Question

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 750 kJ/kg of heat is transferred to air dur- ing the constant-volume heat-addition process. Taking into account the variation of specific heats with temperature, determine (a) the pressure and temperature at the end of the heat- addition process, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle.

Answer 1

Answer:

a) T_3 = 1539 K , P_3 = 3898 KPa

b) w_net,out = 392.4 KJ/kg

c) n_th = 0.523 - 52.3 %

d) mep = 495 KPa

Explanation:

Given:

- Sate 1: P_1 = 95 KPa, T_1 = 27 C

- q_in = 750 KJ/kg  (constant volume process)

- R_air = 0.287 KJ/kgK

- r = 8  (compression ratio)

Find:

a) Pressure and Temperature (P_3 & T_3) at the end of Heat addition process.

Analysis:

Path 1 to 2 (isentropic compression):

T_1 = 300 K   ----------> u_1 = 214.07 KJ/kg  , v_r1 = 621.2

v_r2 = v_r1 * (V_2 / V_1) = 621.2 *(1 / 8) = 77.65

v_r2 = 77.65 -----------> u_2 = 491.2 KJ/kg , T_2 = 673.1 K

Use Ideal Gas Law for states 1 and 2:

P_2 * v_2 / T_2 = P_1 * v_1 / T_1

P_2 = P_1 * (v_1 /v_2) * ( T_2 / T_1 )

P_2 =  95 KPa * (8)*(673.1/300) = 1705 KPa

Path 2 to 3 (constant volume Heat - Addition):

q_in = (u_3 - u_2 )

u_3 = q_in + u_2

u_3 = 750 +491.2 = 1241.2 KJ/kg

u_3 = 1241.2 KJ/kg  ----------> T_3 = 1539 K , v_r3 = 6.588

Use Ideal Gas Law for states 2 and 3:

P_3 * v_3 / T_3 = P_2 * v_2 / T_2

P_3 = P_2 * ( T_3 / T_2 )

P_3 =  1705 KPa*(1539/673.1) = 3898 KPa

Answer: T_3 = 1539 K and P_3 = 3898 KPa

Path 3 to 4 (isentropic expansion):

v_r4 = v_r3 * (V_1 / V_2) = 6.588 *(8) = 52.7

v_r4 = 52.70 -----------> u_4 = 571.69 KJ/kg , T_4 = 774.5 K

Path 4 to 1 (constant volume Heat - Rejection):

q_out = (u_4 - u_1 )

q_out = 571.69 - 214.07

q_out = 357.62 KJ/kg

b) The net work output w_net,out

Analysis:

w_net, out = q_in - q_out

w_net, out = 750 - 357.62

w_net, out = 392.4 KJ/kg

Answer: The net work output w_net,out = 392.4KJ/kg

c) Thermal Efficiency n_th

Analysis:

n_th = w_net, out / q_in

n_th = 392.4 / 750

n_th = 0.523 ~ 52.3%

Answer: = Thermal Efficiency n_th = 52.3%

d) mean effective pressure for the cycle.

Use ideal gas Law at state 1:

v_1 = R*T_1 / P_1 = 0.287 * 300 / 95

v_1 = v_max= 0.906 m^3 /kg

v_min = v_max / r

mep = w_net,out / (v_1-v_2) = w_net,out / v_1*(1 - 1 / r)

mep = 392.4 / 0.906 * (1 - 1/8)

mep = 495.0 KPa

Answer: mean effective pressure for the cycle mep = 495.0 KPa