Parallel Flow Exercise Homework: Find QB Value

[williamcarter]

Homework Statement

I would really appreciate if you could give me some hints regarding what exactly to iii)

Homework Equations

we know they are in parallel
so Q=Q1+Q2+...+Qn
delta hloss=delta hloss1+deltahloss2+...
delta P=delta P1=delta P2=...

The Attempt at a Solution

We know flow is turbulent hence
delta h loss=32*f*L*Q^2/pi^2*g*D^5
where f=fanning friction factor
L=length
Q=vol flowrate
g=gravit accel
D=diameter

We have 3 pipes in parallel
so Q1=QA+QB+QC

I need to show that QB~0.3Q1

 

[Chestermiller]

What are the limiting value of the fanning friction factor at high Reynolds numbers for surface roughnesses of 0.001, 0.007, and 0.035? Using these values, what is the $\Delta P$ for each pipe as a function of Q for that pipe?

 

[williamcarter]

What are the limiting value of the fanning friction factor at high Reynolds numbers for surface roughnesses of 0.001, 0.007, and 0.035? Using these values, what is the $\Delta P$ for each pipe as a function of Q for that pipe?

Well, we know that Q=pi*R^4/8Mew * deltaP/L but I guess that was for laminar flow , so it won't work here.

 

[Chestermiller]

Well, we know that Q=pi*R^4/8Mew * deltaP/L

That's only for laminar flow.

 

[williamcarter]

That's only for laminar flow.

for Turbulent we know delta h loss=32*f*L*Q^2/pi^2*g*D^5
delta h loss=K*Q^2

 

[Chestermiller]

for Turbulent we know delta h loss=32*f*L*Q^2/pi^2*g*D^5
delta h loss=K*Q^2

We will be working with the first formula. So, please answer my questions in post #2.

 

[williamcarter]

We will be working with the first formula. So, please answer my questions in post #2.

I will be using the moody chart to get the fanning friction factors:
for relative rougnhess of 0.001 we have f=0.005
for relative roughness of 0.007 we have f=0.009
for relative roughness of 0.035 we have f=0.015

Now delta P as a function of Q for each pipe
Because it is in parallel delta PA=delta PB=delta PC

delta P=32*ro*f*L*Q^2/pi^2*D^5

 

[Chestermiller]

I will be using the moody chart to get the fanning friction factors:
for relative rougnhess of 0.001 we have f=0.005
for relative roughness of 0.007 we have f=0.009
for relative roughness of 0.035 we have f=0.015

Now delta P as a function of Q for each pipe
Because it is in parallel delta PA=delta PB=delta PC

delta P=32*ro*f*L*Q^2/pi^2*D^5

OK, now, for the friction factors of the 3 pipes and the given dimensions of the three pipes, express $\Delta P/\rho$ in terms of Q for each of the three pipes.

 

[williamcarter]

OK, now, for the friction factors of the 3 pipes and the given dimensions of the three pipes, express $\Delta P/\rho$ in terms of Q for each of the three pipes.

delta P/ro=(32*f*L/pi^2*D^5)*Q^2

and deltaPA=deltaPB=deltaPC because in parallel

for pipe A
delta PA/ro=(32*0.005*L/pi^2*D)*Q^2

 

[Chestermiller]

delta P/ro=(32*f*L/pi^2*D^5)*Q^2

and deltaPA=deltaPB=deltaPC because in parallel

I need you to plug in the numbers.

 

[williamcarter]

I need you to plug in the numbers.

for pipe A
delta PA/ro=(32*0.005*L/3.14^2*D)*Q^2=(0.16*L/9.86*D^2)*Q^2

for pipe B
delta PB/ro=(32*0.009*L/3.14^2*D)*Q^2=(0.288*L/9.86*D^2)*Q^2

for pipe C
delta PC/ro=(32*0.015*L/3.14^2*D)*Q^2=(0.48*L/9.86*D^2)*Q^2

 

[Chestermiller]

for pipe A
delta PA/ro=(32*0.005*L/3.14^2*D)*Q^2=(0.16*L/9.86*D^2)*Q^2

for pipe B
delta PB/ro=(32*0.009*L/3.14^2*D)*Q^2=(0.288*L/9.86*D^2)*Q^2

for pipe C
delta PC/ro=(32*0.015*L/3.14^2*D)*Q^2=(0.48*L/9.86*D^2)*Q^2

The figure gives a value of L and D for each pipe. Plug those in.

 

[williamcarter]

The figure gives a value of L and D for each pipe. Plug those in.

for pipe A
delta PA/ro=(32*0.005*L/3.14^2*D)*Q^2=(0.16*L/9.86*D^2)*Q^2
delta PA/ro=(0.16*10/9.86*(4*10^-2)^2)*Q^2=101.32*QA^2

=101.32*QA^2

for pipe B
delta PB/ro=(32*0.009*L/3.14^2*D)*Q^2=(0.288*L/9.86*D^2)*Q^2
delta PB/ro=(0.288*4/9.86*(6*10^-2)^2)*Q^2=32.42*QB^2

=32.42*QB^2

for pipe C
delta PC/ro=(0.48*6/9.86*(10*10^-2)^2)*Q^2=29.18*QC^2

=29.18*QC^2

 

[Chestermiller]

for pipe A
delta PA/ro=(32*0.005*L/3.14^2*D)*Q^2=(0.16*L/9.86*D^2)*Q^2
delta PA/ro=(0.16*10/9.86*(4*10^-2)^2)*Q^2=101.32*QA^2

=101.32*QA^2

for pipe B
delta PB/ro=(32*0.009*L/3.14^2*D)*Q^2=(0.288*L/9.86*D^2)*Q^2
delta PB/ro=(0.288*4/9.86*(6*10^-2)^2)*Q^2=32.42*QB^2

=32.42*QB^2

for pipe C
delta PC/ro=(0.48*6/9.86*(10*10^-2)^2)*Q^2=29.18*QC^2

=29.18*QC^2

OK. So
$$h_A=h=K_AQ_A^2$$
$$h_B=h=K_BQ_B^2$$
$$h_C=h=K_CQ_C^2$$where $$K_A=101.32$$$$K_B=32.42$$$$K_C=29.18$$
Now you just repeat what you did in the previous part of the problem.

 

[williamcarter]

Now you just repeat what you did in the previous part of the problem.

Thank you, what exactly do you mean by this?

 

[Chestermiller]

OK. So
$$h_A=h=K_AQ_A^2$$
$$h_B=h=K_BQ_B^2$$
$$h_C=h=K_CQ_C^2$$where $$K_A=101.32$$$$K_B=32.42$$$$K_C=29.18$$
Now you just repeat what you did in the previous part of the problem.

$$Q_A=\frac{\sqrt{h}}{\sqrt{K_A}}$$$$Q_B=\frac{\sqrt{h}}{\sqrt{K_B}}\tag{1}$$$$Q_C=\frac{\sqrt{h}}{\sqrt{K_C}}$$
$$Q=Q_A+Q_B+Q_C=Q=\sqrt{h}\left[\frac{1}{\sqrt{K_A}}+\frac{1}{\sqrt{K_B}}+\frac{1}{\sqrt{K_C}}\right]$$
$$\sqrt{h}=\frac{Q}{\left[\frac{1}{\sqrt{K_A}}+\frac{1}{\sqrt{K_B}}+\frac{1}{\sqrt{K_C}}\right]}\tag{2}$$

 

[williamcarter]

$$Q_A=\frac{\sqrt{h}}{\sqrt{K_A}}$$$$Q_B=\frac{\sqrt{h}}{\sqrt{K_B}}$$$$Q_C=\frac{\sqrt{h}}{\sqrt{K_C}}$$
$$Q=Q_A+Q_B+Q_C=Q=\sqrt{h}\left[\frac{1}{\sqrt{K_A}}+\frac{1}{\sqrt{K_B}}+\frac{1}{\sqrt{K_C}}\right]$$
$$\sqrt{h}=\frac{Q}{\left[\frac{1}{\sqrt{K_A}}+\frac{1}{\sqrt{K_B}}+\frac{1}{\sqrt{K_C}}\right]}$$

Thank you

 

[williamcarter]

$$Q_A=\frac{\sqrt{h}}{\sqrt{K_A}}$$$$Q_B=\frac{\sqrt{h}}{\sqrt{K_B}}$$$$Q_C=\frac{\sqrt{h}}{\sqrt{K_C}}$$
$$Q=Q_A+Q_B+Q_C=Q=\sqrt{h}\left[\frac{1}{\sqrt{K_A}}+\frac{1}{\sqrt{K_B}}+\frac{1}{\sqrt{K_C}}\right]$$
$$\sqrt{h}=\frac{Q}{\left[\frac{1}{\sqrt{K_A}}+\frac{1}{\sqrt{K_B}}+\frac{1}{\sqrt{K_C}}\right]}$$

sqrt(h)=sqrt(delta h loss)
what exactly it is sqrt(h)?.I do not have value for it
Have everything except Q and sqrt(h)
Problem told me to show that QB~0.3Qtotal

 

[Chestermiller]

sqrt(h)=sqrt(delta h loss)
what exactly it is sqrt(h)?.I do not have value for it
Have everything except Q and sqrt(h)
Problem told me to show that QB~0.3Qtotal

$\sqrt{h}$ is exactly what you said, the square root of the head loss. What do you get if you eliminate $\sqrt{h}$ between Eqns. 1 and 2 in post #16. C'mon man, it's just algebra. How were you able to solve part ii if you didn't do this?

 

[williamcarter]

$\sqrt{h}$ is exactly what you said, the square root of the head loss. What do you get if you eliminate $\sqrt{h}$ between Eqns. 1 and 2 in post #16. C'mon man, it's just algebra. How were you able to solve part ii if you didn't do this?

Yes, I understood this.However I do not understand how I am supposed to show that QB=0.3Q1 as I don't have numerical values for delta h.

 

[Chestermiller]

Yes, I understood this.However I do not understand how I am supposed to show that QB=0.3Q1 as I don't have numerical values for delta h.

Eliminate h between the two equations. So you don't need to know h.

 

[williamcarter]

Eliminate h between the two equations. So you don't need to know h.

I did it like this

Data:
KA=101.31
KB=32.42
KC=29.18

equating (1) and (2) to get rid of sqrt(h)
gives QB=Q/(1/sqrtKA+1/sqrtKB+1/sqrtKC)*sqrt(KB))

QB=Q/(1/sqrt(101.31)+1/sqrt(32.42)+1/sqrt(29.18)*sqrt(32.42))
QB= Q /2.61
QB=0.38Q