Mistake on TFS Sample Test?

[ValonaBrau]

Question 132:

A centrifugal pump is sized to deliver 200 gpm of liquid with a specific gravity of 0.7 and a total differential head of 60 ft of water. The hydraulic horsepower required is most nearly:

A 2.1

B 2.7

C 3.0

D 4.3

In my reasoning, the pump head is 60 ft OF WATER which is equal to 26 psi. Pump work is pressure times volumetric flow rate, so specific gravity doesn't enter in to the equation. The book answer, however, multiplies the head by specific gravity to get 2.1. If they had specified the head was simply 60 ft, I would agree, but in specifying the head in feet water they are specifying the pressure regardless of the fluid density. Anyone have an opinion?

 

[ME_VT_PE]

(volumetric flow rate * specific weight of fluid * Pump head)/550

edit* if the fluid was water, specific gravity is 1, so its not included...

 

[justin-hawaii]

@ValonaBrau

The term feet of water is used to describe the pressure created by the water elevation.  In order to convert from feet of water to pressure per square inch as you have stated with the 0.433 conversion factor, you must use the density of water.  

For example, 1 cubic inch of water is equal to 

(62.4 lbs/ft3) * (1 ft3/ 1728 cubic inches) = 0.036 lbs/cubic inch

Then if you have 12 of those cubic inches stacked upon each other, then you have 1 foot of water elevation.

(0.036 lbs/cubic inch ) * (12 inch/ft) = 0.433 psi per foot of water

So as you can see, the conversion of 60 feet of water to 26 psi is dependent on the density of water.  When you have a different fluid, with a different  specific gravity, you need to convert the density to the actual density of fluid.  

For example, 1 cubic inch of new fluid is equal to 

(62.4 lbs/ft3) *(0.7) * (1 ft3/ 1728 cubic inches) = 0.0253 lbs/cubic inch

Then if you have 12 of those cubic inches stacked upon each other, then you have 1 foot of the new fluid.

(0.0253 lbs/cubic inch ) * (12 inch/ft) = 0.3033 psi per foot of new fluid

So you can use your HP equation, with PSI and GPM, but with this conversion.  

 

[ValonaBrau]

Justin,

If the problem had stated simply, "feet of head", then I would agree with you. The problem stated "Feet of water", thus the pressure across the pump is enough to support a column of water 60 ft high. The density of the fluid that is doing the pushing is irrelevant . The term "feet of water" is a unit of pressure (eg. see this calculator: https://www.unitconverters.net/pressure-converter.html). The actual head on the pump is 60 / .7 = 85.7 ft.

ME_VT, that equation is incorrect. Power = Force * Velocity. Force = pressure * area, Velocity is Volumetric flow /area. Therefore Power = Pressure * Volumetric Flow. Mass does not appear in the equation. 

 

[ME_VT_PE]

Justin,

If the problem had stated simply, "feet of head", then I would agree with you. The problem stated "Feet of water", thus the pressure across the pump is enough to support a column of water 60 ft high. The density of the fluid that is doing the pushing is irrelevant . The term "feet of water" is a unit of pressure (eg. see this calculator: https://www.unitconverters.net/pressure-converter.html). The actual head on the pump is 60 / .7 = 85.7 ft.

ME_VT, that equation is incorrect. Power = Force * Velocity. Force = pressure * area, Velocity is Volumetric flow /area. Therefore Power = Pressure * Volumetric Flow. Mass does not appear in the equation. 

You should email Doctor Tom. Also, Revoke my passing PE exam score because I used that equation about 27 times during the exam...

 

[ME_VT_PE]

Justin,

If the problem had stated simply, "feet of head", then I would agree with you. The problem stated "Feet of water", thus the pressure across the pump is enough to support a column of water 60 ft high. The density of the fluid that is doing the pushing is irrelevant . The term "feet of water" is a unit of pressure (eg. see this calculator: https://www.unitconverters.net/pressure-converter.html). The actual head on the pump is 60 / .7 = 85.7 ft.

ME_VT, that equation is incorrect. Power = Force * Velocity. Force = pressure * area, Velocity is Volumetric flow /area. Therefore Power = Pressure * Volumetric Flow. Mass does not appear in the equation. 

hydraulic horsepower = volumetric flow rate * the specific weight of the fluid * the Pump or turbine head

which gives you an answer in ft-lb/sec , divide by 550 to convert to HP

 

[ValonaBrau]

I don't know where you are getting that equation. In your equation you would end up with units of Ft^4 / s.

I have the FET supplemental handbook in front of me. It says Wdot = rho * g * H * Q / N. Rho * g * H = Pressure, thus Wdot = P * H /N. Or another way, H would be the actual feet of head, H= Hw / SG = rhoW * Hw / rho -> Wdot = rhoW *g *Hw * Q

 https://www.engineeringtoolbox.com/pumps-power-d_505.html

 

[SacMe24]

If you have MERM 13th edition, the equation they're using is on page 18-9, Table 18.5. It is the very first equation on the top left of the table, straight plug-and-chug:

Q (in gpm) x h (in feet) x SG , all divided by 3956.

The answer comes up to be 2.12 which corresponds to A.

 

[ValonaBrau]

SacMe24, that equation is correct and equivalent to mine, but my point is that you need to use the actual fluid head. The head pressure in the problem is given in feet of WATER. The actual fluid head is 60 / .7 = 85.7 ft which cancels out the SG term in the above equation so 200 gpm * 85.7 / 0.7 = 3.03. 

 

[SacMe24]

SacMe24, that equation is correct and equivalent to mine, but my point is that you need to use the actual fluid head. The head pressure in the problem is given in feet of WATER. The actual fluid head is 60 / .7 = 85.7 ft which cancels out the SG term in the above equation so 200 gpm * 85.7 / 0.7 = 3.03. 

I see what you mean....maybe the problem is poorly worded because I can't think of another way to get the answer they did.

 

[ValonaBrau]

I think the word "water" was left in by mistake and that the author intended it to read as 60 ft of pump head.

 

[SacMe24]

I think the word "water" was left in by mistake and that the author intended it to read as 60 ft of pump head.

As you continue to go through all the refresher material that's out there for your discipline (mine was MDM) , you will find that unfortunately, there are plenty of mistakes and erratas out there...frustrating when you're trying to prepare for such a hard exam. Even the NCEES practice exams have mistakes.

Anyway, when you encounter another one post it here and maybe one of us can help you figure it out.

Good luck with studying !

 

[ME_VT_PE]

I don't know where you are getting that equation. In your equation you would end up with units of Ft^4 / s.

I have the FET supplemental handbook in front of me. It says Wdot = rho * g * H * Q / N. Rho * g * H = Pressure, thus Wdot = P * H /N. Or another way, H would be the actual feet of head, H= Hw / SG = rhoW * Hw / rho -> Wdot = rhoW *g *Hw * Q

 https://www.engineeringtoolbox.com/pumps-power-d_505.html

Get on the bus, I’m taking you to school. You don’t end up with that. Convert GPM to ft3/s, use the specific gravity to find your specific weight, multiply by feet of head and convert to HP. And yes. The term feet of water is confusing, but that’s one correct form of the equation. The other is what SAcme said. 

 

[justin-hawaii]

Justin,

If the problem had stated simply, "feet of head", then I would agree with you. The problem stated "Feet of water", thus the pressure across the pump is enough to support a column of water 60 ft high. The density of the fluid that is doing the pushing is irrelevant . The term "feet of water" is a unit of pressure (eg. see this calculator: https://www.unitconverters.net/pressure-converter.html). The actual head on the pump is 60 / .7 = 85.7 ft.

ME_VT, that equation is incorrect. Power = Force * Velocity. Force = pressure * area, Velocity is Volumetric flow /area. Therefore Power = Pressure * Volumetric Flow. Mass does not appear in the equation. 

Gotcha @ValonaBrau.  Good catch, I missed where it stated "water", I just read the "head".  You can email NCEES and hopefully they include it in the next errata.  They updated their HVAC/R errata based on this last post, but I had to email them and it took from August 2017 til March 2018 . 



Here is the current errata for T&F

https://ncees.org/wp-content/uploads/2016-Mec_Thermal_errata_ALL-2.pdf

Current errata for HVAC/R:  

https://ncees.org/wp-content/uploads/2016-Mec_HVAC_errata_ALL-2.pdf

 

[ValonaBrau]

Ah, I wasn't familiar with the term "specific weight" = rho * g. So yes, I follow your equation now.

 

[Vel2018]

Justin,

If the problem had stated simply, "feet of head", then I would agree with you. The problem stated "Feet of water", thus the pressure across the pump is enough to support a column of water 60 ft high. The density of the fluid that is doing the pushing is irrelevant . The term "feet of water" is a unit of pressure (eg. see this calculator: https://www.unitconverters.net/pressure-converter.html). The actual head on the pump is 60 / .7 = 85.7 ft.

ME_VT, that equation is incorrect. Power = Force * Velocity. Force = pressure * area, Velocity is Volumetric flow /area. Therefore Power = Pressure * Volumetric Flow. Mass does not appear in the equation. 

I guess you should listen to us here, as we have so many discussion prior to this post and we just passed the PE exam 1st try this April.

IMHO, you are reading the problem incorrectly and over thinking it. It is clearly stated in the problem "total differential head" that is your keyword right there, what is this head? This is your hA right? What is the basis of hA in that equation? YES its water. So why do you have to convert this again when its already in terms of FEET which is what you need in the equation. 

Use that equation in the table of MERM and BTW my memorized and favorite one out of the 4 equation in that table for WHP. That is = hA x Q(GPM) x S.G. / 3960

That's  it. And BTW just a tip memorize this number 0.002228 your best buddy for GPM conversion to CFS and so on.

EDIT: I don't even know why they put this kind of problem in the NCEES practice exam, its kinda so plug and chug grade 1 question where I always think before I took the exam  "would they really ask" these kinds of problems in the real exam??? LMAO its sooo "Plug and Chug Bae"