Answer:
a)148.9hp
b)25.2kg
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.
[tex]\alpha g=\frac{Wi}{Wr}[/tex]
where
α=generator efficiency =0.9
Wi=nominial or ideal efficiency=100Kw
Wr=real effiency
solving for Wr
[tex]Wr=\frac{Wi}{\alpha g } =\frac{100Kw}{0.9} =111.11kW[/tex]
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
[tex]111.11 kW \frac{1hp}{0.746Kw} =148.9hp[/tex]
2. Find the engine supply power using the combustion engine efficiency definition
[tex]\alpha m=\frac{Wo}{Ws}[/tex]
α=engine efficiency =0.35
Wo = output power
=111.11Kw
Wi = supply power
solving for Wi
[tex]Wi=\frac{Wo}{\alpha m } =\frac{111.11Kw}{0.35}=317.45Kw[/tex]
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
[tex]m=\frac{Wi}{CP} =\frac{317.45Kw}{45000KJ/g} =7g/s[/tex]
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
[tex]M=(7 \frac{g}{s})1hour\frac{(3600s)}{1hour} =25200g=25.2Kg[/tex]
