Attached are 2 problems dealing with pressure conduit. I don't think CERM has this. I saw it in Testmaster note. This kind of problem looks very basic but it would take time to solve because one needs to apply Bernouilli eq, head loss formula, Moody chart, friction coefficient, minor loss... to solve it. The thing is how to solve it quicker. Cover the solution and try them to see how you'll do it. I'll bring up my way to solve it later.
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IlPadrino
Well-known member
Rather than use an iterative approach to finding f, I'd use some tables to figure out what flow gives me the right pressure drop (head loss). In the case of the first problem, hL=100 (with all the zeros in the energy equation, it should only take about fifteen seconds to realize this).
My first instinct is to hope all the minor losses can be ignored... but to verify, I'd find the equivalent lengths of the entrance, fittings, and exit. Appendix 17.D would be one approach, but you'd have to hope your assumptions match the problem statements of K. It's probably easier to use the equation for Leq=kD/f and just pick a reasonable f (0.019 for developed turbulent flow). This results in an Leq of 79 ft, which is a little more than 10% of the total pipe length of 702 ft. To ignore this might not have caused you to pick the wrong answer, but I wouldn't ignore it considering how easy it was to calculate. HOWEVER, it makes me feel good about choosing a reasonable value for f rather than iteratively solving.
So then... 781 ft of head loss... I have a nomograph and table that gives loss in feet (per 100 feet length of pipe) for a given diameter and flow assuming C=100. Looking up 12.8 ft of loss per 100 ft of 6" pipe gives me a flow of about 1000 gpm or 2.2 cfs... not exactly close to the answer given. I'm hoping this is because e=0.00015 is not analogous to C=100. Thoughts anyone?
I'm tempted to use my good guess (I hope!) for f and just use the equation hL=fLV2/D2g... the answer is close to the iterative solution (2.89 cfs vice 3.13) but I'm still worried about my choice of f. I'd probably use the iterative approach with Re at this point - the Moody Diagram.
Or if could make an assumption and quickly finish... assume we're talking schedule-40 steel pipe and use Appendix 17.C. 100 ft = 43.3 psi and it's per 781 feet, so per 1000 feet (like the table requires) is 55.4 psi. The table yields 1500 gpm (3.34 cfs) which is close to the iterative solution.
I don't think I saw anything this hard on the last exam... hard in the sense that Re depends on v, v depends on f, and f depends on Re - short of an iterative approach, I don't think you can get an exact answer. Still, I think this is a great practice problem!
My first instinct is to hope all the minor losses can be ignored... but to verify, I'd find the equivalent lengths of the entrance, fittings, and exit. Appendix 17.D would be one approach, but you'd have to hope your assumptions match the problem statements of K. It's probably easier to use the equation for Leq=kD/f and just pick a reasonable f (0.019 for developed turbulent flow). This results in an Leq of 79 ft, which is a little more than 10% of the total pipe length of 702 ft. To ignore this might not have caused you to pick the wrong answer, but I wouldn't ignore it considering how easy it was to calculate. HOWEVER, it makes me feel good about choosing a reasonable value for f rather than iteratively solving.
So then... 781 ft of head loss... I have a nomograph and table that gives loss in feet (per 100 feet length of pipe) for a given diameter and flow assuming C=100. Looking up 12.8 ft of loss per 100 ft of 6" pipe gives me a flow of about 1000 gpm or 2.2 cfs... not exactly close to the answer given. I'm hoping this is because e=0.00015 is not analogous to C=100. Thoughts anyone?
I'm tempted to use my good guess (I hope!) for f and just use the equation hL=fLV2/D2g... the answer is close to the iterative solution (2.89 cfs vice 3.13) but I'm still worried about my choice of f. I'd probably use the iterative approach with Re at this point - the Moody Diagram.
Or if could make an assumption and quickly finish... assume we're talking schedule-40 steel pipe and use Appendix 17.C. 100 ft = 43.3 psi and it's per 781 feet, so per 1000 feet (like the table requires) is 55.4 psi. The table yields 1500 gpm (3.34 cfs) which is close to the iterative solution.
I don't think I saw anything this hard on the last exam... hard in the sense that Re depends on v, v depends on f, and f depends on Re - short of an iterative approach, I don't think you can get an exact answer. Still, I think this is a great practice problem!
I saw Testmaster notes had this kind of problem and a few Moody charts so it doesn’t hurt to look at this type of problem.
First, in the solution of prob. 9.212, the calculation of V (6th line down) is incorrect. V should be 16.36 ft/s.
The solution has 2 steps: step 1 is for flow analysis (which involve Bernouilli eq, head loss eq, minor loss eq) and step 2 is for iteration. I develop a formula for these two cases:
H=(V^2/2/G)*(F*L/D+C+K)
This formula is the end product of step 1 analysis.
H is total energy loss in ft and equal to the difference in height of water levels, G is 32.2 for US unit, L and D unit is ft, K is total of minor loss coefficient, C = 1 for flow to open air, C = 0 for flow tank to tank.
Enter this equation into the HP-33s calculator.
Solve for V , assume f = 0.02 first.
With V, calculate Re=VD/ ע then use Re number and ε/D find f again from Moody chart.
Solve for V again with the new f value.
Usually this second value of V is good enough.
One more thing, on the Moody chart of my note I paste the table 17.2 CERM (Specific Roughness) and top 8 rows of Appendix 14A CERM (for finding nu number), so that I don’t fumble around when I search the Re number.
I don’t think the Examiner would give anything more complicate than these two cases.
An alternative is V, D, minor loss K are given, find L; this case the same equation above can be used to solve for L (instead of solve for V). Or maybe D is the unknown with V, L, K are given.
If Q (cfs) is given instead of V (remember 1 gpm = 0.002228 cfs if Q is given in gpm), the equation above will become:
H=(Q^2/0.617/D^4/2/G)*(F*L/D+C+K)
Same procedure applied, but with Q instead of V.
For an easier case, try solving the problem 9.197 with Q=3.8 cfs, same D=6”, H=100’, minor loss K=0.5, f =0.016; find L?
First, in the solution of prob. 9.212, the calculation of V (6th line down) is incorrect. V should be 16.36 ft/s.
The solution has 2 steps: step 1 is for flow analysis (which involve Bernouilli eq, head loss eq, minor loss eq) and step 2 is for iteration. I develop a formula for these two cases:
H=(V^2/2/G)*(F*L/D+C+K)
This formula is the end product of step 1 analysis.
H is total energy loss in ft and equal to the difference in height of water levels, G is 32.2 for US unit, L and D unit is ft, K is total of minor loss coefficient, C = 1 for flow to open air, C = 0 for flow tank to tank.
Enter this equation into the HP-33s calculator.
Solve for V , assume f = 0.02 first.
With V, calculate Re=VD/ ע then use Re number and ε/D find f again from Moody chart.
Solve for V again with the new f value.
Usually this second value of V is good enough.
One more thing, on the Moody chart of my note I paste the table 17.2 CERM (Specific Roughness) and top 8 rows of Appendix 14A CERM (for finding nu number), so that I don’t fumble around when I search the Re number.
I don’t think the Examiner would give anything more complicate than these two cases.
An alternative is V, D, minor loss K are given, find L; this case the same equation above can be used to solve for L (instead of solve for V). Or maybe D is the unknown with V, L, K are given.
If Q (cfs) is given instead of V (remember 1 gpm = 0.002228 cfs if Q is given in gpm), the equation above will become:
H=(Q^2/0.617/D^4/2/G)*(F*L/D+C+K)
Same procedure applied, but with Q instead of V.
For an easier case, try solving the problem 9.197 with Q=3.8 cfs, same D=6”, H=100’, minor loss K=0.5, f =0.016; find L?
Mike_NC
Well-known member
Would this be a AM type WR problem or PM?
I am taking a review class and the professor seems to thing problems with more than a few lines of calculations would be more or less a PM problem. Just wanting some feedback. I am not a WR person, so I am trying to atleast get the 'low hanging fruit' in the WR/Enviro AM section.
I am taking a review class and the professor seems to thing problems with more than a few lines of calculations would be more or less a PM problem. Just wanting some feedback. I am not a WR person, so I am trying to atleast get the 'low hanging fruit' in the WR/Enviro AM section.
jeb6294
Well-known member
I would second that...strongly.
Also, remember that you only have a few answers to choose from. Rather than go through a bunch of complicated and time consuming calculations, sometimes it's faster/easier to simplify an equation down to something more manageable and then start plugging in the possible answers and solve it that way.
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