Respuesta :
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
[tex]Q_{in}[/tex] = [tex]C_p[/tex]×(T₃ - T₂) = 1.006 × (2377.025 - 950.81) = 1,434.77 kJ/kg
T₄ = T₃ × (V₃/V₄)^(k-1) =
Therefore,
[tex]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[/tex]
T₄ ≈ 1064 K
[tex]Q_{out}[/tex] = [tex]-C_v \times (T_4 - T_1)[/tex]
[tex]C_v = C_p/k = 1.006/1.4 \approx 0.7186 \ kJ/kg[/tex]
∴ [tex]Q_{out}[/tex] = 0.7186×(1064 - 295) = 552.6034 kJ/kg
1) The net work = [tex]Q_{in}[/tex] - [tex]Q_{out}[/tex] = 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, [tex]\eta _{th}[/tex], is given as follows;
[tex]\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\%[/tex]
Therefore, the thermal efficiency ≈ 61.5%.
