Engineering, 25.02.2020 23:57 karlacr5117

Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each replaced with polytropic processes having n = 1.3. The compression ratio is 10 for the modified cycle. At the beginning of compression, p1 = 1 bar and T1 = 310 K. The maximum temperature during the cycle is 2200 K. Determine

(a) The heat transfer and work in kJ per kg of air for each process in the modified cycle.
(b) The thermal efficiency.
(c) The mean effective pressure, in bar.

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Answer from: goonnation9311

Calculate the compression ratio (r) given in the question.

$\to r = 10 \\\\\frac{V_1}{V_2}=10\\\\$

In processes 1–2, write the expression for the polytropic process in terms of pressure and volume.

$PV^{n} = constant \\\\P_1V_1^{n} = P_2V_2^{n} \\\\\frac{P_1}{P_2}=(\frac{V_2}{V_1})^n\\\\ \frac{ 1 bar(\frac{100 KPa}{1 bar})}{P_2}=(\frac{1}{10})^{13}\\\\ P_2=1 bar(\frac{ 100\ KPa }{1 \ bar}) 10^{13}\ = 1995.262 kPa$

In processes 1–2, write the expression for the polytropic process in terms of temperature and volume.

$TV^{n-1} = constantT_1V_1^{n-1} = T_2V_2^{n-1} \\\\\frac{T_1}{T_2} =(\frac{V_2}{V_1})^{n-1} \\\\\frac{310 K}{T_2} =(\frac{1}{10})^{13-1} \\\\T_2= 310 (10)^{13-1}= 618.531\ K\\\\$

In terms of temperature and pressure, write the standard formula for the constant volume process (processes 2–3).

$\frac{P}{T} = constant \\\\ \frac{P_2}{T_2} =\frac{P_3}{T_3}\\\\\frac{1995.262\ kPa}{ 618.531 \ K}=\frac{P_3}{2200 K}\\\\ P_3=\frac{1995.262\ kPa ( 2200\ K)}{618.531 \ K} = 7096.78\ kPa \\\\$

Processes 2 to 3 have a constant volume. As a result, V_2 = V_3

The 4 to 1 process has a constant volume. As a result, V_1=V_4

Give the expression for polytropic process 3 to 4 in terms of pressure and volume.

$PV^n = constant\\\\ P_3V_3^n = P_4V_4^n \\\\\frac{P_3}{P_4} =(\frac{V_4}{V_3})^n\\\\\frac{P_3}{P_4} =(\frac{V_1}{V_2})^n\\\\ \frac{7096.78 \ kPa}{P_4} = 10^{13}\\\\P_4=\frac{7096.78\ kPa}{10^{13}} = 355.68\ kPa\\\\$

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.

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Consider a modification of the air-standard Otto cycle in which the isentropic compression and expan

Answer from: hannahchristine457

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 %

Explanation:

(a) Volume compression ratio $\frac{v_1}{v_2} = 10$

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

$\frac{p_1}{p_2} = (\frac{v_2}{v_1} )^n$ Therefore p₂ = p₁ ÷ $(\frac{v_2}{v_1} )^n$ = (1 bar) ÷ $(\frac{1}{10} )^{1.3}$ = 19.953 bar

Similarly for a polytropic process we have

$\frac{T_1}{T_2} = (\frac{v_2}{v_1} )^{n-1}$ or T₂ = T₁ ÷ $(\frac{v_2}{v_1} )^{n-1}$ = $\frac{310}{0.1^{0.3}}$ = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = $\frac{8.3145}{28.9628}$ = 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 = $\frac{R(T_2-T_1)}{n-1}$ = $\frac{0.287(618.531-310)}{1.3 - 1}$ = 295.16 kJ/kg

Q = $(\frac{n-\gamma}{\gamma - 1} )W$ = $(\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg$ = -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 = $\frac{R(T_4-T_3)}{n-1}$ = Where T₄ is given by $\frac{T_4}{T_3} = (\frac{v_3}{v_4} )^{n-1}$ or T₄ = T₃ × $0.1^{0.3}$

= 2200 × $0.1^{0.3}$ T₄ = 1102.611 K

W = $\frac{0.287(1102.611-2200)}{1.3 - 1}$ = -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

$\eta = 1-\frac{T_4-T_1}{T_3-T_2}$ = $1-\frac{1102.611-310}{2200-618.531}$ = 0.499 or 49.9 % Efficient

$p_{m} = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}$ where r = compression ratio and r_p = $\frac{p_3}{p_2}$

However p₃ = $\frac{p_2T_3}{T_2}$ = $\frac{(19.953)(2200)}{618.531}$ =70.97 atm

r_p = $\frac{p_3}{p_2}$ = $\frac{70.97}{19.953} = 3.56$

Therefore $p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9}$ = 9.44 bar

Please find attached generalized diagrams of the Otto cycle

Answer from: Quest

correct me if i’m wrong but i believe it’s false

Answer from: Quest

It would be d because i did like a whole test about this

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Consider a modification of the air-standard Otto cycle in which the isentropic compression and expan...

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