You are Here: Home >< Physics

# Fluids Mechanics question watch

1. Water flows through a horizontal pipe.
The nozzle inlet has a cross sectional area (Ain) of 300mm2; the nozzle outlet has a cross sectional area (Aout) of 50mm2.
The inlet velocity is 5 m/s and the inlet pressure is 300Kpa.
Calculate Reynolds number based on nozzle diameter and water velocity both measured at inlet. State whether flow is laminar or turbulent.Calculate the volume flow rate (Q) and the outlet pressure (Pout).
Assume that there are no energy losses.

Hints: 1) Use Bernoulli’s equation and eliminate relevant terms (the pipe is horizontal so there is no height difference i.e. and hence no potential energy change) 2) For continuity . We can omit the density term from the Continuity equation since it will remain the same on both sides of the equation (30 marks)

This is my Physics Fluids coursework question. I have a sort of an idea how to do it if explained to me properly. Unfortunately I have no idea on what to do thanks to my teachers very strong accent. We have been told to show the working out by the way.

I feel like i'm locked in prison and the answer to this is the key to freedom.
2. (Original post by metimts)
Water flows through a horizontal pipe.
The nozzle inlet has a cross sectional area (Ain) of 300mm2; the nozzle outlet has a cross sectional area (Aout) of 50mm2.
The inlet velocity is 5 m/s and the inlet pressure is 300Kpa.
Calculate Reynolds number based on nozzle diameter and water velocity both measured at inlet. State whether flow is laminar or turbulent.Calculate the volume flow rate (Q) and the outlet pressure (Pout).
Assume that there are no energy losses.

Hints: 1) Use Bernoulli’s equation and eliminate relevant terms (the pipe is horizontal so there is no height difference i.e. and hence no potential energy change) 2) For continuity . We can omit the density term from the Continuity equation since it will remain the same on both sides of the equation (30 marks)

This is my Physics Fluids coursework question. I have a sort of an idea how to do it if explained to me properly. Unfortunately I have no idea on what to do thanks to my teachers very strong accent. We have been told to show the working out by the way.

I feel like i'm locked in prison and the answer to this is the key to freedom.
Post your working so far so we can see what you have done, or at least how you think you would go about it.
3. (Original post by metimts)
Water flows through a horizontal pipe.
The nozzle inlet has a cross sectional area (Ain) of 300mm2; the nozzle outlet has a cross sectional area (Aout) of 50mm2.
The inlet velocity is 5 m/s and the inlet pressure is 300Kpa.
Calculate Reynolds number based on nozzle diameter and water velocity both measured at inlet. State whether flow is laminar or turbulent.Calculate the volume flow rate (Q) and the outlet pressure (Pout).
Assume that there are no energy losses.

Hints: 1) Use Bernoulli’s equation and eliminate relevant terms (the pipe is horizontal so there is no height difference i.e. and hence no potential energy change) 2) For continuity . We can omit the density term from the Continuity equation since it will remain the same on both sides of the equation (30 marks)

This is my Physics Fluids coursework question. I have a sort of an idea how to do it if explained to me properly. Unfortunately I have no idea on what to do thanks to my teachers very strong accent. We have been told to show the working out by the way.

I feel like i'm locked in prison and the answer to this is the key to freedom.
Bernoulli's equation, assuming no energy losses, no heat transfer and constant density can be written in words:

[pressure energy] + [elevation energy] + [kinetic energy] at one point is equivalent to [pressure energy] + [elevation energy] + [kinetic energy] at another point.

http://udel.edu/~inamdar/EGTE215/Pipeflow.pdf

Volume flow rate is simply V = Area * Velocity.

This equation comes from mass flow rate, Q = density * Area * Velocity. With constant density, you divide through by density to obtain a volume flow rate. (Using mass = density * volume).

Mass flow rate itself is obtained from how much "mass" of fluid swept by during a short moment in time. So when you have mass = density * volume, the volume part is split into Area * length, the length coming from the distance the fluid moves in a short period of time (distance = speed * time). Substituting that in, you get mass = area * velocity * time * density. Then you get your mass flow rate by dividing through by time. See the link below:

http://web.mit.edu/16.unified/www/FA...ctures/f06.pdf

Stating whether the flow is laminar or turbulent is a strange question. You can only say that with higher Reynolds numbers, turbulent flow is more likely. This is because Reynolds number is the ratio of the inertial forces to the viscous forces. Inertia in this case really is a poor choice of words in my opinion. Its better to say "momenta" or the "advective (bulk) movement" if you like. The viscous forces represent the ability of the fluid to damp out perturbations. When the Reynolds number is high, the viscous forces have less and less influence on the large scale bulk motion and therefore the fluid may become turbulent because large scale perturbations cannot be smeared out by viscosity.But you can't say for sure a fluid is turbulent because Reynold's number is simply a non-dimensional number which characterizes part of the system. It doesn't include other parts of the system, such as small scale perturbations on a pipe walls on airfoils which might trigger turbulence earlier than may have been expected.

Please note that this does not mean viscosity becomes less important. It simply means the viscosity, that dampening out of perturbations, occurs at a much smaller fluid scale. More information about this can be found in Lumley's book on turbulence, Chapter 1.

In low Reynolds numbers, the viscosity can smear out any large scale perturbations and therefore large scale motion is unlikely to be perturbed from its original motion (set by boundary conditions), i.e. laminar flow.

Viscosity of a fluid itself is molecular in origin and explanations for its origin are found in transport phenomena textbook. Stuff to do with mean free paths and molecular momentum which I have yet to understand.
4. (Original post by metimts)
Water flows through a horizontal pipe.
The nozzle inlet has a cross sectional area (Ain) of 300mm2; the nozzle outlet has a cross sectional area (Aout) of 50mm2.
The inlet velocity is 5 m/s and the inlet pressure is 300Kpa.
Calculate Reynolds number based on nozzle diameter and water velocity both measured at inlet. State whether flow is laminar or turbulent.Calculate the volume flow rate (Q) and the outlet pressure (Pout).
Assume that there are no energy losses.

Hints: 1) Use Bernoulli’s equation and eliminate relevant terms (the pipe is horizontal so there is no height difference i.e. and hence no potential energy change) 2) For continuity . We can omit the density term from the Continuity equation since it will remain the same on both sides of the equation (30 marks)

This is my Physics Fluids coursework question. I have a sort of an idea how to do it if explained to me properly. Unfortunately I have no idea on what to do thanks to my teachers very strong accent. We have been told to show the working out by the way.

I feel like i'm locked in prison and the answer to this is the key to freedom.
Volume flow is constant Q = A1*v1 = A2*v2

z1 + P1/(rho*g) + v12/2g = z2 + P2/(rho*g) + v22/2g

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 27, 2016
The home of Results and Clearing

### 3,523

people online now

### 1,567,000

students helped last year
Today on TSR

### IT'S TODAY!

A-level results chat here

### University open days

1. Bournemouth University
Fri, 17 Aug '18
2. University of Bolton
Fri, 17 Aug '18
3. Bishop Grosseteste University