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9 October, 02:25

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.

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  1. 9 October, 02:38
    0
    = 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%
  2. 9 October, 04:54
    0
    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%
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