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.

Answers: 2

Show answers

Answers

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.

Learn more:

link

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

answer:

explanation:

Answer from: Quest

Iactually don’t know but i know someone you can ask

Other questions on the subject: Engineering

Engineering, 03.07.2019 14:10, BardiFan

Amass of m 1.5 kg of steam is contained in a closed rigid container. initially the pressure and temperature of the steam are: p 1.5 mpa and t 240°c (superheated state), respectively. then the temperature drops to t2= 100°c as the result of heat transfer to the surroundings. determine: a) quality of the steam at the end of the process, b) heat transfer with the surroundings. for: p1.5 mpa and t 240°c: enthalpy of superheated vapour is 2900 kj/kg, specific volume of superheated vapour is 0. 1483 m/kg, while for t 100°c: enthalpy of saturated liquid water is 419kj/kg, specific volume of saturated liquid water is 0.001043m/kg, enthalpy of saturated vapour is 2676 kj/kg, specific volume of saturated vapour is 1.672 m/kg and pressure is 0.1 mpa.

Answers: 3

continue

Engineering, 04.07.2019 18:10, johnthienann58

Thermal stresses are developed in a metal when its a) initial temperature is changed b) final temperature is changed c) density is changed d) thermal deformation is prevented e) expansion is prevented f) contraction is prevented

Answers: 2

continue

Engineering, 04.07.2019 18:20, rjone8429

Asimple rankine cycle uses water as the working fluid. the water enters the turbine at 10 mpa and 480c while the condenser operates at 6 kpa. if the turbine has an isentropic efficiency of 80 percent while the pump has an isentropic efficiency of 70 percent determine the thermal efficiency

Answers: 1

continue

Engineering, 06.07.2019 03:10, kturvey225

Consider a 1.2 meter high and 2 meter wide glass window whose thickness is 6 mm and thermal conductivity is 0.78 w/m °c. the room temperature is maintained at 24 °c, while the outdoor temperature is -5 °c. the convective heat transfer coefficients on the inner and outer surfaces of the window are 10 w/m2 °c and 25 w/m2 °c respectively. (a) draw the thermal resistance network, etermine the steady rate of heat transfer t e glass window (c) determine the temperature of the inner surface of the window

Answers: 3

continue

You know the right answer?

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