# Given certain parameters, find the percentage of energy that becomes

1. Jul 15, 2011

### dboozer

1. The problem statement, all variables and given/known data
The brake fuel conversion efficiency of an egine is 0.3. The mechanical efficiency is 0.8. The combustion efficiency is 0.94. The heat losses to the coolant and oil are 60 kW. The fuel chemical energy entering the engine per unit time, (m_f * Q_HV), is 190 kW. What percentage of this energy becomes
Brake work
Friction work
Heat losses
Exhaust chemical energy
Exhaust sensible energy

2. Relevant equations
?

3. The attempt at a solution
I don't know how to start this problem.

2. Jul 16, 2011

### gsal

energy balance....energy coming in is equal to all the energy going out in various forms...friction, transfered to coolant, potential energy left in the fuel because the system did not get to burn it, etc

hope this helps a bit

3. Jul 17, 2011

### dboozer

Thanks for the help gsal. Your post gave me some ideas, especially the bit about the energy left in the unburned fuel... I'm amazed something as simple as energy balance escaped me.

I looked up some definitions to see if they'd help.

Mechanical Efficiency: Ratio of useful power delivered per cycle to the indicated power per cycle

The gross indicated power per cycle (Pig) is the power delivered to the piston during the compression and expansion strokes only.

$\eta_m = \frac{P_b}{P_{ig}}=1-\frac{P_f}{P_{ig}}$

Fuel Conversion Efficiency: Ratio of work produced per cycle to the amount of fuel energy supplied.

$\eta_f = \frac{P}{\dot{m_f}Q_{HV}}$

Volumetric Efficiency
Volume flow-rate of air into the intake system divided by the rate at which volume is displaced by the piston

$\eta_v = \frac{m_{air}}{\rho_{air}V_d}$

if ρair is the atmospheric air density, then the volumetric efficiency is the pumping performance of the entire system

if ρair is the air density at the inlet manifold, then the volumetric efficiency is the pumping performance of the inlet port and valve only.

Last edited: Jul 17, 2011
4. Jul 17, 2011

### dboozer

Since the combustion efficiency is 0.94 and the energy content of the fuel entering the engine is 190kW, 6% of this energy (11.4 kW) is lost as exhaust chemical energy.

This means that 178.6 kW of fuel energy is delivered to the engine per cycle.

Of this 178.6 kW, 60 kW are lost to the coolant and oil. Only 118.6 kW remain.

Since the brake fuel conversion is 0.3, only 30% of the remaining energy (35.58 kW) is converted to work by the engine. The rest of the energy is used to increase the temperature (83.02 kW is sensible energy).

80% (28.464 kW) of this power is useful, and the remaining 7.116 kW is lost to friction.

Thus
Brake Work = 28.464 kW
Friction Work = 7.116 kW
Heat losses = 60 kW
Exhaust Chemical Energy = 11.4 kW
Exhaust Sensible Energy = 83.02 kW

TOTAL = 28.464 + 7.116 + 60 + 11.4 + 83.02 = 190 kW

To answer the question,
Brake Work = 14.98%
Friction Work = 3.75%
Heat Losses = 31.58%
Exhaust Chemical Energy = 6%
Exhaust Sensible Energy = 43.69%

TOTAL = 100%

Last edited: Jul 17, 2011
5. Jul 17, 2011

Nicely done.

6. Jun 2, 2013

### marellasunny

The fact that the question ask us to find in terms of work is very confusing,given that the fuel energy is given in terms of kW i.e (rate of flow of fuel*heating value of fuel)
Does anyone know how to arrive at work(J) from the power values(kW)?

Also,there is a mistake in the calculations for brake power and friction power.
By definition:
Fuel conversion efficiency(0.3)= Brake power(P_b)/fuel chemical energy(190kW)
=>P_b=57kW

Mechanical efficiency= Brake power(P_b)/Indicated power(P_i)
=>P_i=71.25kW

P_f=14.25kW

and therefore, sensible exhaust energy loss=118.6kW-71.25kW=47.35kW