticle component
Fv= density x gravity x volume
Xc = (A1x1 +- A2x2)/ (A1+-A2)

3, resultant force
Magnitude, F = cert(Fh2 + Fv2)

References:

Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

thank you and have a good day

Lee chee yang_17009078

2020-01-19 15:33:09 UTC

Reflection:
I've learned the Definition of pressure ,Types of pressure (Absolute,Gage, & Vacuum) and Characteristics of pressure ,pressure,P = F/A and (P=ρgh)

Reference:

Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

Hakimi_18000885

shazrielhakimiimran — 2020-01-19 15:39:27 UTC

I have learnt that there are three types of pressure, absolute pressure, gauge pressure and vacuum pressure and also learnt about the characteristics of pressure.
Next, pressure also varies with depth. Increase in depth will also increase the pressure. Besides, The apparatus that was invented to measure atmospheric and gas pressure are manometer and barometer.
Last but not least, I have learnt about static pressure that can be divided into two types which are hydrostatic for liquid and aerostatic for gases. Hydrostatic can be divided into two categories. The first one is hydrostatic pressure on submerged flat surface and the other one is hydrostatic on submerged curved surface.
Reference :
Pearson : College Physics by Hugh D.Young, Philip W. Adams, and Raymond J. Chastain

GuanYou_18001063

2020-01-19 15:44:00 UTC

For the fluid statics,i understand that there are 3 type of pressure that is absolute pressure,gage pressure and vacuum pressure.Those pressure are use according to different condition to solve the problems (by using the formula in the slide).i learnt how to calculate the pressure a different type of fluid with a certain height .

For the second class, there is a lots of formulas need to memorize and use it to calculate the pressure or hydrostatic force.Understanding the question is very important when solving this kind of question and i will try my best and keep practice.

Reference:

Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

Chieng-17009369

2020-01-19 15:47:22 UTC

Reflection :
1) I learn that what is pressure.
Formula of pressure is P=F/A or P=pgh and p is density of the liquid.
2) three types of pressure
i)absolute pressure
ii)gage pressure
iii)vacuum pressure
3)measurement of pressure
i) manometer
ii) barometer
4)vertical and horizontal force

Reference:
Fluid mechanics and hydraulic machines https://www.academia.edu/10119500/A_Text_Book_of_Fluid_Mechanics_and_Hydraulic_Machines_-_Dr._R._K._Bansal

Lau_18000889

2020-01-19 15:50:43 UTC

Reflection:
After this week, I has gained a lot of knowledge about pressure. I learned that there are three types of pressure which are gage, absolute and vacuum pressure. Besides, I also get to recalled both formula of pressure that I learned before which is P = F/A and P=ρgh. So by referring to the second formula I get to know that the greater the depth of water the higher the pressure is. Plus, I also know an apparatus called manometer can use to measure the difference of pressure. Lastly the pressure is acting perpendicularly to the surface. Through the group work I managed to get more familiar with the formula that I learned and also the correct way to solve problems that regarding to this week's topic which is pressure.

References:
Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

Hadzmin_17008211

2020-01-19 15:52:28 UTC

In this week i’ve learnt that pressure can be differentiate in 3 different type which is Absolute, Gauge and Vacuum pressure. Furthermore i’ve learnt more about application of the formula P= (pgh) and F= (pgh x A) where p=density, g=gravity acceleration, h=height and A=Area of contact. The standardise atmospheric pressure is 76mmHg and can be measured by barometer while manometer can gives us the relationship between atmospheric pressure and pressure in the fluid.

References :
Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

Yong Kung Wei_18000873

2020-01-19 15:52:28 UTC

Reflection:
After the another week, i had recalled bck my knowledges about pressure.What are new to me are for pressure are
1) Absolute Pressure
2) Gage Pressure:
Absolute pressure - Atmospheric pressure
3) Vacuum Pressure :
Atmospheric pressure - Absolute pressure
4) pressure= rho*g*h amd I used it to solve a lot of questions making me memorised it
5) Barometer- use to measure atmospheric pressure ( can used water or mercury) if using mercury(Ex: 76mmHg)
6) Manometer - measure preessure different according to the question given, sometime the density different and the depth of fluid also different
7) Hydrostatic Pressure Force= rho*g*h*Area
8) Find the location of the center of force using formula of h_p
and also recalled back how to find the centroid and the moment of inertia od=f an area.
These are all what I learned in a week and really gained a lot of knowledges.
Happy Chinese New Year!

My reference:

Fluid Mechanics.Pearson. by Hibbler: https://b-ok.cc/ireader/2774689

Faris_18001086

2020-01-19 15:52:28 UTC

Reflection:
What i learnt on week 2 is :
1) Types of pressure - gage , absolute & vacuum pressure.
2) Pressure = F/A , P = density x g x h
3) Characteristics of pressure -
i) pressure act perpendicular to the wall or surface area
ii) The values of the pressure acting at any point in a fluid at rest are equal regardless of its direction
iii) The fluid pressure applied to a fluid in a closed vessel is transmitted to all parts at the same pressure value as that applied (Pascal’s law).
4)The measurement devices
5) Example of calculations which are quite tough for me . i will try to do more exercise soon.

Reference :
Dr. R.K Bansal (2010), A Text Book of Fluid Mechanics and Hydraulics Machines (Revised Ninth Edition). New Delhi.
https://www.academia.edu/10119500/A_Text_Book_of_Fluid_Mechanics_and_Hydraulic_Machines_-_Dr._R._K._Bansal

Tuan_18000941

2020-01-19 15:57:54 UTC

Reflection:

1) I know how to apply the knowledge of fluid properties and hydrostatic to problems associated with fluid forces, pressures, stability and buoyancy. After that, the pressure have 3 differences. Its absolute pressure, gage pressure and vacuum pressure. For absolute pressure is to measure relative to absolute vacuum like absolute pressure is zero. And it’s the actual pressure at a given position. For the gage pressure it is the most pressure-measuring devices are calibrated to read zero in the atmosphere. And for the vacuum pressure is pressure below the atmosphere pressure. All of the pressure have 2 formular that can apply, its p gage = p abs –p atm and p vac= p atm –p abs.

2) I also study about pascal’s law. The pressure applied to a confined fluid increases the pressure throughout by the same amount. It have two piston that have the different area . the formular is F1A1 = F2A2.

3) The pressure of depth have in lecture last week.( p = patm +pgh)

i.ve learned about barometer but I have a bit confuse about it. The things that I know is of length or the cross-sectional area of the tube has no effect on the height of the fluid column of a barometer. Its provided that the tube diameter is large enough to avoid surface tension (capillary) effects

4) And the last thing I have learned is manometer. Its have the differences calculate beween fluid. And the pressure of fluid can be calculalte and fix in one calculation.

References :
Hibbler,R.C.(2015) Fluid Mechanics.Pearson. https://b-ok.cc/ireader/2774689

Zulfaqar_17010089

shazrielhakimiimran — 2020-01-19 15:58:18 UTC

From last week I've learned

1. There are three types of pressure which are absolute, gage and vacuum.

2.There's also some characteristics of pressure,
-acts perpendicular to the wall that contacts with
the fluid.
-value of pressure at any point in a fluid at rest
are equal.
-F1/A1 = F2/A2

3. P = force,F/area,A

4. Static pressure can also be divided into two types which are hydrostatic for liquid and aerostatic for gasses.

Reference:
Pearson : College Physics Tenth Edition
Hugh D. Young, Philip W. Adam, Raymond
J. Chastain.

Amin Hakim 18001039

2020-01-19 16:03:00 UTC

Reflection :
During week 2, I've learned about
- Definition of pressure
- P=F/A
- Absolute, Gauge, Vacuum
- P=pgh
- Characteristics of pressure
- Pressure is higher at lower position
- Pressure is equal at a leveled position
- The purpose of using barometer and manometer
- Hydrostatic pressure = pgh x A
- Not to forget, the exercises helped a lot in understanding the behavior of pressure

Aneeq_18001094

2020-01-19 16:39:26 UTC

Reflection
-During this week, I have learned a lot about pressure.
-Pressure is high at low poistion
-Pressure is low at high position
-The formula for pressure is P=F/A
-For hydrostatic pressure, P=pghA

Zhenwei 18001069

2020-01-20 07:26:18 UTC

Yajish Giri (18000834)

2020-01-20 07:28:04 UTC

Reflection:

--> pressure,P = F/A

= p(rho)gh

--> type of pressure

1. absolute (actual pressure)

2. gage (difference between the absolute pessure and local atmospheric pressure)

3. vacuum (below atmospheric)

--> Pascal's Law

--> Pa = Pb

--> hydrostatic (fluid is a liquid)

--> aerostatic (fluid is a gas)

Shakyr_2964

2020-01-20 23:31:59 UTC

For week 2 class I have learnt on how the pressure works in the pipe and also the variation of gas transfer in the various shapes

Closure

carrol_ng82 — 2020-01-27 23:34:54 UTC

Submission after this is not counted.

Alya_18000917

2020-02-22 15:27:33 UTC

REFLECTION
i have learnt about:

1.
˫ Stream flow     
˫ Flowrate/Discharge -
  volume     
˫ mass flowrate – mass              ˫ Q = AV  
    = volume/time

2.   
˫ Continuity equation - 
  no mass can be created/   
  destroyed
˫ ∑Qin-∑Qout =∆s/∆t

3.  
˫Bernouli Equation    
 - Pressure head = P/ρg 
 - Velocity head = v²/2g
 - Elevation head= z 
 - Total head,   H = P/ρg + v²/2g + z

Eriza_18001088

2020-02-22 17:15:50 UTC

Reflection:
I have learnt about

1) Flowrate (volume of fluid that passes through an area)

Q = volume/time (V/t)
Q = cross section area x flow velocity (Av)

2) Mass flowrate (mass of fluid passing through a cross-sectional area)

m ̇= mass/time (m/t)
m ̇= 𝜌Q

3) Continuity equation

∑𝑄𝑖𝑛 −∑𝑄𝑜𝑢𝑡 =∆𝑆/∆𝑡

4) Continuity equation for flow in a pipe

𝑚2 = 𝑚1
𝜌2 𝐴2 𝑉2 = 𝜌1 𝐴1 𝑉1
𝐴2 𝑉2 = 𝐴1 𝑉1 (flow is incompressible)
𝑄2 = 𝑄1

5) Bernoulli equation

P/𝜌g is the pressure head
V2/2g is the velocity or kinetic head
z or h is the elevation head
H is the total head

Haziqah_18000890

2020-02-23 06:34:47 UTC

I have learnt

1) STREAMLINE

> line drawn through the flow field that the velocity is tangent to the streamline at every point

2) TYPES OF FLUID FLOW

3) IMPORTANT TERMS

> Flowrate (volume of fluid that passes through an area/ unit time)

Q = v/t or Av (m³/s)

> Mass flowrate (mass of fluid passing through cross sectional)

m = m/t or ρQ (kg/s)

4) CONSERVATION OF ENERGY

-> CONTINUITY EQUATION

(not fully occupied with fluid)

ΣQ _in - ΣQ _out= ΔS/Δt

(fully occupied with fluid)

Q _in= Q _out

-> BERNOULLI EQUATION

Energy = head

1) pressure head = P/ρg

2) velocity head = V²/2g

3) elevation head = z or h

HGL (hydraulic grade line)

- sum of static pressure & elevation

EGL (energy grade line)

- the total head of the fluid

Syirin_18000899

2020-02-23 06:39:30 UTC

Reflection:
For this week, i have learned about:
1. Fluid flow = motion of flow
2. Streamline: line that drawn through the flow field in a such manner. The local velocity is tangent to the streamline at every point along the line.
3. Fluid flow is a part of fluid mechanics and deals with fluid dynamics.
4. Stream flow: the flow path of fluid particle. During the flow no other particle will cross flow.
5. Flowrate, Q = Volume/time (m3/s)
6. Mass flowrate = mass/time (kg/s)
7. The flow rate entering - the flow rate leaving = the rate of change of storage.
8. It is considered as steady flow when there is no change in storage (if the control volume is rigid & fully occupied by the fluid)
9. ρ2A2V2=ρ1A1V1
10. Bernoulli equation shows a relation between pressure, velocity and elevation in a frictionless flow of constant density.

Diana_18000915

2020-02-23 07:44:13 UTC

Reflection :
| Stream flow |

The flow path of fluid particle.

| Flowrate |

The flow of volume of fluid V through a surface per unit time, Q=v/t (m³/s)

| Mass flowrate |

The mass of fluid passing a point in a system per unit time, ṁ=m/t

| Conservation of Energy - Continuity Equation |
The net mass transfer to or from a control volume during a time interval ∆t is equal to the net change (increase or decrease) in the total mass within the control volume during ∆t, ΣQ_in=ΣQ_out

| Conservation Energy - Bernoulli Equation |

P/ρg : pressure head

V²/2g : kinetic head

z/h : elevation head

P + ½ ρv² + ρgh = constant

Pressure energy, kinetic energy, potential energy

- Steady and balance flow of frictionless fluid and constant density.

Tan Kai Ying_17009034

2020-02-23 12:07:43 UTC

Reflection:
I have learnt:

1. Flowrate
- volume of fluid passes through an area per unit time
- Q=volume/time
-Q=Av

2. Mass Flowrate
- mass of fluid passes through an area per unit time
- m=mass/time

3. Conservation of Mass-Continuity equation
- ΣQ _in - ΣQ _out= ΔS/Δt
- if the control volume is rigid and fully occupied, ΣQ _in = ΣQ _out4. Bernoulli Equation
- the relationship between pressure(P), velocity(V), and elevation(h) for steady flow of a frictionless fluid with a constant density.

- P₁/ρg + v₁²/2g + z₁ =
P₂/ρg + v₂²/2g + z₂+ Σh_L

Dayang_17009173

dayang_17009173 — 2020-02-23 12:51:16 UTC

i have learnt:
1) Flow rate: the volume of fluid which passes per unit time
~Q= volume/time(V/t)
~Q= Av (Area x Flow velocity)

2)Mass flowrate: mass of fluid passing through cross sectional

~m = m/t or ρQ (kg/s)

3) Continuity equation
~ ∑𝑄𝑖𝑛 −∑𝑄𝑜𝑢𝑡 =∆𝑆/∆𝑡

4)Bernoulli equation
~ pressure head =P/𝜌g
~ velocity or kinetic head = V2/2g
~ elevation head = z or h
~ total head = H

Chieng_17009369

2020-02-23 12:53:01 UTC

I have learnt :
1) flow rate
Q=v/t
Q=AV

2) mass flow rate
m=m/t

3) Continuity equation
∑𝑄𝑖𝑛 −∑𝑄𝑜𝑢𝑡 =∆𝑆/∆𝑡

4)Bernoulli equation
pressure head =P/𝜌g
velocity or kinetic head = V2/2g
elevation head = z or h
total head = H

Calvin_18000952

2020-02-26 12:19:32 UTC

I have learnt:
1. Flowrate: Volume of fluid that passes through an area per unit time.

Q = Volume/Time (V/t)
Q = Area*Flow velocity (A*v)

2. Mass flowrate: The mass of fluid passing through a cross-sectional area per unit time.

m = mass/time (m/t)
m = ρQ

3. Conversation of Mass (Continuity Equation): The flow rate entering a control volume minus the flow rate leaving the control volume must be equal to the rate of change of storage.

∑Q_in -∑Q_out =∆S/∆t

4. Bernoulli Equation: ±the relationship between pressure (P), velocity (v) and elevation (h) for steady flow of a frictionless fluid of constant density.

Siang_18001003

2020-02-26 13:05:00 UTC

I have learnt:

1. Flow rate
- volume of fluid that passes through an area per unit time.
2. Mass flow rate
- mass of fluid passing through a cross-sectional area per unit time
3. Conversation of mass
- also known as Continuity equation
- No mass can be created or destroyed
- The flow rate entering a control volume minus the flow rate leaving the control volume must be equal to the rate of change of storage.

4. Bernoulli equation
- pressure head
- velocity or kinetic head
- elevation head

Amira_18001056

amirasaleh — 2020-02-26 13:14:26 UTC

I have learnt that :-
1- Stream flow
The flow path of each of the fluid particle.
No particle will cross flow other particle during the flow.

2- Flowrate/Discharge
Flowrate, is the volume of fluid that passes through an area per unit time.
Q = volume/time (V/t) or Q = cross section area x flow velocity (Av)
Unit: m3/s

3 - Mass flowrate
Mass flowrate, is the mass of fluid passing through a cross-sectional area per unit time
m ̇= mass/time (m/t) or m ̇
Unit: kg/s

Contuinity Equation

∑𝑄𝑖𝑛 −∑𝑄𝑜𝑢𝑡 =∆𝑆/∆𝑡

Bernoulli Equation

· pressure head = P/𝜌g

· velocity or kinetic head = V2/2g

· elevation head = z or h

· total head = H

Closure of week 7

2020-02-26 13:30:08 UTC

Posting after this tab will not be considered.

Hellen_19000903

2020-03-01 15:31:25 UTC

Reflection:
I have learnt about:
The Energy Equation
H = P/ρg + v²/2g + z
where H is the total energy head.
By applying Bernoulli's equation, we get
Total energy per unit weight at 1 = Total energy per unit weight at 2
P1/ρ1g + v1²/2g + z1 = P2/ρ2g + v2²/2g + z2 + h + w - q
where
h = loss per unit weight
w = work done per unit weight
q = energy supplied per unit weight
The equations given were used in solving some applications such as
Venturi meter.

Kai Ying_17009034

2020-03-01 15:47:53 UTC

Reflection:
I have learnt

1. Energy Equation
H= P/ρg + v²/2g + z

2. Total Energy per unit Weight 1 = Total Energy per unit Weight 2
P₁/ρ₁g + v₁²/2g + z₁ =
P₂/ρ₂g + v₂²/2g + z₂ + h + w - q

3. Different types of pipe such at Venturi Meter.

Aneeq_18001094

2020-03-01 15:53:50 UTC

Energy eqn
E= P/rho(g) + v²/2g + z
I also learned how to apply the eqn in different kinds of pipe and tube.
Total Energy is also equal on both sides.

Calvin_18000952

2020-03-01 16:00:40 UTC

I have learnt:
1. The Energy Equation:
H = P/ρg + v²/2g + z

2. Total Energy per unit weight is the same on both sides.
(P/ρg + v²/2g + z) at 1 = (P/ρg + v²/2g + z + h + w - q) at 2

Kai Chuen_19000374

2020-03-01 16:32:30 UTC

I have learnt:
1. The Energy Equation:
H = P/ρg + v²/2g + z

2. The energy on the left side is equivalent to the right side
(P/ρg + v²/2g + z) at 1 = (P/ρg + v²/2g + z + h + w - q) at 2

3. The Venturi meter

Wanah_18000797

2020-03-01 17:37:56 UTC

Reflection:

I have learnt
1. Total energy per unit weight at 1 = Total energy per unit weight at 2
(P1/ρ1g + v1²/2g + z1 = P2/ρ2g + v2²/2g + z2 + h + w - q)

2. the differences when calculating the pump and venturi meter

3. the energy equation
H = P/ρg + v²/2g + z

lee chee yang 17009078

2020-03-01 18:16:35 UTC

1. energy equation
H = P/ρg + v²/2g + z
2.Total Energy per unit weight is the same on both sides.
(P/ρg + v²/2g + z) at 1 = (P/ρg + v²/2g + z + h + w - q) at 2

Beh Jie Lin_18001093

2020-03-01 18:41:56 UTC

i have learnt about:
1. The Energy Equation which is
H = P/ρg + v²/2g + z

2. Total Energy per unit weight is the same on both sides which is
(P/ρg + v²/2g + z) at 1 = (P/ρg + v²/2g + z + h + w - q) at 2

3. the pump and venturi meter

Siang_18001003

2020-03-01 19:55:41 UTC

I have learnt:
1. The Energy Equation

H = P/ρg + v2/2g + z
2. Venturi Meter

Eriza_18001088

2020-03-01 20:12:47 UTC

Reflection:
I have learnt about

1. Energy equation
H = P/ρg + v2 /2g + z

2. Total energy per unit weight at 1 = Total energy per unit weight at 2
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

Haziqah_18000890

2020-03-01 22:43:16 UTC

Reflection:
I have learnt:

> The Mechanical energy

(no energy supply)
- H = P/ρg +  v²/2g +  Z

- z₁+ P₁/ρg  + v₁²/2g = 
z₂ +P₂ /ρg  + v₂²/2g

(pump)
- z₁+ P₁/ρg  + v₁²/2g = 
z₂ +P₂ /ρg  + v₂²/2g + h + w - q

> Steady flow energy eq

ΔE = ΔQ - ΔW

> Venturi tube

Q =( A₁/(m² - 1)^1/2) sqrt(2gh( ρman/ρ-1))

Guan_18001063

2020-03-02 00:39:45 UTC

i understand how to use Energy equation which is
H = P/ρg + v²/2g + Z

and

Total energy per unit weight at 1
is equal to
Total energy per unit weight at 2
P1/ρ1g + v1²/2g + z1 =
P2/ρ2g + v2²/2g + z2 + h + w - q

KungWei_18000873

2020-03-02 01:41:19 UTC

I have learnt
1) In a Bernoulli's equation, there are potential energy, kinetic energy, and pressure energy involved.
2) The flow exiting into the air is steady, incompressible, and irrotational only that we can apply for Bernoulli's equation.
3) Bernoulli's Equation
- along a streamline
P1/ρg + ((v1^2)/2g) +z1 = P2/ρg + ((v2^2)/2g) + z2
4)
Total energy per unit weight 1 = Total energy per unit weight 2
P₁/ρ₁g + v₁²/2g + z₁ =
P₂/ρ₂g + v₂²/2g + z₂ + h + w - q

Chieng_17009369

2020-03-02 02:12:32 UTC

Reflection :
1. Energy Equation
H= P/ρg + v²/2g + z
2. Total Energy per unit weight 1 is the same with total Energy per unit weight 2
(P/ρg + v²/2g + z) at 1 = (P/ρg + v²/2g + z + h + w - q) at 2

Lau_18000889

2020-03-02 02:27:02 UTC

Reflection:
I have learnt
1. Energy equation
Formula :
H = P/ρg + v2 /2g + z
2. Total energy per unit weight at 1 is the same as total energy per unit weight at 2
Formula:
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

Pei En_17010224

2020-03-02 07:02:29 UTC

For this week, I had learnt
1) Total energy per unit weight at 1= Total energy per unit weight at 2
P1/p1g + v1^2 /2g + z1 = P2/p2g + v2^2 /2g + z2 + h + w - q
2) Energy equation
H=P/pg + v^2 /2g + z

safea_18000912

2020-03-02 13:41:21 UTC

the energy equation formula is:
H = P/ρg + v2 /2g + z;
where H is the total energy head.

the total energy per unit weight at position 1 is the same as total energy per unit weight at position 2, which is given by the formula:
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q;
where h = loss per unit weight, w = work done per unit weight, q = energy supplied per unit weight.

Faris_18001086

2020-03-02 17:14:22 UTC

Reflection:

Flowrate= volume of fluid that passes through an area

Q = volume/time (V/t)

Q = cross section area x flow velocity (Av)

Mass flowrate = mass of fluid passing through a cross-sectional area

m ̇= mass/time (m/t)

m ̇= 𝜌Q

Continuity equation

∑𝑄𝑖𝑛 −∑𝑄𝑜𝑢𝑡 =∆𝑆/∆𝑡

Continuity equation for flow in a pipe

𝑚2 = 𝑚1

𝜌2 𝐴2 𝑉2 = 𝜌1 𝐴1 𝑉1

𝐴2 𝑉2 = 𝐴1 𝑉1

𝑄2 = 𝑄1

Bernoulli equation

Relationship between pressure , velocity ,and elevation for steady flow of a fluid with a constant density.

- P1/ρg + v12/2g + z1 = P2/ρg + v22/2g + z2 + ΣhL

Faris_18001086

2020-03-02 17:30:34 UTC

1. Energy Equation
H = P/ρg + v2 /2g + z

2. Total energy per unit weight at 1 = Total energy per unit weight at 2 :
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

Alya_18000917

2020-03-03 01:53:44 UTC

i have learnt about :

1.
Total energy per unit weight at 1 = Total energy per unit weight at 2 + loss per unit weight + work done per unit weight – energy supplied per unit weight

•  P₁/ρ₁g + v₁² /2g +  z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

2. Energy differential of the system = Energy supplied to the system – energy leaving the system

•∆E = ∆Q-∆W

Diana_18000915

2020-03-03 15:16:33 UTC

Reflection :

The total energy per unit weight remains constant with frictionless fluid along a streamline.

Total energy between two point,

P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂

Energy could have been supplied by introducing a PUMP,

P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂ + h + w - q

Energy differential = Energy supplied - Energy leaving

ΔE = ΔQ - ΔW

Energy grade line (EGL),

P/ρg + v²/2g + z

The line that represents the total head of the fluid.

Hydraulic grade line (HGL),

P/ρg + z

The line that represents the sum of the static pressure and the elevation heads.

Dynamic head,

V²/2g

The difference between the heights of EGL and HGL

Syirin_18000899

2020-03-03 15:45:02 UTC

Reflection:
What i have learned on week 8:

1. Bernoulli's equation states that
points 1 and 2 lie on a streamline,
the fluid has constantly total energy per unit weight,
the flow is steady, and
there is no friction.

Total head, H = P/ρg + v2 /2g + z

Total energy per unit weight at 1 = Total energy per unit weight at 2
Assuming no energy has been supplied

P1/ρ1g + v1^2 /2g + z1 = P2/ρ2g + v2^2 /2g + z2

FOR PUMP

Total energy per unit weight at 1 = Total energy per unit weight at 2 + Loss per unit weight + Work done per unit weight - Energy supplied per unit weight

P1/ρ1g + v1^2 /2g + z1 = P2/ρ2g + v2^2 /2g + z2 + h + w - q

Application :
Venturi Tube
Plot Tube

Fatih_18001089

2020-03-04 16:33:17 UTC

Stream Flow
flow path of each of fluid particle

Flowrate
Q = Volume (V) / time (t)
Q = Cross section area (A) * flow velocity (V)

Mass Flowrate
m = mass (m) / time (t)
m = pQ

Continuity Equation
∑𝑄𝑖𝑛 − ∑𝑄𝑜𝑢𝑡 = ∆𝑆/∆𝑡
S -> volume of storage
* if the control volume is rigid & fully occupied by fluid, ∆𝑆= 0
called steady flow.
∑𝑄𝑖𝑛 = ∑𝑄𝑜𝑢𝑡

Continuity equation : PIPE
𝑚 2 = 𝑚 1
𝜌 2 𝐴 2 𝑉 2 = 𝜌 1 𝐴 1 𝑉 1
𝐴 2 𝑉 2 = 𝐴 1 𝑉 1
𝑄 2 = 𝑄 1

Bernoulli Equation
P₁/ρg + v₁²/2g + z₁ =
P₂/ρg + v₂²/2g + z₂+ Σh_L

HGL : P/ρg + z
EGL : P/ ρg + v²/2g + z
HGL - EGL : v²/2g

Zulfaqar_17010089

2020-03-04 16:37:12 UTC

1.Total energy per unit weight at 1 = Total energy per unit weight at 2
(p1/p1g + v1^2/2g + z1 = p2/p2g + v2^2/2g + z2 + h + w - q)
2.The energy equation
H = P/pg + v^2/2g + z
Where H is the total energy head

Fatih_18001089

2020-03-04 17:06:10 UTC

Mechanical enregy of a flowing fluid
* pressure generates a force, as fluid flows, cross section will move forward and work will be done.

W = F * d
steady flow of frictionless fluid along streamline
H = P/ρg + v2 /2g + z

Between any two points
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂

PUMP EQUATION :
no energy been supplied
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

Hadzmin_17008211

2020-03-05 02:25:00 UTC

so far for this week i've learnt that

Total energy per unit weight at 1 = total energy per unit weight at 2 + loss per unit weight + Work done per unit weight - Energy supplied per unit weight :

p1/p1g + v1^2/2g + z1 = p2/p2g + v2^2/2g + z2 + h + w - q

Hakimi_18000885

2020-03-05 02:40:38 UTC

during week 8, i have learn about bernoulli equation with loss.
P1/pg + V^2/2g + z1 = P2/pg + V^2/2g + z2 + h + w - q

dayang_17009173

dayang_17009173 — 2020-03-05 06:56:55 UTC

Reflection :
Total energy per unit weight at 1 = Total energy per unit weight at 2 + loss per unit weight + work done per unit weight – energy supplied per unit weight

•  P₁/ρ₁g + v₁² /2g +  z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

2. Energy differential of the system = Energy supplied to the system – energy leaving the system

•∆E = ∆Q-∆W

Tuan Amirul Ikram Bin Tuan Mohd Azhar

2020-03-05 16:46:02 UTC

18000941

i've learn for the this lecture is :

1) Flowrate (volume of fluid that passes through an area)

Q = volume/time (V/t)
Q = cross section area x flow velocity (Av)

2) Mass flowrate (mass of fluid passing through a cross-sectional area)

m ̇= mass/time (m/t)
m ̇= 𝜌Q

3) Continuity equation

∑𝑄𝑖𝑛 −∑𝑄𝑜𝑢𝑡 =∆𝑆/∆𝑡

4) Continuity equation for flow in a pipe

𝑚2 = 𝑚1
𝜌2 𝐴2 𝑉2 = 𝜌1 𝐴1 𝑉1
𝐴2 𝑉2 = 𝐴1 𝑉1 (flow is incompressible)
𝑄2 = 𝑄1

5) Bernoulli equation

P/𝜌g is the pressure head
V2/2g is the velocity or kinetic head
z or h is the elevation head
H is the total head

Tuan Amirul Ikram Bin Tuan Mohd Azhar 18000941

2020-03-05 16:49:19 UTC

On week 8 i've learn about mechanical energy of a flowing fluid. The pressure generates a force, as fluid flows, cross section will move forward and work will be done.

W = F * d
steady flow of friction less fluid along streamline
H = P/ρg + v2 /2g + z

Between any two points
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂

next i also learn about pump equation. for the formula for the no energy supplied is:

P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

no energy been supplied
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

Syahmi_18001009

mohdsyahmisukri — 2020-03-07 04:53:53 UTC

Reflection :
mechanical energy of a flowing fluid
1. total energy per unit weight at 1 = total energy per unit weight at 2
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂

2. This also could be apply to a pump. the formula is :P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂+ h + w - q

3. The application of the mechanical energy of flowing fluid are Venturi tube and Pitot tube

Eriza_18001088

2020-03-07 11:21:25 UTC

Reflection:
I have learnt about

1) Momentum = mass(m) x velocity(v)
2) Mass flow rate, m = 𝜌Av ̇
3) Continuity of mass flow equation
𝜌₁A₁v₁(v₂-v₁)=m(v₂-v₁)=Mass flow per unit time x change of velocity
4) Force, F = m(v₂-v₁)
5) Rate of change of momentum of fluid in x and y direction

6) Resultan force F_R= sqrt ( F_x² + F_y²)

7) Momentum eq. 2D flow along a streamline F = F₁+ F₂+ F₃= 𝑚(𝑣_𝑜𝑢𝑡−𝑣_𝑖𝑛)
8) Force exerted upon flat plate
F=𝜌𝐴 (𝑣−𝑢)(𝑣−𝑢)cos𝜃

Haziqah_18000890

2020-03-08 02:30:05 UTC

Reflection
I have learnt:

> Momentum equation

(*Momentum = mass x velocity*)

Change of rate of momentum

ρ₂A₂V₂= ρ₁A₁V₁

Continuity of mass flow rate

ρ₁A₁V₁(v₂- v₁)= ṁ (v₂-v₁)

F = ṁ(v₂-v₁)

3 component of force

- F1 = reaction force (always -ve)

- F2 = gravitational force (z=0)

- F3 = Force outside the control volume

Flat plate

F = PA(v-u)(v-u)cosθ

# inclination reduce, force reduce

# 0 = Force (max)

# 90’ = Force (min)

Curve plate

Fx = ṁ(v₂cosθ-v₁cosθ)

Fy = ṁ(v₂sinθ -v₁sinθ)

Fr = sqrt( fx²+ fy² )

θ = tan^-1(fy/fx)

Chieng_17009369

2020-03-08 09:27:56 UTC

I have learnt that :

1. momentum = mass* velocity
2.mass flow rate, m = 𝜌Av
3.Continuity of mass flow equation
𝜌₁A₁v₁(v₂-v₁)=m(v₂-v₁)
4. momentum equation in 2D
F=m(vin-vout)
5. force in flat plate
F=𝜌𝐴 (𝑣−𝑢)(𝑣−𝑢)cos𝜃

Kung Wei_18000873

2020-03-08 09:45:48 UTC

Reflection:
1) Momentum is the product of its mass, m and its velocity, v.

Momentum = mv
2) ρ1 A1 v1 (v2-v1 )= m(v2-v1 )
= Mass flow per unit time x Change of velocity
(for one dimensions)
3) For two dimensions
- find velocity in x and y direction
- multiply with the mass per unit time
- calculate resultant fore by using Fx and Fy.
4) Momentum equation can be used to analyse cases such as the force exerted by a jet striking on
a) flat surface F=ρA (v-u)(v-u)cos⁡θ
b) curved surface
- same equation as flat surface, but find the Vx,Vy , and then Fx,Fy, finally find the resultant force and its direction.

Lau_18000889

2020-03-08 10:41:29 UTC

Relection:
I have learnt that how to calculate the momentum of fluid flow .
1) Fluid mass flow rate :
- m = 𝜌Av
2) Continuity of mass flow equation:
- ρ1 A1 v1 (v2-v1 )= m(v2-v1 )
3) Components of fluid forces
- F1 , any forces that acting on the boundaries of the control volume.
- F2 , forces exerted by gravity
- F3 , any forces that exerted by the fluid outside the boundaries of control volume.
4) Force of the fluid

-F = m(v2-v1)
Force for plate
- F = ρA (v-u)(v-u)cos⁡θ
5) Force in 2-dimesion
- Fx = m(v2cos⁡θ-v1cos⁡θ)
- Fy = m(v2sin⁡θ-v1sin⁡θ)
- Fr = sqrt(Fy^2+Fx^2)
- θ = tan-1(Fy/Fx)

Kai Ying_17009034

2020-03-08 12:00:11 UTC

Reflection:
I have learnt

1. Momentum = mass * velocity

2. Mass flow rate = ρAv

3. Flow is 1D
-Force = m(v₂-v₁)

4. Flow in 2D
-F_x= m ( v₂cos Φ - v₁cos θ )
-F_y= m ( v₂sin Φ - v₁sin θ )
-Resultant Force = (F_x² + F_y²)^1/2
5. Flow in 3D
-F_z = m( v_z²-v_z¹ )

6. Total Force = F = F₁+F₂+F₃ = m(V_out-V_in)

7. 3 component forces
-F₁ = Force exerted by any solid body within the control volume
-F₂ = Force exerted by body forces (gravity) which acting on z-direction.
-F3 = Force exerted by the fluid outside the control volume (pressure gradient)

8. On a flat plate
-Force, F=ρA (v-u)(v-u) cos ⁡θ

9. On a curve plate
-F_x= m ( v₂cos Φ - v₁cos θ )
-F_y= m ( v₂sin Φ - v₁sin θ )
-Resultant Force = (F_x² + F_y²)^1/2- Angle = tan^-1 (F_y/F_x)

Calvin_18000952

2020-03-08 12:57:18 UTC

I have learnt:
1. Momentum: The product of its mass, m and its velocity, v.

Momentum = mv

2. Momentum Equation for Two Dimensional Flow along a Streamline

F (x,y)
= Rate of change of momentum of fluid in (x,y) direction.
= Mass per unit Time x Change of velocity in (x,y) direction

3. Three Components of Forces

F1 =
Force exerted on the fluid in the control volume by any solid body within the control volume.
F2 =
Force exerted on the fluid in the control volume by body forces such as gravity which acting on z-direction.
F3 =
Force exerted in the given direction on the fluid in the control volume by the fluid outside the control volume

4. Force Exerted by a Jet Striking a Flat Plate

5. Force Due to the Deflection of a Jet by a Curved Lane

Hellen_19000903

2020-03-08 14:07:55 UTC

Reflection:
I have learnt:
MOMENTUM

Defined as the product of the mass and velocity of a particle.
Momentum = mu

where m is the mass and u is the velocity
The momentum equation for a

2D flow along the streamline

F = (F_x² + F_y²)^1/2

where
Fx = rate of change of momentum in the x direction
= m(v₂cos(x) - v₁cos(y))
Fy = rate of change of momentum in the y direction
= m(v₂sin(x) - v₁sin(y))

F = (F_x²+ F_y² + F_z²)

where Fz = rate of change of momentum in the z direction

The momentum equation is applied
1.) Where force exerted by a jet strikes a flat plate

F = ρA(v - u)(v-u)cos (x)

2.) Where force is exerted due to the deflection of a plate by a curved vane

R = (R_x² + R_y²)^1/2

Noting that an inclination angle for the resultant force need to be found. Such as when inclined at x direction, the inclination angle is the inverse of tan(R_y/R_x)

Pei En_17010224

2020-03-08 14:42:27 UTC

I had learnt
1. Momentum=mass*velocity
2. Change of rate of momentum
P1A1V1=P2A2V2
3. Continuity of mass flow
P1A1V1(V2-V1) =m(V2-V1)
F=m(V2-V1)
4. Flat plate
F=PA(v-u) (v-u) cos@
5. Curve plate
Fx=m(V2 cos@-V1 cos@)
Fy=m(V2 sin@-V1 sin@)
F=sqrt(Fx^2 +Fy^2)

Siang_18001003

2020-03-08 16:07:31 UTC

I have learnt about

1. Momentum
F = mv

2. Momentum Equation for Two Dimensional Flow along a Streamline

3. Fluid mass flow
m ̇=ρA(v-u) cos⁡θ

4. The three component forces

F1 = Force exerted in the given direction on the fluid in the control volume by any solid body within the control volume or coinciding with the boundaries of the control volume.

F2 = Force exerted in the given direction on the fluid in the control volume by body forces such as gravity which acting on z-direction.

F3 = Force exerted in the given direction on the fluid in the control volume by the fluid outside the control volume e.g. pressure gradient.

5. FORCE DUE TO THE DEFLECTION OF A JET BY
A CURVED VANE

Jie Lin_18001093

2020-03-08 20:07:26 UTC

i've learnt about:
1. momentum= mass x velocity
2. for 1 dimension:
ρ1 A1 v1 (v2-v1 )= m(v2-v1 )
= Mass flow per unit time x Change of velocity

for 2 dimensions:
i.find velocity in x and y direction
ii. multiply with the mass per unit time
iii. calculate resultant fore by using Fx and Fy

4.Momentum equation:
can be used to analyse cases such as the force exerted by a jet striking on
flat surface &curved surface

Hadzmin_17008211

2020-03-09 01:27:16 UTC

for this week, i have learnt that ;
momentum = mass x velocity
unit : kgm/s
for fluid mass flow ;
m = pa(v-u)cos(theta)

i also learnt the partition of x direction and y direction of a fluid flow
-F_x= m(v2cos phi - v₁cos theta )
-F_y= m(v2sin phi - v₁sin theta )
-Resultant Force = (F_x² + F_y²)^1/2

Kai Chuen_19000374

2020-03-09 02:36:13 UTC

For this week I have learned

1. Momentum= mass times velocity

2. Flow in 1D

a. Force= m(v2-v1)

3. Flow in 2d

a. Fx = m ( v2 cos Φ - v1 cos θ)

b. Fy = m ( v2 sin Φ - v1 sin θ)

4. Flow in 3D

a. Fz= m(v2^2-v1^2)

5. 3 component forces

a. Force 1 = Force exerted by any solid body within the control volume

b. Force 2 = Force exerted by body forces (gravity) which acting on z-direction.

c. Force 3 = Force exerted by the fluid outside the control volume (pressure gradient)

2020-03-09 08:09:02 UTC

add Week 4 add Week 5

leecheeyang17009078

2020-03-09 08:09:28 UTC

i have learnt that ;
momentum = mass x velocity
unit : kgm/s
for fluid mass flow ;
m = pa(v-u)cos(theta)
Force= m(v2-v1)

Flow in 2d

. Fx = m ( v2 cos Φ - v1 cos θ)

Fy = m ( v2 sin Φ - v1 sin θ)

Flow in 3D

Diana_18000915

2020-03-09 13:46:01 UTC

Reflection :

Three components in Bernoulli equation - pressure head, velocity head, elevation head.

Momentum of a particle or object is defined as the product of its mass, m and its velocity, v.

(M = mv)

Conservation of momentum that utilized the Newton’s second law for a flowing fluid leads to the derivation of momentum equation.

The total forces of fluid acting in the flow direction are equal to the rate of change of momentum of the fluid flow.

Flat plate

One direction along a streamline :
ṁ =ρ₁A₁v₁₌ρ₂A₂v₂
F = ṁ(v₂-v₁)

Curved plate (Force of Flowing Fluid on Pipe Bends)

Two direction along a streamline :
F_y = ṁ(v₂sinφ -v₁sinθ) ,
F_x = ṁ(v₂cosφ -v₁cosθ)

Resultant Force

F² = F_x² + F_y² at an angle

θ= tan^-1F_y/F_x from horizontal

Syirin_18000899

2020-03-09 13:55:16 UTC

Reflection:
Definition of momentum : the product of mass and velocity for a moving body. (Momentum = mv)

When the fluid flow is assumed to be steady and nonuniform in nature,

𝜌2.𝐴2.𝑣2=𝜌1.𝐴1.𝑣1=𝑚 ̇, (fluid mass flow rate)

𝐹=𝑚 ̇(𝑣2−𝑣1 )

Fx = 𝑚 ̇(v2.cosφ -v1.cos𝜃)

Fy = 𝑚 ̇(v2.sinφ -v1.sin𝜃)

By combining the Fx and Fy, we can get the resultant force.

Fr = √Fx²+ Fy²
tan 𝜃 = Fy/Fx

Total force exerted on the fluid in a control volume in a given direction = Rate of change of momentum in the given direction of the fluid passing through the control volume

The three component forces:

F1 = solid body within the control volume

F2 = gravitational forces acting on z direction

F3 = the control volume by the fluid outside the control volume

For exerted upon the plate equal to the rate of momentum destroyed to the plate,
m ̇=ρA(v-u) cos⁡θ

F=𝜌𝐴 (𝑣−𝑢)(𝑣−𝑢)cos𝜃

Alya_18000917

2020-03-11 07:31:10 UTC

 Momentum= mass x velocity
         = mv

Change of rate of momentum

ṁ =ρ1A1v1=ρ2A2v2

F = ṁ(v2-v1)

Curve plate

Fx = ṁ(v2cosθ-v1cosθ)
Fy = ṁ(v2sinθ -v1sinθ)
Fr = sqrt( fx2 + fy2 )
θ = tan-1(fy/fx)

Flate plate

 F=𝜌𝐴 (𝑣−𝑢)(𝑣−𝑢)cos𝜃

Momentum equation 2D flow along a streamline

F = F1+ F2+ F3= 𝑚(𝑣𝑜𝑢𝑡−𝑣𝑖𝑛)

safea_18000912

2020-03-13 04:43:18 UTC

Momentum is the product of its mass, m and its velocity, v.
formula is given by: Momentum(p) = mv

F (x,y)
= Rate of change of momentum of fluid in (x,y) direction.
= Mass per unit Time x Change of velocity in (x,y) direction

there are three Components of Forces, f1, f2 and f3.

F1 is the force exerted on the fluid in the control volume by any solid body within the control volume.
F2 is the force exerted on the fluid in the control volume by body forces such as gravity which acting on z-direction.
F3 is the force exerted in the given direction on the fluid in the control volume by the fluid outside the control volume

i also learnt and understood about force xxerted by a jet striking a flat plate and the fprce uue to the deflection of a jet by a curved lane.

thank you.

Haziqah_18000890

2020-03-14 01:45:33 UTC

Reflection:

I have learnt

> There are 2 method of analysis

- Langrange fluid flow (tracking the particles along the way)

- Euler fluid flow (fix and capture)

> The flow condition
- By using Re = vdρ/ μ

- Laminar flow (Re < 2000)

- Transition (2000 < Re < 4000)

- Turbulent (Re > 4000)

> Head loss

Bernoulli’s
- z₁+ P₁/ρg  + v₁²/2g = 
z₂ +P₂ /ρg  + v₂²/2g + Σh 
- Σk = kf + Σkm

General eq
- h_L = kv²/2g
- Σh_L = (k_f + Σk_m)v²/ 2g

> Two types of head losses

- Friction head loss
H_f = k_fv²/2g

1.formula (laminar)
   f = 64/ Re
2.moody diagrams (turbulent & transition)

- Minor losses (due to pipe fittings & valve)
1. Loss coefficient          h_I = k_mv²/2g
2. Equivalent pipe length    L_ef = L + L_eq

P_l = f _L/d (ρv2)/2 - kg/m²s-

h_l = P_l/pg -m-

Eriza_18001088

2020-03-14 03:42:24 UTC

Reflection :
1) Reynolds number, Re = vdρ/ μ

2) to identify flow

-Laminar Flow: Re < 2000 
-Transition Flow: 2000 ≤ Re ≤ 4000 
-Turbulent Flow: Re > 4000

3) Head loss, h_L=kv²/2g
4) Bernoulli's eq (Head loss)

z₁+ P₁/ρg  + v₁²/2g = 
z₂ +P₂ /ρg  + v₂²/2g + Σh

5) Total head loss, Σh_L=(k_f + Σk_m) v²/2g

Siang_18001003

2020-03-14 04:11:39 UTC

Reflection :
1) Method of analysis
- Lagrange fluid flow
- Euler fluid flow
2) Reynolds number, Re = vdρ/ μ
3) Type of flow

-Laminar Flow: Re < 2000 
-Transition Flow: 2000 ≤ Re ≤ 4000 
-Turbulent Flow: Re > 4000

4)Head loss, h_L=kv²/2g

5)Bernoulli's eq (Head loss)

z₁+ P₁/ρg  + v₁²/2g = 
z₂ +P₂ /ρg  + v₂²/2g + Σh

6) Total head loss, Σh_L=(k_f + Σk_m) v²/2g

Syahmi_18001009

mohdsyahmisukri — 2020-03-14 06:54:46 UTC

Reflection :

1. momentum = mass x velocity

2. mass flow rate, m = density1 x a1 x v1 = density2 x a2 x v2

3. For curve plate, there are two direction:
Fx = m(v2 cos(phi) - v1 cos(theta) )
Fy = m(v2 sin(phi) - v1 sin(theta) )
then,we need to find the resultant force.
Fr = sqrt ( Fx^2 + Fy^2 )
for angle :- tan (theta) = Fy/Fx

Syahmi_18001009

mohdsyahmisukri — 2020-03-14 07:04:07 UTC

Reflection :

1. To know the types of flow, we need to find Reynold numbers,Re

Re = (v x d x density)/viscocity

Laminar Flow : Re < 2000
Transition Flow : 2000 <= Re <= 4000
Turbulent Flow : Re > 4000

2. to find pressure drop

Pressure drop = (f x L x density x v^2) / 2D

3. to find head loss

Head loss = pressure drop / (density x g )

Lee Chee Yang_17009078

2020-03-14 09:28:59 UTC

1) To identify flow by using Re = vdρ/ μ

-Laminar Flow: Re < 2000 
-Transition Flow: 2000 ≤ Re ≤ 4000 
-Turbulent Flow: Re > 4000
2) Bernoulli's  (Head loss)z₁+ P₁/ρg  + v₁²/2g = 
z₂ +P₂ /ρg  + v₂²/2g + Σh
3) Head loss, h_L=kv²/2g
4)The flow condition By using Re = vdρ/ μ

Pei En_17010224

2020-03-15 02:06:56 UTC

I have learnt
1. Flow rate, Q=volume/time(v/t)
Q=Av
2. Mass flow rate, m=mass/time(m/t)
m=pQ
3. Continuity equation for flow rate in pipe
m1=m2
P1A1V1=P2A2V2
A1V1=A2V2(flow is incompressible)
Q1=Q2
4. Bernoulli equation
P/rho*g
V2/2g is the velocity or kinetic head
Z or h is the elevation of head
H is the total head

KungWei_18000873

2020-03-15 03:22:28 UTC

Through this week,I have learnt
1) There are three flow conditions which are Laminar flow, Transition flow and Turbulent flow.
2)
Laminar Flow: Re < 2000

Transition Flow: 2000 Re ≤4000

Turbulent Flow: Re > 4000

3) In order to calculate the Reynolds number to determine the flow conditions, the formula is used :
Re = vdρ/ μ

4) Calculate for pressure drop
(f*L*p*v^2)/2d
-f for laminar flow = 64/Re

5) Head loss
h = pressure drop/pg

6) f can also be determined by referring to the Moody Diagram.

Lau_18000889

2020-03-15 03:38:20 UTC

Reflection:
I have learnt that
1) There are two methods of analysis which are
- Lagrange fluid flow
- Euler fluid flow
2) The flow condition of fluid can be determined by the Reynolds number.
- Laminar flow
Re < 2000
- Transition flow
2000 < Re < 4000
- Turbulent flow
Re > 4000
3) The formula to calculate Reynolds number is
- Re = vdρ/ μ
4) Friction factor can be determined from the moody diagram or calculate by using formula.
- f for laminar flow = 64/Re
5)Pressure drop formula
-Pl = (f)(L/d)((pv^2)/2)
6) Total head loss formula
- Hl = Pl/ρg

Aneeq_18001094

2020-03-15 04:30:42 UTC

I have learnt

1) STREAMLINE

> line drawn through the flow field that the velocity is tangent to the streamline at every point

2) TYPES OF FLUID FLOW

3) IMPORTANT TERMS

> Flowrate (volume of fluid that passes through an area/ unit time)

Q = v/t or Av (m³/s)

> Mass flowrate (mass of fluid passing through cross sectional)

m = m/t or ρQ (kg/s)

4) CONSERVATION OF ENERGY

-> CONTINUITY EQUATION

(not fully occupied with fluid)

ΣQ _in - ΣQ _out= ΔS/Δt

(fully occupied with fluid)

Q _in= Q _out

-> BERNOULLI EQUATION

Energy = head

1) pressure head = P/ρg

2) velocity head = V²/2g

3) elevation head = z or h

HGL (hydraulic grade line)

- sum of static pressure & elevation

EGL (energy grade line)

- the total head of the fluid

Aneeq_18001094

2020-03-15 04:32:37 UTC

Momentum is the product of its mass, m and its velocity, v.
formula is given by: Momentum(p) = mv

F (x,y)
= Rate of change of momentum of fluid in (x,y) direction.
= Mass per unit Time x Change of velocity in (x,y) direction

there are three Components of Forces, F1, F2 and F3.

F1 is the force exerted on the fluid in the control volume by any solid body within the control volume.
F2 is the force exerted on the fluid in the control volume by body forces such as gravity which acting on z-direction.
F3 is the force exerted in the given direction on the fluid in the control volume by the fluid outside the control volume

i also learnt and understood about force xxerted by a jet striking a flat plate and the fprce uue to the deflection of a jet by a curved lane.

Aneeq_18001094

2020-03-15 04:33:38 UTC

1) Re = vdρ/ μ

-Laminar Flow: Re < 2000 
-Transition Flow: 2000 ≤ Re ≤ 4000 
-Turbulent Flow: Re > 4000
2) Bernoulli's  (Head loss)z₁+ P₁/ρg  + v₁²/2g = 
z₂ +P₂ /ρg  + v₂²/2g + Σh
3) Head loss, h_L=kv²/2g
4)The flow condition By using Re = vdρ/ μ

dayang_17009172

dayang_17009173 — 2020-03-15 08:43:18 UTC

- Momentum(p)= mass(m) x velocity(v)
- Mass flow rate, m = 𝜌Av ̇
- 𝜌₁A₁v₁(v₂-v₁)=m(v₂-v₁)=Mass flow per unit time x change of velocity
- Force, F = m(v₂-v₁)
- Rate of change of momentum of fluid

- Resultan force F_R= sqrt ( F_x² + F_y²)

- Momentum eq. 2D flow along a streamline F = F₁+ F₂+ F₃= 𝑚(𝑣_𝑜𝑢𝑡−𝑣_𝑖𝑛)
- Force exerted upon flat plate
F=𝜌𝐴 (𝑣−𝑢)(𝑣−𝑢)cos𝜃

dayang_17009173

dayang_17009173 — 2020-03-15 08:45:15 UTC

- find Reynold numbers,Re
>Re = (v x d x density)/viscocity

Laminar Flow : Re < 2000
Transition Flow : 2000 <= Re <= 4000
Turbulent Flow : Re > 4000

- to find pressure drop
>Pressure drop = (f x L x density x v^2) / 2D

- to find head loss
>Head loss = pressure drop / (density x g )

JieLin_18001093

2020-03-15 10:31:00 UTC

I have leart:

- Langrange fluid flow (tracking the particles along the way)

- Euler fluid flow (fix and capture)

For The flow condition

- Re = vdρ/ μ

- Laminar flow (Re < 2000)

- Transition (2000 < Re < 4000)

- Turbulent (Re > 4000)

For Head loss

Bernoulli’s

- z1 + P1/ρg + v1²/2g =

z2 +P2 /ρg + v2²/2g + Σh

- Σk = kf + Σkm

Kai Chuen_19000374

2020-03-15 13:27:39 UTC

This week I have
learnt that
1) Reynolds number is
able to tell the condition fluid flow, whether it is smooth or turbulent
a. Laminar flow
Re < 2000
b. Transition flow
2000 < Re < 4000
c. Turbulent flow
Re > 4000

d. equation is: Re = vdρ/ μ

2) Friction factor can be found from the moody diagram or calculate by using formula.
f for laminar, flow = 64/Re

3) Total head loss formula
Hl = Pl/ρg

4) 2 methods of analysis of fluid flow
a. Lagrange fluid flow
b. Euler fluid flow

5) Pressure drop formula
Pl = (f)(L/d)((pv^2)/2)

Hellen_19000903

2020-03-15 14:17:19 UTC

Reflection:
I have learnt:

1. Two methods of analysis
i. Lagrange method: a particle of a fluid is tracked as it moves(motion and its properties)
ii. Euler method : at a certain area/space and time, the fluid is analysed (properties-pressure, density and velocity)
2. Flow condition
Reynolds number,

R_e = vdρ /μ

where
v = average velocity
d = diameter
ρ = water density
μ = water viscosity

This formulae helps to determine whether the flow is

Laminar Flow : Re < 2000
Transition Flow: 2000 ≤ Re ≤ 4000
Turbulent flow : Re > 4000

3. Head losses
There are two types:
i. Friction loss
ii. Minor losses
Σh_L = (k_f+ Σk_m) v²/2g
where k_f is friction loss and k_mis minor losses.
From Darcy-Weisbach:
k_f = 4fL/d
f = friction factor
This is obtained using:

f = 16/R_e

for Laminar flow

and Moody diagram for turbulent flow
Minor losses are due to pipe fittings and valves

Diana_18000915

2020-03-15 14:53:29 UTC

Reflection :

Determine movement of flow by ;

Lagrange method: fluid particles are “tagged”. The flow properties are determined by tracking the motion in time.

Euler method: p, ρ, v are written as functions of space and times. The flow is determined by the analysing the behaviour of the functions.

Flow condition ;

Laminar flow : parallel layers (Re<2000)

Transition flow : not in parallel layers (2000≤Re≤4000)

Turbulent : mixed flow (Re>4000)

Re = vdρ/ μ

Where average velocity v, glass tube diameter d, water density ρ and water viscosity μ.

Syirin_18000899

2020-03-15 14:56:57 UTC

Reflection:

Lagrange Method : Pieces of the fluid are "tagged". The flow properties are determined by tracking the motion and properties of the particles as they move in time.

Euler method : The fluid properties p, ρ, v are written as functions of space and times.
The flow is determined by the analysing the behaviour of the functions.

Flow condition of particles
Laminar : travel in parallel layers
Turbulent : not travel in parallel layes
Transitional : move in mixed flow, laminar + turbulent

Re = vdρ / μ

Laminar Flow : Re<2000
Transition Flow : 2000≤Re≤4000
Turbulent Flow : Re>4000

Head losses, hf
Bernoulli equation (from continuity and energy equation)

friction factor, f = 64/Re

pressure drop, ΔP = (fLρv^2)/2D

head loss, hf = f (L/D) x (v^2/2g)

Kai Ying_17009034

2020-03-15 15:25:05 UTC

Reflection :
I have learnt

1. Methods to study moment of flow
- Lagrange : The flow properties are determined by tracking the motion and properties of the particles as they move
- Euler : The flow is determined by the analyzing the behaviour of the functions

2. Flow condition
- Reynolds number, Re = vdρ/μ

- Laminar Flow : Re < 2000
----Moves along the direction of flow

-Transition Flow : 2000 <= Re <= 4000
----flow transits from laminar to turbulent at the centre of pipe.

-Turbulent Flow : Re > 4000
----only average motion of flow i parallel to pipe axis

3. Head loss
- h_L = kv²/2g

4. Friction factor
- f= 64/Re (For laminar flow)
- based on Moody Diagram

5. Relationship of Pressure drop & Head loss
- Pressure drop, ΔP_L =
f(L/D)(ρv²_avg/2)

- head loss, h_L = ΔP_L / ρg

Pei En_17010224

2020-03-16 12:23:14 UTC

I have learnt
1. Lagrange method: pieces of the fluid are "tagged". The flow properties are determined by tracking the motion and properties of the particles as they move in time.
2. Euler method: the fluid properties p, rho, v are written as functions of space and times. The flow is determined by the analysing the behaviour of the functions.
3. Reynolds number, Re=vdp/mu
Laminar flow:Re<2000
Transition flow:2000<=Re<=4000
Turbulent flow:Re>4000
4. Head loss
Bernoulli equation from continuity equation
5. f=64/Re
Pressure drop, P=fLpv^2 /2D
Head loss, h=pressure drop/pg

Calvin_18000952

2020-03-23 09:37:27 UTC

I have learnt:
1. Euler's Approach
Attach thermometer to the top of chimney, point 0 . Record T as a function of time. As different smoke particles pass through O , the temperature changes. Gives T(x0, y0, z0, t) . More thermometers to get T(x, y, z, t) .

2. Lagrange Approach
Thermometers are attached to a particle, A . End up with TA = TA(a) . Can have many particles and track T for all of them. If we also know, position of each particle of function of time, can translate Lagrange information into Euler information.

3. Reynolds number, Re:
Re = vdρ / μ

Laminar Flow: Re < 2000

Transition Flow: 2000 <= Re <= 4000

Turbulent Flow: Re > 4000

4. Two Types of Head Losses
- Friction Loss, the largest head loss

- Minor Losses

h = kv^2 / 2g

Guan_18001063

2020-03-24 04:42:23 UTC

I have learnt :

-The moment equation which is mass multiply with velocity

- moment = mv

- For continuity of mass flow equation

-p1A1v1(v2-v1) = m(v2-v1)

-mass flow per unit time x change of velocity

-mass flow rate = pAv

-For two dimension we would like to use cos & sin function to find the

Velocity in x & y direction

- Fx = m(v2cos θ - v1cos θ)

- Fy = m(v2sin θ - v1sin θ)

- Resultant force = sqrt (Fx^2 + Fy^2)

-Force exerted by a jet striking a flat plate

-F = pA(v-u)(v-u)cosθ

-For curve surface

-we need to find the velocity in x & y direction using sin & cos

And get the Fx & Fy

Guan_18001063

2020-03-24 06:01:39 UTC

I have learnt

Lagrange method = Focus on an individual particle and track it during a fluid flow, determine the characteristic.

Euler method = Analysis the behaviour of the particles as they pass through a space at a particular point

-Reynolds number ,Re can use to determine the flow condition

- Laminar Flow: Re < 2000

-Transition Flow: 2000 <= Re <= 4000

-Turbulent Flow: Re > 4000

-Re = vdp / μ

friction factor (laminar flow) can be determine by equaltion

f = 64/Re and also from the moody diagram

pressure drop, ΔP = f*(L/D)*(pV^2/2）

head loss, hf = pressure drop/pg

Chieng_17009369

2020-04-01 03:18:56 UTC

Reflection:
1. two methods of analysis
- Lagrange method
- Euler method
2. Reynolds number,Re = vdρ/μ
a. Laminar flow
Re < 2000
b. Transition flow
2000 < Re < 4000
c. Turbulent flow
Re > 4000
3. Total head loss formula
- Hl = Pl/ρg
4.Pressure drop formula
-Pl = (f)(L/d)((pv^2)/2)

Fatih_18001089

2020-04-12 03:39:26 UTC

Momentum = m * v
- fluid stream posses momentum
- the change of momentum will
occur if there is change in velocity

Mass flow rate (m) = 𝜌Av ̇
- 𝜌₁A₁v₁(v₂- v₁) = m (v₂- v₁)

Increase of momentum per uit time, causedbya force, F
F = m (v₂- v₁)
For two dimensional streamline,
Fx = m(v2cos θ - v1cos θ)

Fy = m(v2sin θ - v1sin θ)
Fr = (Fx2 + Fy2) ^ 1/2

Force exerted by jet to flat plate:
F = pA(v-u)(v-u)cosθ

For curved surface we need to resolve to X&Y direction :
Fx = m(v2cos θ - v1cos θ)

Fy = m(v2sin θ - v1sin θ)
Fr = (Fx2 + Fy2) ^ 1/2

Fatih_18001089

2020-04-12 04:25:04 UTC

Lagrange method --> pieces of fluid ''tagged''
Euler method --> written as function of space and time

To determine flow condition, depending on Reynolds number, Re:

Re = vdρ/μ
Laminar : Re < 2000
- travel in parallel
Transition : 2000 < Re < 4000
- do not travel in parallel
Turbulent : Re > 4000
- travel in mixed flow, laminar and turbulent

Pressure drop :
Pl = (f) (L/d) ( (pv^2)/2 )

Head loss:
- Friction loss, largest head loss
- Minor losses

Σh_L = (k_f+ Σk_m) v²/2g
hL = kv2 / 2g
hf = k_f v2 / 2g

kf = friction loss coefficient which are :

- Hagen - Poiseuille

kf = 64μL/(ρvd^2 )

- Blasius

kf = L/(21.55 dv^(0.6) )

- Hazen - Williams

kf = (133.9L)/(C^(1.85) d^(1.165) v^(0.15) )

- Darcy- Weisbach
kf=4fL/d

values of f can be determined using:
- Formula
- Moody diagram

Hadzmin_17008211

2020-04-12 12:38:33 UTC

On this week, I found out that

1. Flow condition :
- Turbulent flow
-Laminar flow
Transition flow

2. Laminar Flow ~> Re < 2000

Transition Flow ~> 2000 Re ≤4000

Turbulent Flow ~> Re > 4000

3. Re = vdρ/ μ

4. Pressure drop :

(f*L*p*v^2)/2d

-laminar flow (f) = 64/Re

5) Head loss :

h = pressure drop/pg

Amira_18001056

amirasaleh — 2020-04-12 15:10:27 UTC

In week 8, I've learnt that :
The Energy Equation
H = P/ρg + v²/2g + z

Total energy per unit weight at 1 = Total energy per unit weight at 2
P₁/ρ₁g + v₁² /2g + z₁ = P₂/ρ₂g + v₂² /2g + z₂

Amira_18001056

amirasaleh — 2020-04-12 15:15:12 UTC

1. Momentum = mass x velocity
2. Mass flow rate, m = 𝜌Av
3. Continuity of mass flow equation = 𝜌₁A₁v₁(v₂-v₁) = m(v₂-v₁)
4. Flow in 2D :-
F_x= m ( v₂cos Φ - v₁cos θ )
F_y= m ( v₂sin Φ - v₁sin θ )
Resultant Force = (F_x² + F_y²)^1/25. Flow in 3D : F_z = m( v_z²-v_z¹ )
6. force in flat plate
F=𝜌𝐴 (𝑣−𝑢)(𝑣−𝑢)cos𝜃

Amira_18001056

amirasaleh — 2020-04-12 15:21:30 UTC

I have learnt that :-

• Lagrange method: Pieces of the fluid are “tagged”.
The flow properties are determined by tracking the motion and properties of the particles as they move in time.

• Euler method: The fluid properties p, ρ, v are written as functions of space and times.
The flow is determined by the analysing the behaviour of the functions.

• Reynolds number, Re:
Laminar Flow: Re < 2000 Transition Flow: 2000 < Re > 4000
Turbulent Flow: Re > 4000

• Two types of head losses: Friction Loss, the largest head loss

Minor Losses

Zulfaqar_17010089

2020-04-12 15:40:01 UTC

I have learnt
1.Momentum = mv
as in mass times with velocity

2.Mass flow rate = ρAv
3.Continuity of mass flow equation = change of mass of flow rate = 𝜌₁A₁v₁(v₂-v₁) = m (v₂-v₁)

4.Knowing the eqn is = rate of change, it can be separated into x-direction and y-direction.

5. F_x= m ( v₂cos Φ - v₁cos θ ) F_y= m ( v₂sin Φ - v₁sin θ )

6.Upon separating, to combine these :

F_R = (F_x² + F_y²)^1/2

Zulfaqar_17010089

2020-04-12 15:54:06 UTC

I have learnt that

1.There are two methods to study the movement of flow:
i. Langrange Method
ii.Euler Method
2. The flow condition of a fluid is determined by the Reynold's Number with the range of :

Laminar Flow:
Re < 2000
Transition Flow:
2000 < Re > 4000
Turbulent Flow:
Re > 4000

using the formula,R_e

vdρ/μ

3.Noting that there are
i.Head Loss
ii.Friction Head Loss
iii.Minor Loss

Posting after this tab will not be considered.

carrol_ng82 — 2020-04-13 07:52:05 UTC

carrol_ng82 — 2020-04-13 07:52:10 UTC

Posting after this tab will not be considered.

carrol_ng82 — 2020-04-13 07:52:15 UTC

Posting after this tab will not be considered.

carrol_ng82 — 2020-04-13 07:52:20 UTC

Posting after this tab will not be considered.