Question

6–23 an automobile engine consumes fuel at a rate of 22 l/h and delivers 55 kw of power to the wheels. if the fuel has a heating value of 44,000 kj/kg and a density of 0.8 g/cm3, determine the efficiency of this engine. answer: 25.6 percent

Answer 1

Explanation & answer:

Given:

Fuel consumption, C = 22 L/h

Specific gravity = 0.8

output power, P  =  55 kW

heating value, H = 44,000 kJ/kg

Solution:

Calculate energy intake

E = C*P*H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

= 215111.1 j/s

Calculate output power

P = 55 kW

= 55000 j/s

Efficiency

= output / input

= P/E

=55000 / 215111.1

= 0.2557

= 25.6% to 1 decimal place.

Answer 2

The efficiency of this engine will be "25.6 %".

Given values are:  

Fuel consumption,

• C = 22 L/h  

Specific gravity,

• 0.8  

Output power,

• P  =  55 kW  

Heating value,

• H = 44,000 kJ/kg  

The energy intake will be:  

→ $E = C\times P\times H$  

      $= \frac{(22)}{ (3600 )} \times (1000)\times (0.8)\times (44000)$

      $= 215111.1 \ J/s$      

   The output power will be:  

→ $P = 55\ kW$  

      $= 55000 \ J/s$  

Efficiency will be:  

$= \frac{Out.}{Inp.}$    

$= \frac{55000}{215111.1}$  

$= 0.2557$  

or,

$= 25.6$ (%)

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