Question

A standby generator powered by and internal combustion engine has a rated capacity of 100kW. Assuming that the generator is 90% efficient how much horsepower must the engine deliver to the generator at full load? Assuming that the thermal efficiency of the engine is 35% how much diesel fuel will the engine consume in one hour at full load?

Answer 1

a)148.9hp

b)25.2kg

Step-by-step explanation:

Hello!

To solve this exercise follow the steps below.

1. Find the power the generator needs to operate at 100Kw,

This is done by remembering that efficiency in a generator is the ratio between real power and ideal power.

`\alpha g=(Wi)/(Wr)`

where

α=generator efficiency =0.9

Wi=nominial or ideal efficiency=100Kw

Wr=real effiency

solving for Wr

`Wr=(Wi)/(\alpha g ) =(100Kw)/(0.9) =111.11kW`

the power that the combustion engine must deliver is 111Kw for the generator to deliver 100Kw, Now use conversion factor to know the value in horsepower

`111.11 kW (1hp)/(0.746Kw) =148.9hp`

2. Find the engine supply power using the combustion engine efficiency definition

`\alpha m=(Wo)/(Ws)`

α=engine efficiency =0.35

Wo = output power

=111.11Kw

Wi = supply power

solving for Wi

`Wi=(Wo)/(\alpha m ) =(111.11Kw)/(0.35)=317.45Kw`

3. use the calorific power of diessel (CP=45KJ /g) to find the mass flow, remember that the supply power is the product of the mass flow and the caloric power

Wi=m(CP)

m=mass flow

solving for m

`m=(Wi)/(CP) =(317.45Kw)/(45000KJ/g) =7g/s`

4. The mass flow is the ratio between the mass consumed and the time (1h) in this way we can find the diessel consumed in one hour at full load

M=diessel mass consumed

m=M/t

M=mt

`M=(7 (g)/(s))1hour((3600s))/(1hour) =25200g=25.2Kg`