# NCEES 2016 pro 119



## Flluterly (Oct 19, 2017)

Can anyone share the thinking process for this problem? I review the solution and sort of get the answer, but not clear. Appreciate any feedback.


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## Slay the P.E. (Oct 19, 2017)

Its just evaluating knowledge of the behavior of piping in parallel. If two parallel branches (the one with the bypass valve and the one with AHU-3) have a common start and end point, then the pressure drop across each of the two branches is the same.


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## Flluterly (Oct 19, 2017)

Thank you for your reply!


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## JHW 3d (Oct 20, 2017)

Slay the P.E. said:


> Its just evaluating knowledge of the behavior of piping in parallel. If two parallel branches (the one with the bypass valve and the one with AHU-3) have a common start and end point, then the pressure drop across each of the two branches is the same.


The diagram sucks (or I suck), but doesn’t parallel flow want all branches to have equal head loss (not just pressure loss)? 

Can someone post the solution?


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## Slay the P.E. (Oct 20, 2017)

When solving parallel pipe flow problems, the condition is that the pressure at the point where the branches split (say, point A) is the same for all the branches and the pressure at the point where the branches meet (say, point B) is also the same for all branches.

You then use Bernoulli across each of the branches knowing that the quantity (pA-pB) is the same for all branches. Each branch could have different friction head loss, different minor head loss, different flow rates, etc.


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## Flluterly (Oct 20, 2017)

Solution:

the minimum required differential pressure is equal to the sum of the losses through a control valve and AHU. the 10 ft of water pressure loss through the AHU must be converted to psi. 

Coil pressure drop= 10 ftX0.433 =4.33 psi

adding this value to the design control valve loss of 5 psi yields 9.33 psi


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## Flluterly (Oct 20, 2017)

Slay the P.E. said:


> When solving parallel pipe flow problems, the condition is that the pressure at the point where the branches split (say, point A) is the same for all the branches and the pressure at the point where the branches meet (say, point B) is also the same for all branches.
> 
> You then use Bernoulli across each of the branches knowing that the quantity (pA-pB) is the same for all branches. Each branch could have different friction head loss, different minor head loss, different flow rates, etc.


For the parallel pipe flow system, the friction head loss should be equal in each branch?


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## Slay the P.E. (Oct 20, 2017)

No. 

What is equal is the change in hydraulic grade line:


So, if you have points A and B connected by two branches, say branch 1 and branch 2, then you would write Bernoulli from A to B along branch 1 (in this kind of problem it is more convenient to write Bernoulli in terms of flow rate _Q_ rather than velocity, _V_):

​
and along branch 2:

​
You eliminate the left side of these two equations and you get one equation relating the two flow rates, Q1and Q2. Typically one knows the total flow rate entering the network, so the other equation would be from knowing what _Q1+Q2_ is.

But anyway, the first term in the square bracket in the previous two equations is the friction head loss across each branch. Each branch can have a different friction head loss. Similarly with the minor losses.

This is essentially what they did in problem 516, except instead of minor loss coefficient K, they used the equivalent length of each fitting.


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## Flluterly (Oct 20, 2017)

Thanks. The head loss in each breach is same. Am I right?


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## Slay the P.E. (Oct 20, 2017)

Yes, the sum of friction and minor losses would have to be the same.


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## Flluterly (Oct 20, 2017)

Thank you again for your help


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