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At the beginning of the compression process of an air-standard Otto cycle, p1 = 1 bar and T1 = 300 K. The compression ratio is 8.5 and the heat addition per unit mass of air is 1400 kJ/kg. Determine (a) the net work, in kJ/kg, (b) the thermal efficiency of the cycle, (c) the mean effective pressure, in bar, (d) the maximum temperature of the cycle, in K.

Respuesta :

Answer:

a) w_cycle = 721.16 KJ/kg

b) n = 0.515

c) mep = 949.2 KPa

d) T_3 = 2231.3 K

Explanation:

Given:

-State 1: T_1 =  300K , P_1 = 1 bar

- Heat added in q_in = 1400 KJ/kg

- compression ratio r = 8.5

Find:

a) The net work, in kJ per kg of air.

b) The thermal efficiency of the cycle.

c) The mean effective pressure, in kPa

d) The maximum temperature in the cycle, in K

Solution:

Assumptions:

1) The air in the piston-cylinder is the closed system.

2) The compression and expansion processes are adiabatic, ΔP = 0.

3) All processes are internally reversible

4) The air is modeled as an ideal gas

5) Kinetic and potential energy effects are negligible

State 1:

T_1 =  300 K , P_1 = 1 bar = 100 KPa

From Air property tables:

u_1 = 214.07 kJ/ kg, V_r,1 = 621.2 m^3 / kg

State 2:

For isentropic compression

V_r,2 = V_r,1 * (V_2 / V_1) = 621.2 / 8.5 = 73.082 m^3 / kg

Hence,

T_2 = 688.2 K, u_2 = 503.06 kJ/ kg

State 3:

The specific internal energy  u_3  is found by using the energy balance for process 2-3.

m * (u_3 - u_2) = Q_in

u_3 = (Q_in / m) + u_2

u_3 = q_in + u_2

u_3 = 1400 + 503.06 = 1903.6 KJ/kg

Hence,

T_3 = 2231.3 K, V_r,3 = 1.9192 m^3 / kg

State 4:

For the isentropic expansion

:

V_r,4 = V_r,3 * (V_4 / V_3 ) =  V_r,3 * (V_1 / V_2 ) = 1.9192*8.5 = 16.3132 m^3 / kg

Hence,

T_4 = 1154.3 K , u_4 = 892.91 KJ/kg

part a

To find net work done of cycle w_cycle:

w_cycle = q_cycle = q_in + (u_1 - u_4)

w_cycle = 1400 + (214.07 - 892.91)

w_cycle = 721.16 KJ/kg

part b

The thermal efficiency is

:

n = w_cycle / q_in = 721.16 / 1400 = 0.515

part c

The displacement volume is  V_1 - V_2 =  m(v_1 - v_2 ), so the mean effective pressure is  given by:

mep = w_cycle / (v_1 - v_2) = w_cycle *v_1/ (1 - 1/r)

Evaluate v_1, use ideal gas law:

v_1 = R*T_1 / P_1 = 0.2869*300/100 = 0.861 m^3/kg

mep = w_cycle *v_1/ (1 - r) = 721.16 * 0.861 / (1 - 1/8.5) = 949.2 KPa

part d

the maximum temperature of the cycle, in K

From state point 3 , we got T_3 = 2231.3 K

Ver imagen shahnoorazhar3
onyebuchinnaji

Answer:

(a) The net work is 721.16 KJ/Kg

(b) The thermal efficiency of the cycle is 51.51%

(c) The mean effective pressure is 9.492 bar

(d) the maximum temperature of the cycle is 1154.3 K

Explanation:

Given: P₁ = 1 bar = 100kPa, T₁ = 300K, V₁/V₂ = 8.5, Q/m = 1400 kJ/kg

To solve this problem the following Assumptions will be made

  • The air in the piston-cylinder assembly is the closed system.
  • The compression and expansion processes are adiabatic.
  • All processes are internally reversible
  • The air is modeled as an ideal gas
  • Kinetic and potential energy effects are negligible

If P₁ = 100kPa, T₁ = 300K, Then U₁ =214.07 kg/kg, Vr₁ = 621.2

For isentropic compression:

[tex]V_r_2 = V_r_1\frac{V_2}{V_1} = \frac{621.2}{8.5} = 73.082[/tex]

Then, T₂ = 688.2K, U₂ = 503.06 kJ/kg

To calculate specific internal Energy (U₃), we apply energy balance process

m(U₃-U₂) = Q -W, ⇒ m(U₃-U₂) = Q -0

m(U₃-U₂) = Q, ⇒ U₃ = Q/m + U₂

U₃ = 1400 + 503.06 = 1903.06 kJ/kg

Thus, T₃ = 231.3 K, Vr₃ = 1.9192

For the isentropic expansion:

[tex]V_r_4= V_r_3\frac{V_4}{V_3} = V_r_3\frac{V_1}{V_2} = (1.9192)(8.5) = 16.3132[/tex]

Finally, the maximum tempertaure; T₄ = 1154.3 K, U₄ = 892.91 kJ/kg

(a) To find the net work, note that [tex]W_{cycle} = Q_{cycle}[/tex]

[tex]\frac{W_{cycle}}{m} = \frac{Q}{m} -(U_4 -U_1)[/tex] ⇒ 1400 - (892.91 - 214.07 ) = 721.16 KJ/Kg

(b) The thermal efficiency η

η = [tex]\frac{W_{cycle}}{m} /\frac{Q}{m} = \frac{721.16}{1400} = 0.5151 = (0.5151 X100)[/tex] = 51.51%

c) The displacement volume is V₁-V₂ m(V₁-V₂)so the mean effective pressure is given by:

[tex]M_e_p = \frac{W_{cycle}/m}{V_1-V_2}= \frac{W_{cycle}/m}{V(1-V_2/V_1)}[/tex]

To obtain V₁

[tex]V_{1} = \frac{RT_1}{P_1} = \frac{\frac{8.314}{28.97} (\frac{KJ}{KgK} )}{100kPa}[\frac{1kPa}{10^3 N/M^2} ][\frac{10^3 N.M}{IkJ} ]=0.861 m^3/kg[/tex]

Therefore,

[tex]M_{ep} = \frac{721.12 kJ/Kg}{0.861\frac{m^3}{kg}[1-\frac{1}{8.5}]}[\frac{10^3 N.M}{1KJ}][\frac{1KPa}{10^3 N/M^2}]= 949.216 kPa[/tex]

The mean effective pressure, In bar, 949.216 kPa = 9.492 bar