NCEES TF 2016 #121 - Mechanical - Engineer Boards
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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.

Edited by c0m0

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##### Share on other sites If you post the problem, I could better help you.

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11 minutes ago, Audi driver, P.E. said:

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

Thanks. Please let me know if this picture works. #### Share this post

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

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Another vote for C.

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6 hours ago, c0m0 said:

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 Eis 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)

Edited by Slay the P.E.
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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?

Edited by c0m0

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10 hours ago, c0m0 said:

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.

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