Answer:
The answers to the question are
(1) Process 1 to 2
W = 295.16 kJ/kg
Q = -73.79 kJ/kg
(2) Process 2 to 3
W = 0
Q = 1135.376 kJ/kg
(3) Process 3 to 4
W = -1049.835 kJ/kg
Q = 262.459 kJ/kg
(4) Process 4 to 3
W=0
Q = -569.09 kJ/kg
(b) The thermal efficiency = 49.9 %
(c) The mean effective pressure is 9.44 bar
Explanation:
(a) Volume compression ratio [tex]\frac{v_1}{v_2} = 10[/tex]
Initial pressure p₁ = 1 bar
Initial temperature, T₁ = 310 K
cp = 1.005 kJ/kg⋅K
Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle
For a polytropic process we have
[tex]\frac{p_1}{p_2} = (\frac{v_2}{v_1} )^n[/tex] Therefore p₂ = p₁ ÷ [tex](\frac{v_2}{v_1} )^n[/tex] = (1 bar) ÷ [tex](\frac{1}{10} )^{1.3}[/tex] = 19.953 bar
Similarly for a polytropic process we have
[tex]\frac{T_1}{T_2} = (\frac{v_2}{v_1} )^{n-1}[/tex] or T₂ = T₁ ÷ [tex](\frac{v_2}{v_1} )^{n-1}[/tex] = [tex]\frac{310}{0.1^{0.3}}[/tex] = 618.531 K
The molar mass of air is 28.9628 g/mol.
Therefore R = [tex]\frac{8.3145}{28.9628}[/tex] = 0.287 kJ/kg⋅K
cp = 1.005 kJ/kg⋅K Therefore cv = cp - R = 1.005- 0.287 = 0.718 kJ/kg⋅K
1). For process 1 to 2 which is polytropic process we have
W = [tex]\frac{R(T_2-T_1)}{n-1}[/tex] = [tex]\frac{0.287(618.531-310)}{1.3 - 1}[/tex]= 295.16 kJ/kg
Q =[tex](\frac{n-\gamma}{\gamma - 1} )W[/tex] = [tex](\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg[/tex] = -73.79 kJ/kg
W = 295.16 kJ/kg
Q = -73.79 kJ/kg
2). For process 2 to 3 which is reversible constant volume heating we have
W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg
W = 0
Q = 1135.376 kJ/kg
3). For process 3 to 4 which is polytropic process we have
W = [tex]\frac{R(T_4-T_3)}{n-1}[/tex] = Where T₄ is given by [tex]\frac{T_4}{T_3} = (\frac{v_3}{v_4} )^{n-1}[/tex] or T₄ = T₃ ×[tex]0.1^{0.3}[/tex]
= 2200 ×[tex]0.1^{0.3}[/tex] T₄ = 1102.611 K
W = [tex]\frac{0.287(1102.611-2200)}{1.3 - 1}[/tex]= -1049.835 kJ/kg
and Q = 262.459 kJ/kg
W = -1049.835 kJ/kg
Q = 262.459 kJ/kg
4). For process 4 to 1 which is reversible constant volume cooling we have
W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg
W=0
Q = -569.09 kJ/kg
(b) The thermal efficiency is given by
[tex]\eta = 1-\frac{T_4-T_1}{T_3-T_2}[/tex] =[tex]1-\frac{1102.611-310}{2200-618.531}[/tex] = 0.499 or 49.9 % Efficient
(c) The mean effective pressure is given by
[tex]p_{m} = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}[/tex] where r = compression ratio and [tex]r_p[/tex] = [tex]\frac{p_3}{p_2}[/tex]
However p₃ = [tex]\frac{p_2T_3}{T_2}[/tex] =[tex]\frac{(19.953)(2200)}{618.531}[/tex] =70.97 atm
[tex]r_p[/tex] = [tex]\frac{p_3}{p_2}[/tex] = [tex]\frac{70.97}{19.953} = 3.56[/tex]
Therefore [tex]p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9}[/tex] = 9.44 bar
Please find attached generalized diagrams of the Otto cycle
Calculate the compression ratio (r) given in the question.
[tex]\to r = 10 \\\\\frac{V_1}{V_2}=10\\\\[/tex]
In processes 1–2, write the expression for the polytropic process in terms of pressure and volume.
[tex]PV^{n} = constant \\\\P_1V_1^{n} = P_2V_2^{n} \\\\\frac{P_1}{P_2}=(\frac{V_2}{V_1})^n\\\\ \frac{ 1 bar(\frac{100 KPa}{1 bar})}{P_2}=(\frac{1}{10})^{13}\\\\ P_2=1 bar(\frac{ 100\ KPa }{1 \ bar}) 10^{13}\ = 1995.262 kPa[/tex]
In processes 1–2, write the expression for the polytropic process in terms of temperature and volume.
[tex]TV^{n-1} = constantT_1V_1^{n-1} = T_2V_2^{n-1} \\\\\frac{T_1}{T_2} =(\frac{V_2}{V_1})^{n-1} \\\\\frac{310 K}{T_2} =(\frac{1}{10})^{13-1} \\\\T_2= 310 (10)^{13-1}= 618.531\ K\\\\[/tex]
In terms of temperature and pressure, write the standard formula for the constant volume process (processes 2–3).
[tex]\frac{P}{T} = constant \\\\ \frac{P_2}{T_2} =\frac{P_3}{T_3}\\\\\frac{1995.262\ kPa}{ 618.531 \ K}=\frac{P_3}{2200 K}\\\\ P_3=\frac{1995.262\ kPa ( 2200\ K)}{618.531 \ K} = 7096.78\ kPa \\\\[/tex]
Processes 2 to 3 have a constant volume. As a result, [tex]V_2 = V_3[/tex]
The 4 to 1 process has a constant volume. As a result, [tex]V_1=V_4[/tex]
Give the expression for polytropic process 3 to 4 in terms of pressure and volume.
[tex]PV^n = constant\\\\ P_3V_3^n = P_4V_4^n \\\\\frac{P_3}{P_4} =(\frac{V_4}{V_3})^n\\\\\frac{P_3}{P_4} =(\frac{V_1}{V_2})^n\\\\ \frac{7096.78 \ kPa}{P_4} = 10^{13}\\\\P_4=\frac{7096.78\ kPa}{10^{13}} = 355.68\ kPa\\\\[/tex]
In processes 3–4, write the expression for the polytropic process in terms of temperature and volume.
Please find the attached file for the steps.
Learn more:
brainly.com/question/14748651
