6MS Thermal Question No 49

[spacebanjo]

Hello,

I'm having trouble understanding the solution to 6MS problem 49. See problem and solution here http://imgur.com/a/8wsse.

In particular:

1) Why can I not use the hydraulic power equations for each branch. IE for the main branch I compute BHP required for the total flow and the pressure drop between reservoir and the 'T'. For branch A and B I compute BHP required for the respective individual flows through each branch and the respective pressure drop through the branch. I then sum these three BHP to get the answer (this method will result in an answer higher than those listed as multiple choice options).

2) I don't understand the 6MS solution approach where "The pressure drops in the two branches are combined using the equation 'P-drop1*P-drop2/(P-drop1+P-drop2)'. "

Consider hypothetically pipe flow through a system with two branches that each have a pressure drop of 100psi and each have a flow-rate of 100gpm. In this balanced system we are moving 200 gpm through it at a total pressure drop of 100psi. However if we use this parallel pipe formula 6MS suggests we would calculate this as moving 200gpm through the system at a total pressure drop of 50psi. This doesn't make sense to me.

What am I missing here?

Thanks for the help!

 

[tbova]

I agree with you, something about this doesn't seem right to me either.  Particularly, the combination of the pressure drops.  I've never seen it done quite that way and intuitively it seems incorrect.  

If this were a real case I would size the pump for the total flow with the higher of the total head required by either stream.

 

[Audi Driver P.E.]

It is a typical (and appropriate) way to calculate flow in parallel.  And it is the same formula used in electrical flow, btw.  If the flow/drops were in series, then it is simply additive.  In your hypothetical, your total drop is 50psi, not 100psi..

 

[spacebanjo]

It is a typical (and appropriate) way to calculate flow in parallel.  And it is the same formula used in electrical flow, btw.  If the flow/drops were in series, then it is simply additive.  In your hypothetical, your total drop is 50psi, not 100psi..

How is it 50? I am missing the "why" of this. 

 

[spacebanjo]

Ok another 'counter example'.

Suppose I have a pipe that branches to two locations as in the problem. One pipe has a flow of 1000gpm at a pressure drop of 1000psi. The other pipe has a flow of 1gpm at a pressure drop of 1psi. Using the 6MS equation: dp1*dp2/(dp1+dp2)=1psi. They would be saying for the equivalent 1001gpm we just have to consider 1psi of drop -- a very small pump is required! Even though one of the pipes has 1000gpm dropping 1000psi which would require a large pump. How can it be?

 

[P-E]

Hello,

I'm having trouble understanding the solution to 6MS problem 49. See problem and solution here http://imgur.com/a/8wsse.

In particular:

1) Why can I not use the hydraulic power equations for each branch. IE for the main branch I compute BHP required for the total flow and the pressure drop between reservoir and the 'T'. For branch A and B I compute BHP required for the respective individual flows through each branch and the respective pressure drop through the branch. I then sum these three BHP to get the answer (this method will result in an answer higher than those listed as multiple choice options).

2) I don't understand the 6MS solution approach where "The pressure drops in the two branches are combined using the equation 'P-drop1*P-drop2/(P-drop1+P-drop2)'. "

Consider hypothetically pipe flow through a system with two branches that each have a pressure drop of 100psi and each have a flow-rate of 100gpm. In this balanced system we are moving 200 gpm through it at a total pressure drop of 100psi. However if we use this parallel pipe formula 6MS suggests we would calculate this as moving 200gpm through the system at a total pressure drop of 50psi. This doesn't make sense to me.

What am I missing here?

Thanks for the help!

You're missing lighter fluid.  

 

[Audi Driver P.E.]

Ok another 'counter example'.

Suppose I have a pipe that branches to two locations as in the problem. One pipe has a flow of 1000gpm at a pressure drop of 1000psi. The other pipe has a flow of 1gpm at a pressure drop of 1psi. Using the 6MS equation: dp1*dp2/(dp1+dp2)=1psi. They would be saying for the equivalent 1001gpm we just have to consider 1psi of drop -- a very small pump is required! Even though one of the pipes has 1000gpm dropping 1000psi which would require a large pump. How can it be?

Well, I don't know exactly how small you think a pump is that can produce 1001 gpm at 1 psi, but yes. That is the case (keeping in mind, of course that what the 6ms problem is applying the equation to is only a part of the total pressure drop across both legs).  And if you think it is silly that it is so, for this example, consider: just how do you suppose you could induce the flow to actually be split as you suggest it is?  Perhaps working out the respective pipe diameters, lengths, type of flow, etc. might be instructive, in that respect.

But before you attempt any of that, it might be worthwhile to note that this type of problem (while actual) is atypical for an exam of the NCEES type.  I looked in my undergraduate fluid mechanics course text by Crowe and while there isn't a good explanation included, I see that there IS a section problem that deals with this very issue at the end of chapter 10 (problem 10.23 in what is the 4th edition), and while the section on parallel flow in that chapter deals with the much more commonly seen problem(s), apparently Crowe figured a student could figure it out.  As I noted, the type of problem 6MS is giving is encountered more frequently in an electrical context where resistors in parallel are what is considered.  Same same, however... you know, since electron flow is modeled as a fluid in 100 level electrical circuit analysis courses.

The key for the TFS exam will be how to properly use and understand the Darcy-Weishbach equation in conjunction with Bernouli's, so try to focus your study on more productive applications and topics.  Most folks on this board don't think too highly of 6MS.  I never have owned it or worked its problem set but I am starting to understand their opinion.

 

[tbova]

The electrical circuit analogy is used often.  However, pressure drop not equivalent to resistance.  Pressure drop is analogous to voltage drop.  If you wanted to figure this problem out using an effective resistance method you could do so, but the equation they gave doesn't do that.

delta_P = Q * R

with delta_p in psi, Q in gpm, and R in psi/gpm and

1/R_tot = 1/R_a + 1/R_b

If I do the problem that way, I get a total head required of 403.8 ft.

For the first example you gave:

R_1 = R_2 = 100 psi / 100 gpm = 1 psi/gpm

1/R_tot = 1/R_1 + 1/R_2 = 1/1 + 1/1 = 2

R_tot = 1/2 psi/gpm

deta_P_tot = Q_tot * R_tot

deta_P_tot = 200 gpm * 1/2 psi/gpm

deta_P_tot = 100 psi

 

[JHW 3d]

This one kind of rubbed me the wrong way for a minute. Then it dawned on me: The hydraulic/electrical analogy is for combining flow resistances in parallel and series, not pressures.  resistance (R) = pressure (p)/flow (Q)

I assert that this problem solution is wrong. As it turns out (I wasn't sure and did it both ways), it doesn't matter if you calculate power using the equivalent resistance method (Power = p*Q/1714, p = Q*R,  where R = Ra*Rb/(Ra + Rb)), or just add the hydraulic power (using MERM Table 18.5) for each branch, since you already know everything about the losses and flows through each branch. Total power required is 130 hp using either method.

Parallel flow problems are usually set up so you either: don't know the friction (or head loss, or pressure drop), don't know the flow, or don't know either -- and you have to guess, then iterate. MERM states: The flow divides in such a manner as to make the head loss in each branch the same. For whatever reason, in this problem they gave you both branch flows and pressure drops to begin with. Which would be fine (and easy), except they proceeded to solve the problem wrong.

 

[JHW 3d]

Apparently, the 6MS people are having second thoughts about this problem, too (from the errata):

 

[spacebanjo]

Thanks for the feedback -- I agree with JHW 3d and tbova. I did try to look up the errata and for whatever reason that portion of their site was down. I am amazed a problem like this survived from the 1st edition to the 2nd edition and only was caught after the fact.

In summary I finished the 6MS problems and they were a good review just for the sake of doing more problems. Many of the problems are realistic but 20-30% of the problems require obscure or unrealistic references; I think it is self-evident which one these are as they require with no work around references no one will bring to the exam (API682, Popes Rules of Thumbs for Mech Engineers, the full text of ASME BPVC, etc).

Once I realized this and didn't let it stress me out I felt better about the 6MS --no need for lighter fluid .

 

[P-E]

Thanks for the feedback -- I agree with JHW 3d and tbova. I did try to look up the errata and for whatever reason that portion of their site was down. I am amazed a problem like this survived from the 1st edition to the 2nd edition and only was caught after the fact.

In summary I finished the 6MS problems and they were a good review just for the sake of doing more problems. Many of the problems are realistic but 20-30% of the problems require obscure or unrealistic references; I think it is self-evident which one these are as they require with no work around references no one will bring to the exam (API682, Popes Rules of Thumbs for Mech Engineers, the full text of ASME BPVC, etc).

Once I realized this and didn't let it stress me out I felt better about the 6MS --no need for lighter fluid .

I'd burn it anyhow.  That also releaves stress.