Respuesta :
Answer:
a) T₃ = 1818.8 K
b) η = 0.614 = 61.4%
c) MEP = 660.4 kPa
Explanation:
a) According to Table A-2 of The ideal gas specific heat of gases, the properties of air are as following:
At 300K
The specific heat capacity at constant pressure = [tex]c_{p}[/tex] = 1.005 kJ/kg.K,
The specific heat capacity at constant volume = [tex]c_{v}[/tex] = 0.718 kJ/kg.K
Gas constant R for air = 0.2870 kJ/kg·K
Ratio of specific heat k = 1.4
Isentropic Compression :
[tex]T_{2}[/tex] = [tex]T_{1}[/tex] [tex](v1/v2)^{k-1}[/tex]
= 300K ([tex]16^{0.4}[/tex])
[tex]T_{2}[/tex] = 909.4K
P = Constant heat Addition:
[tex]P_{3}v_{3} / T_{3} = P_{2} v_{2} /T_{2}[/tex]
[tex]T_{3}=v_{3}/v_{2}T_{2}[/tex]
2[tex]T_{2}[/tex] = 2(909.4K)
= 1818.8 K
b) [tex]q_{in}[/tex] = [tex]h_{3}-h_{2}[/tex]
= [tex]c_{p}[/tex] ([tex]T_{3}[/tex] - [tex]T_{2}[/tex])
= (1.005 kJ/kg.K)(1818.8 - 909.4)K
= 913.9 kJ/kg
Isentropic Expansion:
[tex]T_{4}[/tex] = [tex]T_{3}[/tex] [tex](v3/v4)^{k-1}[/tex]
= [tex]T_{3}[/tex] [tex](2v_{2} /v_{4} )^{k-1}[/tex]
= 1818.8 K (2 / 16[tex])^{0.4}[/tex]
= 791.7K
v = Constant heat rejection
[tex]q_{out}[/tex] = μ₄ - μ₁
= [tex]c_{v} ( T_{4} - T_{1} )[/tex]
= 0.718 kJ/kg.K (791.7 - 300)K
= 353 kJ/kg
η[tex]_{th}[/tex] = 1 - [tex]q_{out}[/tex] / [tex]q_{in}[/tex]
= 1 - 353 kJ/kg / 913.9 kJ/kg
= 1 - 0.38625670
= 0.6137
= 0.614
= 61.4%
c) [tex]w_{net}._{out}[/tex] = [tex]q_{in}[/tex] - [tex]q_{out}[/tex]
= 913.9 kJ/kg - 353 kJ/kg
= 560.9 kJ/kg
[tex]v_{1} = RT_{1} /P_{1}[/tex]
= (0.287 kPa.m³/kg/K)*(300 K) / 95 kPa
= 86.1 / 95
= 0.9063 m³/kg = v[tex]_{max}[/tex]
[tex]v_{min} =v_{2} = v_{max} /r[/tex]
Mean Effective Pressure = MEP = [tex]w_{net,out}/v_{1} -v_{2}[/tex]
= [tex]w_{net,out}/v_{1}(1-1)/r[/tex]
= 560.9 kJ/kg / (0.9063 m³/kg)*(1-1)/16
= (560.9 kJ / 0.8493m³) (kPa.m³/kJ)
= 660.426 kPa
Mean Effective Pressure = MEP = 660.4 kPa
The temperature after the addition process is 1724.8k, the thermal efficiency of the engine is 56.3% and the mean effective pressure is 65.87kPa
Assumptions made:
- The air standard assumptions are made
- The kinetic and potential energy changes are negligible
- The air in the system is an ideal gas with variable or different specific heat capacity.
a) The temperature after the addition process:
Considering the process 1-2, Isentropic expansion
at
[tex]T_1=300k\\u_1=214.07kJ/kg\\v_o_1=621.3\\v_o_2=\frac{v_2}{v_1} *v_o_1[C.R=16]=v_2/v_1\\v_o_2=(v_2/v_1)v_o_1=1/16*621.2=38.825[/tex]
From using this value, v[tex]_o_2[/tex]=38.825, solve for state point 2;
[tex]T_2=862.4k\\h_2=890.9kJ/kg[/tex]
Considering the process 2-3 (state of constant heat addition)
[tex]\frac{p_3v_3}{t_3}=\frac{p_2v_2}{t_2} \\\\T_3=\frac{P_3V_3T_2}{V_2} \\T_3=(\frac{V_3}{V_2}) T_2\\\frac{v_3}{v_2}=2\\T_3=2(862.4)=1724.8k\\[/tex]
NB: p[tex]_3[/tex]≈p[tex]_2[/tex]
b) The thermal efficiency of the engine is
Q[tex]_i_n[/tex]=h[tex]_3-h_2[/tex] = 1910.6-890.9=1019.7kJ/kg
Considering process 3-4,
[tex]v_o_4=\frac{v_A}{v_2}\\ v_o_3 =\frac{V_a}{V_2}*\frac{v_2}{v_3}\\v_o_3=\frac{16}{2}*4.546\\v_o_3=36.37;v_4=659.7kJ/kg[/tex]
Q[tex]_o_u_t=v_4-u_1=659.7-214.07=445.3kJ/kg[/tex]
nth = [tex]1-\frac{Q_o_u_t}{Q_i_n}=1-\frac{445.63}{1019.7}=0.5629*100=56.3%[/tex]%
The thermal efficiency is 56.3%
W[tex]_n_e_t[/tex]=[tex]Q_i_n-Q_o_u_t=574.07kJ/kg[/tex]
[tex]v_1=\frac{RT_1}{p_1}=\frac{0.287*300}{95}=0.906m^3/kg\\v_2=v_1/16=0.05662m^3/kg\\[/tex]
Therefore, the mean effective pressure of the system engine is
[tex]\frac{W_n_e_t}{v_1-v_2}=675.87kPa[/tex]
The mean effective pressure is 65.87kPa as calculated above
Learn more about mean effective pressure
https://brainly.com/question/19309495
