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an ideal otto cycle has a compression ratio of 8. at the begining of the compression process, air is at 95 kpa and 27 c, and 750kj/kg of heat is transferred to air during the constant volume heat addition process Taking into account the variation of specific heats with temperature, determine
(a) the pressure and temperature at the end of the heataddition process,
(b) the net work output,
c) the thermal efficiency, and
(d) the mean effective pressure for the cycle.

Respuesta :

onyebuchinnaji

Answer:

(a) the pressure and temperature at the end of the heat addition process is 1733.79 K and 4392.26 Kpa respectively

(b) the net work output is 423.54 KJ/Kg

(c) the thermal efficiency is 56.5%

(d) the mean effective pressure for the cycle cannot be determined without initial volume of the process

Explanation:

Assumptions:

  • changes in kinetic and potential energy is negligible
  • air is an ideal gas with constant specific heats

The properties of air at room temperature;

Cp = 1.005 KJ/kg.K, Cv = 0.718KJ/kg.K, R = 0.287KJ/kg.K and K = 1.4

Part a:

For isentropic compression:

[tex]T_2=T_1[\frac{V_1}{V_2}]^{K-1}[/tex]

Where;

T₁ = (27+273)K =300K

V₁/V₂ = 8

[tex]T_2=300[8]^{1.4-1} = 300[8]^{0.4} = 689.22K[/tex]

Based on the assumption above;

Q₁ₙ = U₃ -U₂

For an ideal gas with constant specific heats, the change in internal energy in terms of the change in temperature is shown below

U₃ -U₂ = Cv(T₃-T₂)

Thus; Q₁ₙ = Cv(T₃-T₂)

T₃ = (Q₁ₙ/Cv) + T₂

T₃ = (750/0.718) + 689.22 K = 1733.79 K

From general gas equation, we find the second stage pressure

[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]

[tex]P_2 =P_1[\frac{T_2}{T_1}][\frac{V_1}{V_2}] = 95 Kpa[\frac{689.22}{300}](8)[/tex]

P₂ = 1746.024Kpa

To obtain the pressure at stage 3

[tex]P_3 =P_2[\frac{T_3}{T_2}][\frac{V_2}{V_3}] = 1746.024 Kpa[\frac{1733.79}{689.22}](1)[/tex]

P₃ = 4392.26 Kpa

Part b:

To obtain net work output we consider overall energy balance on the cycle

[tex]Q_{out} = U_4-U_1 = C_v(T_4-T_1)[/tex]

For is isentropic  expansion

[tex]T_4=T_3[\frac{V_3}{V_4}]^{K-1} = 1733.79 [\frac{1}{8}]^{0.4} = 754.68K[/tex]

[tex]Q_{out} = 0.718(754.68-300) = 326.46 KJ/Kg[/tex]

To solve for net work output:

[tex]Q_{net} = Q{in}- Q_{out}[/tex] = (750 - 326.46) KJ/Kg = 423.54 KJ/Kg

part c:

To calculate the thermal efficiency, we use net work output per input work

η = 423.54/750

η =  0.565 = 56.5%

part d:

Mean effective pressure for the cycle (MEP)

[tex]MEP = \frac{Q_{net}}{V_1-V_2}[/tex] = [tex]\frac{Q_{net}}{V_1(1-\frac{1}{r})} = \frac{423.54}{V_1(1-\frac{1}{8}) }[/tex]

MEP = [tex]\frac{Q_{net}}{V_1(1-\frac{1}{r})} = \frac{423.54}{V_1(0.875)}[/tex]

Thus mean effective pressure cannot be determined without initial volume of the process.

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