Respuesta :
Answer:
a) T₂ is 701.479 K
T₃ is 1226.05 K
T₄ is 2350.34 K
T₅ is 1260.56 K
b) The net work of the cycle in kJ is 2.28 kJ
c) The power developed is 114.2 kW
d) The thermal efficiency, [tex]\eta _{dual}[/tex] is 53.78%
e) The mean effective pressure is 1038.25 kPa
Explanation:
a) Here we have;
[tex]\frac{T_{2}}{T_{1}}=\left (\frac{v_{1}}{v_{2}} \right )^{\gamma -1} = \left (r \right )^{\gamma -1} = \left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}[/tex]
Where:
p₁ = Initial pressure = 95 kPa
p₂ = Final pressure =
T₁ = Initial temperature = 290 K
T₂ = Final temperature
v₁ = Initial volume
v₂ = Final volume
[tex]v_d[/tex] = Displacement volume =
γ = Ratio of specific heats at constant pressure and constant volume cp/cv = 1.4 for air
r = Compression ratio = 9.1
Total heat added = 4.25 kJ
1/4 × Total heat added = [tex]c_v \times (T_3 - T_2)[/tex]
3/4 × Total heat added = [tex]c_p \times (T_4 - T_3)[/tex]
[tex]c_v[/tex] = Specific heat at constant volume = 0.718×2.821× 10⁻³
[tex]c_p[/tex] = Specific heat at constant pressure = 1.005×2.821× 10⁻³
v₁ - v₂ = 2.2 L
[tex]\left \frac{v_{1}}{v_{2}} \right =r \right = 9.1[/tex]
v₁ = v₂·9.1
∴ 9.1·v₂ - v₂ = 2.2 L = 2.2 × 10⁻³ m³
8.1·v₂ = 2.2 × 10⁻³ m³
v₂ = 2.2 × 10⁻³ m³ ÷ 8.1 = 2.72 × 10⁻⁴ m³
v₁ = v₂×9.1 = 2.72 × 10⁻⁴ m³ × 9.1 = 2.47 × 10⁻³ m³
Plugging in the values, we have;
[tex]{T_{2}}= T_{1} \times \left (r \right )^{\gamma -1} = 290 \times 9.1^{1.4 - 1} = 701.479 \, K[/tex]
From;
[tex]\left (\frac{p_{2}}{p_{1}} \right )^{\frac{\gamma -1}{\gamma }}= \left (r \right )^{\gamma -1}[/tex] we have;
[tex]p_{2} = p_{1}} \times \left (r \right )^{\gamma } = 95 \times \left (9.1 \right )^{1.4} = 2091.13 \ kPa[/tex]
1/4×4.25 = [tex]0.718 \times 2.821 \times 10^{-3}\times (T_3 - 701.479)[/tex]
∴ T₃ = 1226.05 K
Also;
3/4 × Total heat added = [tex]c_p \times (T_4 - T_3)[/tex] gives;
3/4 × 4.25 = [tex]1.005 \times 2.821 \times 10^{-3} \times (T_4 - 1226.05)[/tex] gives;
T₄ = 2350.34 K
[tex]\frac{T_{4}}{T_{5}}=\left (\frac{v_{5}}{v_{4}} \right )^{\gamma -1} = \left (\frac{r}{\rho } \right )^{\gamma -1}[/tex]
[tex]\rho = \frac{T_4}{T_3} = \frac{2350.34}{1226.04} = 1.92[/tex]
[tex]T_{5} = \frac{T_{4}}{\left (\frac{r}{\rho } \right )^{\gamma -1}}= \frac{2350.34 }{\left (\frac{9.1}{1.92 } \right )^{1.4-1}} =1260.56 \ K[/tex]
b) Heat rejected = [tex]c_v \times (T_5 - T_1)[/tex]
[tex]Therefore \ heat \ rejected = 0.718 \times 2.821 \times 10^{-3}\times (1260.56 - 290) = 1.966 kJ[/tex]
The net work done = Heat added - Heat rejected
∴ The net work done = 4.25 - 1.966 = 2.28 kJ
The net work of the cycle in kJ = 2.28 kJ
c) Power = Work done per each cycle × Number of cycles completed each second
Where we have 3000 cycles per minute, we have 3000/60 = 50 cycles per second
Hence, the power developed = 2.28 kJ/cycle × 50 cycle/second = 114.2 kW
d)
[tex]Thermal \ efficiency, \, \eta _{dual} = \frac{Work \ done}{Heat \ supplied} = \frac{2.28}{4.25} \times 100 = 53.74 \%[/tex]
The thermal efficiency, [tex]\eta _{dual}[/tex] = 53.78%
e) The mean effective pressure, [tex]p_m[/tex], is found as follows;
[tex]p_m = \frac{W}{v_1 - v_2} =\frac{2.28}{2.2 \times 10^{-3}} = 1038.25 \ kPa[/tex]
The mean effective pressure = 1038.25 kPa.
