Answer to Question #215329 in Molecular Physics | Thermodynamics for Wavie

Question

The following data refer to a four cylinder internal combustion engine operating on a four stroke cycle at 4400 rev/min: volume compression ratio 7.6:1; cylinder diameter 78mm; stroke 105mm; brake power 38 kW; indicated power 47.5 kW; air consumption 21 kg/kg fuel; fuel consumption 13.6 kg/hour; energy release of 1 kg of fuel, 41800 kJ. Determine,

(i) the mechanical efficiency

(ii) the mean effective pressure

(iii) the volumetric efficiency (referred to 1 atm = 1.01325 bar, 0°C)

(iv) the brake thermal efficiency

(v) the brake thermal efficiency ratio relative to the air-standard cycle.

Assume R = 0.287 kJ/kg K.

Answer 1

Solution

(i)The mechanical efficiency

n_m=$\frac{b.p}{i.p}$

b.p is the brake power

i.p is the indicated power

n_m=$\frac{38}{47.5}=0.8$

(ii)the mean effective pressure

We know;

i.p=$\frac{P_mLAN}{60000×2}×n$

P_m is the mean effective pressure N/m²

L is the stroke m

A is the cross section area of cylinder m²

N is the speed in RPM

n is number of cylinders

47.5=$\frac{P_m×0.105×4.7784×10^-3×4400}{60000×2}×4$

P_m=645496.3666N/m²

Answer

P_m=6.455 bar

(iii)the volumetric efficiency (referred to 1atm=1.01325bar,0°c)

Calculate the volume induced per minute

V_i=$\frac{mRT}{p}$ =$\frac{13.6×21×0.287×273}{60×10^2×1.01325}$ =3.68m³/min

Calculate the volume swept per minute,

$V_s=4×\frac14π×78^2×10^-6×0.105×\frac{4400}{2}$

V_s=4.4m³/min

Volumetric efficiency n_v

n_v=$\frac{3.68}{4.4}=0.836$

Answer

0.836 or 83.6%

(iv)Brake thermal efficiency n_b

n_b=$\frac{work output}{energy supplied}$

n_b=$\frac{47.5×3600}{13.6×41800}=0.302$

Answer

0.302 or 30.2%

(v) the brake thermal efficiency relative to air standard cycle.

The efficiency of air standard cycle of comparison is give by;

n=1-$\frac{1}{r^{\gamma-1}}$ =$1-\frac{1}{7.6^0.4}=0.555$

The ratio with brake power

Efficiency ratio=$\frac{0.302}{0.555}=0.5441$

Answer

0.544