# Homework Help: Fluid Dynamics–Find pipe diameter,head loss & actual mean velocity.

1. Jul 17, 2013

### Moranovich

1. The problem statement, all variables and given/known data
a) Calculate the theoretical diameter of pipe, nominal pipe diameter and mean velocity of nominal pipe diameter.

Maximum velocity = 1.8m/s
Q(vol. flow rate) = 0.01M^3-1

b) Find the Reynolds number for the process and then the head loss due to friction, state any assumptions made.

2. Relevant equations

D=sqrrt 4*q/Pi*U
q=volumetric flow rate
D=theoretical diameter

Q=V*A
V=Q/A
V=mean velocity
A=cros sect area
Q=vol. flow rate

Re=um*d*P/U
Where um=mean velocity
d = diameter of pipe
p = density
U = dynamic viscosity of fluid
Re= Reynolds number

Head loss due to friction Hf=32*U*L*um/P*g*d^2
U = dynamic viscosity of fluid
L length of pipe from the pump = 27m
P= density taken from previous questions I have done for oil= 860Kgm-3
g = gravity 9.81m/s
d = diameter of theoretical pipe =120mm

3. The attempt at a solution

I have worked out the theoretical diameter to be 120mm from the first equation assuming that the mean velocity is roughly half the maximum velocity ie 0.9m/s

So sqrrt 4*q/Pi*U = sqrrt 4*0.01/3.142*0.9=0.12m
Then choosing from the table nominal pipe 100mm the actual velocity for the nominal pipe is 1.27m/s using this eq.

Q=V*A
V=Q/A
V=0.01/7.85*10^-3=1.27m/s

To work out the Reynolds number I have used the density and dyn. viscosity from previous questions (860kgm & 0.032)
Re=0.9*0.12*860/0.032=2902.5

To work out the head loss I have used these density and dyn. Viscosity figures again and worked out from the diagram that (from the pump) the length of pipe is 27M.
So Hf=32*0.032*27*0.9/860*9.81*0.12^2=0.205M

So head loss due to friction is 0.205M.

I am not sure if I should use these assumptions i.e density and viscosity that are not given in the question. Am I on the right track or away on a crazy tangent??!!

Any help would be much appreciated – thank you in advance.

Cheers

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2. Jul 17, 2013

### Staff: Mentor

You're on the right track, but you used the pressure drop equation for laminar flow, and the flow is turbulent (for the viscosity and density you assumed).

3. Jul 19, 2013

### Moranovich

Thanks for your comment Chestermiller much appreciated.

So I have now used the following Darcy equation for turbulent flow
Hf=4LCfUm^2/2gd

(Cf obtained from Moody diagram)

So = 4*27*0.004*0.9^2 / 2*9.81*0.12
Hf=0.143m of fluid.

Is this the correct method??

Cheers.

4. Jul 19, 2013

### Staff: Mentor

Yes. This looks right. (But, I haven't checked your arithmetic)

5. Oct 24, 2013

### ajoyce

Hi there,

I am looking to understand the methods used in questions c) and d)

Can you provide any insight into how i can answer a question similar to these

6. Jan 23, 2017

### eastend111

Why is the mean velocity roughly half of the maximum velocity? Does this apply to laminar/turbulent/transitional flows?