Answer:
= 25.6%
Explanation:
Given that,
Fuel consumption, C = 22 L/h
Specific gravity = 0.8
output power, P = 55 kW
heating value, H = 44,000 kJ/kg
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%
Given Information:
Output Power = P = 55 kW
Fuel Consumption Rate = ν = 22 L/h
Fuel Heating Value q = 44,000 kJ/kg
Fuel Density = ρ = 0.8 g/cm³
Required Information:
Efficiency of the engine = η = ?
Answer:
Efficiency of the engine = 25.56%
Explanation:
The efficiency of the engine is given by
η = P/Q
Where P is the output power and Q is the input heat supply
The heat supply is given by
Q = ρνq
Where ν is fuel consumption rate and q is the heating value of the fuel and ρ is density
First convert density from g/cm³ to kg/L
0.8*1000L/1000kg = 0.8 kg/L
Q = 0.8*22*44,000
Q = 7.74x10⁵ kJ/h
Convert kJ/h to kW (kW and kJ/s are equivalent units)
Q = 7.74x10⁵/3600
Q = 215.11 kW
Finally, now we can find the efficiency
η = P/Q
η = 55/215.11
η = 0.2556
η = 25.56%
Therefore, the efficiency of this engine is 25.56%
