Question

The displacement volume of an internal combustion engine is 3 liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by P1=95kPa, T1=22oC, and V1 = 3.17 liters.
Determine the net work per cycle, in kJ, the power developed by the engine, in kW, and the thermal efficiency, if the cycle is executed 1000 times per min.

Answer 1

Answer:

1) The power developed by the engine is 14705.7739 kW

2) The thermal efficiency is approximately 61.5%

Explanation:

The given parameters are;

P₁ = 95 kPa

T₁ = 22°C

V₁ = 3.17 liters

The cutoff ratio = 2.5

Displacement volume = 3 liters

The number of times the cycle is executed per minute = 1000 times per minute

We have;

The displacement volume = V₁ - V₂ = 3 l

V₁ = 3.17 l

V₂ = 3 - 3.17 = 0.17 l

Compression ratio = V₁/V₂ = 3.17/0.17 ≈ 18.65

P₂/P₁ = P₂/(95 kPa) =  (V₁/V₂)^(k) = 18.65^1.4

P₂ = (95×18.65^(1.4)) ≈ 5710.5 kPa

T₂/T₁ = (V₁/V₂)^(k - 1)

T₂/(295 K)= (18.65)^(1.4 - 1)

T₂ = 295 * (18.65)^(1.4 - 1) = 950.81 K

The cutoff ratio = V₃/V₂ = 2.5

T₃ = T₂ × V₃/V₂  = 2.5 * 950.81 K = 2377.025 K

$Q_{in}$ = $C_p$×(T₃ - T₂) = 1.006 × (2377.025 - 950.81) = 1,434.77 kJ/kg

T₄ = T₃ × (V₃/V₄)^(k-1) =

Therefore,

$T_4 = T_3 \times \left (\dfrac{r_c}{r} \right )^{k - 1} = 2377.025 \times \left( \dfrac{2.5}{18.65} \right )^{1.4 - 1} \approx 1064 \ K$

T₄ ≈ 1064 K

$Q_{out}$ = $-C_v \times (T_4 - T_1)$

$C_v = C_p/k = 1.006/1.4 \approx 0.7186 \ kJ/kg$

∴ $Q_{out}$ = 0.7186×(1064 - 295) = 552.6034 kJ/kg

1) The net work = $Q_{in}$ - $Q_{out}$ = 1,434.77 kJ/kg - 552.6034 kJ/kg ≈ 882.17 kJ/kg

The number of cycle per minute = 1000 rpm

The number of cycle per minute = 1000 rpm/60 = 16.67 cycles per second

The power developed by the engine = The number of cycles per second × The net work of the engine

Therefore;

The power developed by the engine = 16.67 cycles/second  × 882.17 kJ/kg

The power developed by the engine = 14705.7739 kW

2) Efficiency, $\eta _{th}$, is given as follows;

$\eta _{th} = \dfrac{Q_{in}-Q_{out}}{Q_{in}} \times 100 = 1 - \dfrac{Q_{out}}{Q_{in}} \times 100= 1 - \dfrac{552.6034}{1434.77}\times 100 \approx 61.5\%$

Therefore, the thermal efficiency ≈ 61.5%.