Question

An 8-cylinder 4-stroke Diesel engine with a compression ratio of 17 and displacement volume of 6.5 liters is considered. We analyze a case where the temperature is 350 K and the pressure is 150 kPa before the start of compression. The fuel blend used in this engine has a lower heating value of 40 MJ/kg and the air-fuel ratio of each cycle of the engine is 28. Use heat capacities and ratio of heat capacity values at 900 K, with air as the assumed working gas.

Determine:
a. The maximum temperature of the cycle.
b. The cut-off ratio of the cycle, that is, the volume ratio V3/V2.
c. The net work per kg of air per cylinder.
d. The thermal efficiency of the engine.
e. The Mean Effective Pressure.
f. The specific fuel consumption.
g. The break power in kW and hp at the engine speed of 1000 rpm.

Answer 1

a. 2581.807 K

b. 2.783

c. 5.32 kJ

d. 37.41 %

e. 839.8 kPa

f. 240.59 g/kWh

g. 44.34 kW

Step-by-step explanation:

Here, we have

`T_(2)=T_(1)\left ((v_(1))/(v_(2)) \right )^(k-1)`

Properties of air at 900 K are as follows

`c_p`= 1.1209 kJ/kgK)
`c_v`= 0.8338 kJ/kgK) k = 1.344

T₂ = T₁×17^(1.344-1)= 927.56 K

P₂ = P₁×(v₁/v₂)^k = 150 kPa × 17^1.344 = 6757.944 KPa

r = (vc - vd)/vc = 17

v_d = 6.5 liters = 0.0065 m³

∴ v_c = 0.00040625 m³

v₁ = v_c +v_d = 0.00040625 + 0.0065 = 0.00691 m³

The mass of gas in the cylinder = P₁V₁/(RT₁)

= 150·0.00691 /(0.287·350) = 0.010313 kg

The fueled burnt in cycle →

A F = mₐ/m f = (m - m f)/m f = (0.010313 - m f)/m f = 28

m f = 3.556×10⁻⁴ kg

Q_(in) = m f· q H V·ηc = 3.556×10⁻⁴ × 40000 = 14.2 kJ

Q_(in) = m c_v(T₃-T₂) = 14.2 kJ = 0.010313×0.8338 (T₃-927.56 )

T₃ = 2581.807 K (Maximum temperature)

b. The cut-off ratio is T₃/T₂ = 2581.807/927.56 = 2.783

V₂ = V₁/r = 0.0065/17 = 3.824 ×10⁻⁴ m³

V₃ = 2.783( 3.824 ×10⁻⁴ ) = 1.06 ×10⁻³ m³

v₄ = v₁

P₃ = P₂

T₄ = T₃×(V₃/V₄)^(k-1) = 2581.807 (1.06 ×10⁻⁴/0.0065)^0.344 = 1385.43 k

P₄ = P₃(V₃/V₄)^(k) = 6757.944 (1.06 ×10⁻⁴/0.0065)^1.344 = 593.756 kPa

Process 4 - 1

Constant volume heat rejection

Q(out) = m·c_v(T₄ - T₁) = 0.010313 × 0.8338 × (1385.43 - 350) = 8.9 kJ

Giving the net out put as

c. The net work per kg of air per cylinder is

W(net) = Q(in) - Q(out) = 14.2 - 8.9 = 5.32 kJ

d. The thermal efficiency is given by

η(th) = W(net)/Q(in) = 5.32/14.2 = 0.3741 =37.41 %

e. The Mean Effective Pressure (MEP)

MEP = W(net)/(V₁-V₂) = 5.32 /(0.0065-3.824 ×10⁻⁴) = 839.8 kPa

f. The specific fuel consumption

s f c = m f/W(net) = 3.556×10⁻⁴/5.32 = 6.683×10⁻⁵ kg/kJ = 240.59 g/kWh

g. The power for the engine speed of 1000 rpm is given by

W'(net) = W(net) ×n'/2 = (5.32 kJ/cycle)(1000(rev/min)/(2 rev/cycle)(1 min/60 s) = 44.34 kW