Diesel Engine driving a generator

AI Thread Summary
A diesel engine with an output of 60 hp operates at 31% efficiency and drives a generator with 79% efficiency, supplying a load at 240 V. The generator output is calculated to be approximately 35.35 kW, and the current drawn is about 147.27 A. The fuel consumption calculation involves converting the engine's power output to BTUs and then to gallons of light fuel oil (LFO), leading to a confusion about the correct conversion factors. The discussion clarifies that the engine's output is the generator's input, and the output power should be used for current calculations. The final fuel consumption needs to be expressed in gallons per day, which requires additional steps in the calculations.
talaroue
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Homework Statement



A diesel engine which drives a generator has an output of 60 hp operating at a n=31%. Then generator has a n=79% and supploes a load circuit at 240 V. The diesel engine uses light fuel oil (LFO) with a calorific value of 19,600 BTU/lb.
Calculate:
1. The generator output in kW
2. The current in the generator
3. the fuel consumption in gallons per day

(weight of LFO=8.3lb/gal; 1BTU=1055J)


Homework Equations


P=IV


The Attempt at a Solution


I am not sure if this is correct, its out of the book so I don't know.

For the first part i simple turned the 60hp into W knowing that 1 hp=745.7 W, so that gave me 44,742 W. I then multiplied that number by .79(the effecicency of the generator to get 35,346.2W which is equal to 35.35 kW. Is this correct?

For the second part I simpled used P=IV plugging in P=35.35kW, and V=240 to get 147.27 A. I believe this part is right aslong as the first part is right.

The third and final part i used what I learned in chemistry, stochiometry in a way,

35,346.2 W *(1 BTU/1055 W)*(1 lb/19,600 BTU)*(1 gal/8.3 lb)=2.06x10^-4 gals

everything except gals cancels, but the question asks gals/day but i don't know how I would do that...

Thanks for your help!
 
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any ideas?
 
I am just confusing myself more and more.
 
Why can't I understand circuits :-/ thank god my major is Civil.
 
talaroue said:

Homework Statement



A diesel engine which drives a generator has an output of 60 hp operating at a n=31%. Then generator has a n=79% and supploes a load circuit at 240 V. The diesel engine uses light fuel oil (LFO) with a calorific value of 19,600 BTU/lb.
Calculate:
1. The generator output in kW
2. The current in the generator
3. the fuel consumption in gallons per day

(weight of LFO=8.3lb/gal; 1BTU=1055J)


Homework Equations


P=IV


The Attempt at a Solution


I am not sure if this is correct, its out of the book so I don't know.

For the first part i simple turned the 60hp into W knowing that 1 hp=745.7 W, so that gave me 44,742 W. I then multiplied that number by .79(the effecicency of the generator to get 35,346.2W which is equal to 35.35 kW. Is this correct?

For the second part I simpled used P=IV plugging in P=35.35kW, and V=240 to get 147.27 A. I believe this part is right aslong as the first part is right.

The third and final part i used what I learned in chemistry, stochiometry in a way,

35,346.2 W *(1 BTU/1055 W)*(1 lb/19,600 BTU)*(1 gal/8.3 lb)=2.06x10^-4 gals

everything except gals cancels, but the question asks gals/day but i don't know how I would do that...

Thanks for your help!

In your equation: 35,346.2 W *(1 BTU/1055 W)*(1 lb/19,600 BTU)*(1 gal/8.3 lb)=2.06x10^-4 gals

Where did (1 BTU/1055 W) come from?
 
Your diesel engine puts out 60 hp but it is only 31% efficient. Therefore the input energy to the engine is 193.5 hp. That is the energy the oil must supply to get 60 hp output.

1 hp=33475 btu/hr, 1 lb of oil provides 19,600 btu's, 1 gal of LFO weighs 8.3 lb, there are 24 hours in a day (assuming the engine 24 hours)

This information will give you gallons of LFO used per day
 
RTW69: Where did you find that 1 hp=33475 btu/hr?

OmCheeto: I thought that it said 1 BTU/1055J but really it said 1 BTU=1055 J so I could just multiple change for BTU to J just by multipling. I think I am going to go with RTW69 said and see where that takes me.
 
Were my other parts correct for this problem then?
 
talaroue said:
RTW69: Where did you find that 1 hp=33475 btu/hr?

OmCheeto: I thought that it said 1 BTU/1055J but really it said 1 BTU=1055 J
It is the same thing in the context of your equation.
so I could just multiple change for BTU to J just by multipling.

Yes. But you converted 1 BTU = 1055 J to 1 BTU = 1055 W.
1 Joule does not equal 1 Watt.
I think I am going to go with RTW69 said and see where that takes me.
My conversion table says 1 hp = 42.4 btu/min, which does not yield 33475 btu/hr.
 
  • #10
I think I was wrong before can someone tell me if I am right now? (i am making another post right now)
 
  • #11
Part A: The generator output in kW

If the engine has a input of 193.55 hp that means the output of the generator has to be 193.55 correct? so then I just do 193.55*745.7 W/1000=144.33 kW?

Part B: The current in the gnerator load ciruit

If the generator has an output of 193.55 with a effeicency of 79% the input is 245 hp, so then I use P=VI solve for I=P/V I have 245 hp*745.7W/240 V=761.24 Amps

Part C: The fuel consumption

193.55hp*(33475BTU/hr)*(1 lb/ 19,600 BTU)* (1 gal/8.3 lb)*(24 hr/1 day)=955.85 gal/day
 
  • #12
or instead of the 193.55 hp should it be the input of the generator which is 245 hp?
 
  • #13
talaroue said:
Part A: The generator output in kW

If the engine has a input of 193.55 hp that means the output of the generator has to be 193.55 correct?

No. The initial problem stated:
A diesel engine which drives a generator has an output of 60 hp

ie. the output of the diesel engine is 60 hp.

The output of the generator will be less since it is only operating at 79% efficiency.
 
  • #14
1 Boiler horsepower (Bohp)=33,475 Btu/hr. It is a term used to rate boilers and is the energy needed to evaporate 34.5 lbs of water at 212 degrees F in one hour. You should probably use electric horsepower instead. I electric horsepower is 746 watts. 1 watt is 3.4129 Btu/hr or 1 electric horsepower is 2545 btu/hr. Sorry for the confusion.
 
  • #15
The output of the engine=the input of the generator? is this correct
 
  • #16
When I find the current in the generator load circuit do I use the power output? or input?
 
  • #17
talaroue said:
The output of the engine=the input of the generator? is this correct

Yes

talaroue said:
When I find the current in the generator load circuit do I use the power output? or input?

Output
 
  • #18
Ok, and why is that?
 
  • #19
talaroue said:
Ok, and why is that?

Why what? Yes or output?
 
  • #20
Why is it the output? Sorry about that
 
  • #21
talaroue said:
Why is it the output? Sorry about that

No. I am sorry. Your answers to parts 1 and 2 in your first post were correct. I guess I should have pointed that out.

Following my hints in posts #5 and #9, and RTW's first two sentences of post #6, will help solve part 3.
 

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