Net Positive Suction Head

During a test on a water pump, no cavitation is detected when the atmospheric pressure is 15 psi, the water temperature is 80°F, and the discharge is 13 gal/sec. The inlet galvanized pipe diameter is 6 in, and the pump is situated 16 ft above the suction reservoir. What is most nearly the net positive suction head at standard atmospheric pressure conditions?

A 8.2 ft

B 11 ft

C 14 ft

D 17 ft

I worked on this problem for about an hour this afternoon. I eventually looked at the solution for this problem and it still didn't make sense.

Can somebody try this problem and let me know how they solved the problem.

Thanks,

Matt

Matt,

Here's how I'd approach the problem...

1. Net Positive Suction Head Available (NPSHA ) = hatm + hz(s) - hf(s) - hvp

[You'll find this equation in the CERM as 18.30(a)]

2. To find atmospheric head, hatm, use the equation for fluid pressure: p (psi) = specific wt. (lb/ft3) * hatm (ft) / 144 (in2/ft2). This should get you hatm = 34.6ft

3. The static suction head, hz(s), is given in the problem statement as -16ft

4. To find friction head, hf(s), use the Hazen-Williams formula for pipe flow: hL = (4.72*L*Q^1.85) / (C^1.85*D^4.87) where L=pipe length(ft), Q=flowrate(ft3/s), C=Hazen-Williams coefficient (I used 140 but maybe 120 is better), and D=diameter of pipe(ft). This should get you hf(s) = 0.7ft, which keeps me from worrying about C being 120 or 140.

5. Vapor pressure head, hvp, comes straight from the CERM appendix 14.A and is 1.17ft for 80 degrees.

So the result of equation 1: NPSHA = 34.6ft + (-16ft) - 0.7ft - 1.2ft = 16.7ft.

I'd have chosen D.

IlPadrino said:

Matt,
Here's how I'd approach the problem...

1. Net Positive Suction Head Available (NPSHA ) = hatm + hz(s) - hf(s) - hvp

[You'll find this equation in the CERM as 18.30(a)]

2. To find atmospheric head, hatm, use the equation for fluid pressure: p (psi) = specific wt. (lb/ft3) * hatm (ft) / 144 (in2/ft2). This should get you hatm = 34.6ft

3. The static suction head, hz(s), is given in the problem statement as -16ft

4. To find friction head, hf(s), use the Hazen-Williams formula for pipe flow: hL = (4.72*L*Q^1.85) / (C^1.85*D^4.87) where L=pipe length(ft), Q=flowrate(ft3/s), C=Hazen-Williams coefficient (I used 140 but maybe 120 is better), and D=diameter of pipe(ft). This should get you hf(s) = 0.7ft, which keeps me from worrying about C being 120 or 140.

5. Vapor pressure head, hvp, comes straight from the CERM appendix 14.A and is 1.17ft for 80 degrees.

So the result of equation 1: NPSHA = 34.6ft + (-16ft) - 0.7ft - 1.2ft = 16.7ft.

I'd have chosen D.

Click to expand...

Hey, Matt, here is how I would solve..

I arrived with the same answer as IlPadrino, D, but with different approach.

I started just from the governing energy equation

(p1/gamma) + z1 + (v^2/2g)1 = (p2/gamma) + z2 + (v^2/2g)2 + hL.

HL is neglected as opposed to the above approach. Reason? Head loss is mainly a fucntion of length, and pipe material as well as fluid. You are given no information about that so I am not going to assume on that.

So, if you keep the datum on the surface of reservoir, Z1 = 0; and velocity term is also zero at the surface of reservoir. THerefore, you have only presure term that is 15psi.

On the right hand side z2 = 16ft (=0.3048*16m) and you can calculate the v (=Q/A, all info is given). Therefore, what is left is just the pressure head. If you plug all these and solve for pressure you will get the result. I worked in SI unit and got 5.303m and if I convert that into ft I got 17.4ft. So, answer is D.

That's my take. Can you please post the approach that you saw in the original question?

[No message]

Tido: I don't think your approach is a good solution because you've ignored vapor pressure head which is fundamental to the issue of cavitation. I agree the friction head is small in this case, but it can't generally be ignored.

Matt: The solution you posted is similar to mine. I think the Hazen-Williams formulas are easier to solve for friction head so I'd stick with that. Do you have any questions about my or the posted solution?

^^^ I agree with IlPadrino and would add that it is VERY important to try to not get fancy with problems on the exam.

In terms of breaking this problem down for solution, my thinking:

1. We know we are dealing with NPSH - what are the controlling equation(s)?

2. Based on those equation(s), what terms are provided explicitly?

3. Based on those equation(s), what terms are provided implicityly (e.g. can be looked up, calculated, derived from what is given)?

There will be a very broad number of problems and methods for solving those problems. In general, those problems should be 'relatively' easy to solve (e.g. solution within six minutes) so you should always be on the lookout for the lowest common denominator for the solution. While simplifications are useful, make sure they are always valid - this exam will test your knowledge of the validity for making simplifying assumptions. A standard, methodical approach to how you solve problems will be essential to being able to successfully navigate the exam.

:2cents:

JR

jregieng said:

^^^ I agree with IlPadrino and would add that it is VERY important to try to not get fancy with problems on the exam.
In terms of breaking this problem down for solution, my thinking:

1. We know we are dealing with NPSH - what are the controlling equation(s)?

2. Based on those equation(s), what terms are provided explicitly?

3. Based on those equation(s), what terms are provided implicityly (e.g. can be looked up, calculated, derived from what is given)?

There will be a very broad number of problems and methods for solving those problems. In general, those problems should be 'relatively' easy to solve (e.g. solution within six minutes) so you should always be on the lookout for the lowest common denominator for the solution. While simplifications are useful, make sure they are always valid - this exam will test your knowledge of the validity for making simplifying assumptions. A standard, methodical approach to how you solve problems will be essential to being able to successfully navigate the exam.

:2cents:

JR

Click to expand...

All good points. Your Item 1. can often be the biggest obstacle and it's solved either by fundamental understanding (which can be too late to achieve if you're studying for an exam in the next few months) or conditioning through many sample problems (you'll see a problem that looks familiar and work from there). Your Item 3. might add "quickly" and reminds me why tables and nomographs are so valuable. In this example, the CERM has a friction head nomograph (sorry... I can't remember which appendix) and I think it will get you an answer that is "good enough".

I think this problem is fairly classified as a six-minute problem if you knew the equations for NPSHA and friction head.

Matt: The solution you posted is similar to mine. I think the Hazen-Williams formulas are easier to solve for friction head so I'd stick with that. Do you have any questions about my or the posted solution?

IlPadrino said:

Tido: I don't think your approach is a good solution because you've ignored vapor pressure head which is fundamental to the issue of cavitation. I agree the friction head is small in this case, but it can't generally be ignored.
Matt: The solution you posted is similar to mine. I think the Hazen-Williams formulas are easier to solve for friction head so I'd stick with that. Do you have any questions about my or the posted solution?

Click to expand...

I undestand your concern, IlPadrino. Point well taken. But remember the CERM equation you point out in #1 actually ignores the velocity head, which in this case is, 0.37 ft. But I know 1.2ft is greater than 0.37ft.

owillis28 said:

Matt: The solution you posted is similar to mine. I think the Hazen-Williams formulas are easier to solve for friction head so I'd stick with that. Do you have any questions about my or the posted solution?I couldn't find any table in the book that listed a vapor pressure of 0.507 psi. The table that I used in the CERM (I don't have my books with me right now) listed the vapor pressure head of 1.17 ft. After doing a little bit of reading, I realized that 1.17ft * 0.433 psi/ft = 0.507 psi.

I was able to get the first two terms in the expanded NPSH equation shown in the solution for atmospheric head (hatm) and vapor pressure head (hvap).

I am still a little confused on the Swamee-Jain equation for finding the friction factor. I tried using the Moody diagram but didn't get there answer. Also, the n2 (????) in the denominator is something I am still looking into where the found the head loss.

Thanks

owillis

Click to expand...

Matt,

I find it's easier to keep the losses in feet rather than psi. Regarding your question about n2, it is pi not n, and is a result of substituting A = pi*r2 = pi*D2/4. When I get home tonight, I'll look at the CERM to find the friction factor. Did you follow the friction loss calculation from the Hazen-Williams formula?

IlPadrino said:

Matt,
I find it's easier to keep the losses in feet rather than psi. Regarding your question about n2, it is pi not n, and is a result of substituting A = pi*r2 = pi*D2/4. When I get home tonight, I'll look at the CERM to find the friction factor. Did you follow the friction loss calculation from the Hazen-Williams formula?

Click to expand...

I feel kind of stupid now that I look at my last response about the n2. I was looking at the rest of the solution while I was feeding my 2 month old early this morning and I guess I was a "little out of it".

I tried to go back through and understand the friction loss calc since I didn't get the same answer from the Moody diagram. I will try and look at it tonight. If you have any further discussion/explanation from the solution I posted earlier, I would greatly appreciate it.

I WOULD REALLY LIKE TO PASS THIS TIME.!!!!!!!!!!!!!!!!!!!!!

owillis

owillis28 said:

I feel kind of stupid now that I look at my last response about the n2. I was looking at the rest of the solution while I was feeding my 2 month old early this morning and I guess I was a "little out of it".
I tried to go back through and understand the friction loss calc since I didn't get the same answer from the Moody diagram. I will try and look at it tonight. If you have any further discussion/explanation from the solution I posted earlier, I would greatly appreciate it.

I WOULD REALLY LIKE TO PASS THIS TIME.!!!!!!!!!!!!!!!!!!!!!

owillis

Click to expand...

owillis,

OK... here's how I'd explain the friction factor. For laminar flow, f=64/Re; it is *not* dependent on e/D (relative roughness). For fully turbulent flow, f is dependent on e/D only; it is *not* dependent on Re. Remember that flow is laminar when Re is less than 2100, is turbulent when Re is more than 4000, and is in the critical zone when between the two. The Moody Diagram (Friction Factors for Any Kind and Size of Pipe) shows this graphically: to the left of the laminar flow line, is gives the equation f=64/Re, in the turbulent zone on the right, the lines are flat which means f is not dependent on Re.

If you want to calculate Re for a pipe flowing full: Re=VD/v where V is mean velocity, D is diameter of pipe, and v is kinematic viscosity of fluid. In this case, v=0.930x10-5 ft2/sec (appendix 14.A) and you should get Re=93,440 which is turbulent. Relative roughness can be calculated by looking up e in a table (design column in appendix 17.A); e/D=0.0005ft/0.5ft=0.001.

From here, you can use the Moody Diagram: Choose the curve where e/D=0.001 and follow it to the left until you get to Re=1E5 (close enough, yeah?), and then head straight to the left until you get to the left axis and then read f. You should get an f of somewhere between 0.018 and 0.019. I think it's easier to use Appendix 17.B because it's quicker and less likely to result in a mistake. In this case, you'd quickly get f=0.0185 for Re=1E5 (or you could interpolate but it's probably not worth the added accuracy) and e/D=0.001.

Or... you could use the Swamee-Jain equation... but now you're going to be pushing the 6 minute solution!

Once you have f, it's pretty easy to use the Darcy-Weisbach formula for head loss, hL=(fL/D)*(V2/2g).

Now... I think you'll agree that the Hazen-Williams equation I used in my solution is *much* quicker. You only need to find C using Appendix 17.A. If you don't have all the forms of the Hazen-Williams equations (solved for V in terms of both R and D, Q in ft3/sec and mgd, and hL let me know and I'll try to scan my one-page summary for you.

This is long-winded, but I think you'll find it's about as far in depth as you'll need to go for the PE exam. Please give me some feedback on my explanation.

Tido said:

I undestand your concern, IlPadrino. Point well taken. But remember the CERM equation you point out in #1 actually ignores the velocity head, which in this case is, 0.37 ft. But I know 1.2ft is greater than 0.37ft.

Click to expand...

Tido,

The CERM doesn't ignore the velocity head... it provides two equations. One (which I used) is based on conditions at the fluid surface at the top of an open fluid source (e.g. tank or reservoir); there is no kinetic energy at this point. The other is based on the conditions at the immediate entrance to the pump; there is both potential and kinetic head loss (and so there are terms for static (pressure) head and velocity head at the pump suction). In this second case, the reduced pressure head already accounts for the friction losses and the atmospheric head.

Wow! Suction head is really...suck huh?

cantaloup said:

Wow! Suction head is really...suck huh?

Click to expand...

LOL