# Question from PPI Practice Problems



## Viper5 (Aug 19, 2016)

I am using the 13th edition of the Mechanical Engineering Practice Problem book to help prepare for the upcoming PE.  I am confused about one of the questions.  Namely, Section 29, Prob 4, part f.  This solution is eventually obtained by rearranging Eqn 28.72. where BHP = IHP - FHP.

What's throwing me off is how IHP is calculated.  I thought it was simply a conversion using densities of the two altitudes by use of Eqn 28.73. 

So when I solved, I used [email protected] = ([email protected]/[email protected])*([email protected] from part e) = 7762 hp        &gt;      book answer of 7087 hp using method of (mdot * delta work). 

Please explain what my solution is not accounting for because it's driving me crazy.  I used the same method as used from a Section 28, problem 8, part d.


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## Audi Driver P.E. (Aug 19, 2016)

Could you post the problem?


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## Viper5 (Aug 19, 2016)

Section 29 Prob 4

A gas turbine operating on the Brayton cycle with an 8:1 pressure ratio is located 7000ft (2100m) altitude.  The conditions at that altitude are 12 psia and 35F (82 kPa and 2 C).  While consuming 0.609 lbm/hp-hr (100kg/GJ) of fuel and 50,000 cfm (23,500 L/s) of air, the turbine develops 6000 bhp (4.5 MW).  The turbine efficiency is 80%, and the compressor efficiency is 85%.  The fuel has a lower heating value of 19,000 Btu/lbm (44 MJ/kg).  The fuel mass is small compared to the air mass.  There is no pressure loss in the combustor.  The turbine receives combustor gases at 1800F (980C).  The air inlet filter area is 254 ft^2 (22.9 m^2).  The turbine is moved to sea level where the conditions are 14.7 psia and 70F (101.3 kPa and 21 C).  The combustion efficiency and combustor temperature remain the same.

a) At 7000 ft, the air density is most nearly

b) At 7000 ft, the air mass flow rate is most nearly

c) At 7000 ft, the idea fuel rate is most nearly

d) At 7000 ft, the combustor efficiency is most nearly

e) At 7000 ft, the friction horsepower is most nearly

*f) At sea level, the new brake horsepower is most nearly &lt;------- This one*

g) At sea level, the new brake specific fuel consumption is most nearly

Section 28 Prob 8

When the atmospheric conditions are 14.7 psia and 80F (101.3 kPa and 27C), a diesel engine with metered fuel injection has the following operating characteristics.

BHP = 200 bhp (150kW)                   BSFC = .48 lbm/hp-hr (81kg/GJ)                 air-fuel ratio = 22:1                     mechanical efficiency = 0.86

The engine is moved to an altitude where the atmospheric conditions are 12.2 psia and 60F (84 kPa and 16C).  The running speed is unchanged.

part d) The new net power is most nearly


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## Viper5 (Aug 24, 2016)

I know it's a bit wordy, but would still appreciate help on this problem.  Thanks.


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## Audi Driver P.E. (Aug 25, 2016)

First of all, I believe your methodology is correct.  What I am having trouble figuring out is what you are using for values in you calculation for f).  It looks like you say your answer for e) is FHP=7762, which just looking at it seems way too high.  But it also looks like you could be saying that IHP2=7762 and if that is the case, then you have to factor in the friction losses into the ultimate BHP2.


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## Viper5 (Aug 25, 2016)

FHP (which can assumed to be constant at all altitudes) is calculated from FHP = IHP - BHP  (all @ 7000 ft).  This is what part e) is asking.  IHP is calculated by mdot*delta work (@ 7000ft).  BHP is given which leaves FHP to be calculated as 778hp.

For part f) the BHP (@0 ft) is to be calculated which translates to figuring out what the new IHP is.  What I did was use

IHP (@0ft) = density(@0ft) / density(@7000ft) * IHP (@7000ft)

This yields IHP (@0ft) = 7762hp.  The book just used mdot*delta work (@0ft) like was done in part e) and calculated IHP to be 7087 hp.  So basically by using my method, IHP is about 700hp larger.  Just trying to figure out why.


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## Audi Driver P.E. (Aug 25, 2016)

Viper5 said:


> FHP (which can assumed to be constant at all altitudes) is calculated from FHP = IHP - BHP  (all @ 7000 ft).  This is what part e) is asking.  IHP is calculated by mdot*delta work (@ 7000ft).  BHP is given which leaves FHP to be calculated as 778hp.
> 
> For part f) the BHP (@0 ft) is to be calculated which translates to figuring out what the new IHP is.  What I did was use
> 
> ...


Like I said, IHP=7762hp  But that is not what is being asked for in f).  In f), they want BHP which is equal to IHP-FHP.  So, if you take your IHP=7762 and subtract your calculated FHP of 778hp, you get?  Unless you are saying the book answer is using an IHP of 7087 in their caclculation for BHP (to obtain the "final answer")?


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## Viper5 (Aug 25, 2016)

Audi driver said:


> Like I said, IHP=7762hp  But that is not what is being asked for in f).  In f), they want BHP which is equal to IHP-FHP.  So, if you take your IHP=7762 and subtract your calculated FHP of 778hp, you get?  Unless you are saying the book answer is using an IHP of 7087 in their caclculation for BHP (to obtain the "final answer")?


I get BHP = IHP - FHP = 7762 - 778 = 6984hp

Book answer = 7087 - 778 = 6309 hp


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## Audi Driver P.E. (Aug 25, 2016)

Sorry I'm not being much help.  I don't know what is accounting for the difference.  To my mind, your solution process is correct.  I am curious, can you post a photo of their solution for all these a) - f) ? (instead of typing it all out)


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## Viper5 (Aug 27, 2016)




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## Viper5 (Sep 2, 2016)

Can somebody please explain the difference?


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