NCEES TF 2016 #121

[c0m0]

The problem solution of calculating BHP does not include sg of acetone. The fluid considered is basically water. 

Maybe I am missing certain assumptions pre-made. Please help. 

Thanks in advance. 

 

[Audi Driver P.E.]

If you post the problem, I could better help you.

 

[c0m0]

If you post the problem, I could better help you.

Thanks. Please let me know if this picture works.

 

[Audi Driver P.E.]

I got C using table 18.5 on 18-8 in 13th Ed. MERM.

 

[Ramnares P.E.]

Another vote for C.

 

[Slay the P.E.]

The problem solution of calculating BHP does not include sg of acetone. The fluid considered is basically water. 

Maybe I am missing certain assumptions pre-made. Please help. 

Thanks in advance. 

You don't need the SG of the fluid because they are giving you the volumetric flow rate (in gpm).

As @Audi driver, P.E. has pointed out, the table 18-5 can be a real time saver here. The equation in the table that has Q (volumetric flow rate in gpm) does not use SG. Here's why:

Look at equation 17.59. You can re-write it as EA = hA g where EA is the energy added by the pump per unit mass. So, the hydraulic horse power P will be EA times the mass flow rate, m-dot:

P = ( m-dot ) x (EA)​

P = ( m-dot ) x hA x g​

and you can get hA by applying the Bernoulli equation from the pump suction to the discharge, in other words, use equation 18-9, hA =  ΔP/(ρg) which we can insert above:

P = ( m-dot ) x (ΔP/ρ)​

Since we don't have mass flow rate, then use Q = m-dot/ρ in the above to get:

P = Q ΔP​

If you restrict Q to be in [gpm] and ΔP in [psi] then you need to introduce a factor 1,714 in the denominator for P to be in [hp]. This is how they obtain that (Q ΔP)/1,714 equation in table 18-5. If you follow the solution provided in the NCEES sample exam you will notice it is exactly the same steps I've written here (what MERM calls EA is what they are calling w)

 

[c0m0]

Got. Thanks.

Just to clarify it. Since in this case Q (gpm) and P(psi) are given, one we can use Tab 18.5 - Line 2 : (Delta(P)* Q)/1714.

Therefore ignore the rest of distractors. Correct?

 

[Slay the P.E.]

Got. Thanks.

Just to clarify it. Since in this case Q (gpm) and P(psi) are given, one we can use Tab 18.5 - Line 2 : (Delta(P)* Q)/1714.

Therefore ignore the rest of distractors. Correct?

Correct.

 

[usernamerequired]

I am having the same problem as the OP. I do see what you're saying Slay but given the reference material in the manual, it can be misleading. When would you use the BHP equation at the bottom? What is the difference between WHP and BHP, as explained for this problem? Also given that it clearly states "for water:" in the manual also confuses me when used in this problem.

 

[Slay the P.E.]

I am having the same problem as the OP. I do see what you're saying Slay but given the reference material in the manual, it can be misleading. When would you use the BHP equation at the bottom? What is the difference between WHP and BHP, as explained for this problem? Also given that it clearly states "for water:" in the manual also confuses me when used in this problem.

View attachment 28422

The “fluid power” (also known as "theoretical power" or "water horsepower" if the working fluid is water) is the power the pump provides to the fluid.
The "shaft" or "brake" horsepower is the power the motor provides to the pump. The shaft power will be slightly larger than the fluid power because irreversibilities within the pump consume energy. This effect is quantified by the pump efficiency, as follows:
Brake Power = (Fluid Power)/η_pump.

The problem at hand asks for the shaft power, so you calculate the water power and then divide that by the pump efficiency.


You are right, the fact that (Q ΔP)/1,714 is under the "for water" heading in the handbook is misleading and confusing. This form of the fluid power equation is actually applicable to any fluid, not just water. The (Q Δh)/3,960 form is applicable only for water, but if you use (Q Δh SG)/3,960, then you can use it for any fluid. See how they use this version with SG in the definition of Brake Power further down in that page.

PS: There is also "purchased" power, which is the power the utility provides to the motor. The purchased power will be slightly larger than the shaft power due to irreversibilities within the motor consuming energy. This effect is quantified by the motor efficiency as follows:
Purchased Power = (Brake Power)/η_motor = (Fluid Power)/[(η_pump)(η_motor)]
(see the bottom of Page 237 in the handbook)

We have this sketch in our book illustrating the relationship between purchased, brake, and fluid power: