Fluid Dynamics–Find pipe diameter,head loss & actual mean velocity.

[Moranovich]

1. Homework Statement
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.

Please see attached diagram***

2. Homework 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

 

[Chestermiller]

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).

 

[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.

 

[Chestermiller]

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.

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

 

[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

 

[eastend111]

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

Thanks in advanced