Question

(1) Problem 9-33 (show the solution process in detail)

9-33 An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 750 kJ/kg of heat is transferred to air during 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. Answers: (a) 3898 kPa, 1539 K, (b) 392.4 kJ/kg, (c) 52.3 percent, (d) 495 kPa

(2) Problem 9-33 but assume constant specific heats.

Please help with problem 2 only. Thank you

Answer 1

Solution:- Otto Cycle P-v and T-s diagram are below

given :-Compression Ratio V₁/V₂=r= 8

P₁ = 95 kpa =95 x 10³Pa

T₁ = 27 + 273 = 300K

Q = 750 KJ/Kg

= 1.4 For Air

a) now we can find out the temperature

(T₂/T₁) = (V₁/V₂) ^{- 1}

T₂/300 = (8)^1.4-1

T₂ = 689.21 K

T₂ = 416.21^oc

Heat transfer Q = C_v (T₃ - T₂)

C_v = constant = 1

750 = 1 ( T₃ - 689.21)

now we get temperature at the end of heat addition

T₃ = 1539K

Now for the pressure at end of heat addition

(T₂/T₁) = (P₂/P₁)^-1/

(689.21/300) = (P₂ /(95x10³))^0.4/1.4

1.26 x 95 x 10³ = P₂

_{therefore, pressure at end of heat addition} P₂ = 120484.801 Pa

p₂ = 120.48 KPa

So we can find pressure at and of heat adddition by

P₃/P₂ = T₃/T₂

P₃ = 3898 Kpa

b) Net Work Output = Efficiency x Heat addition

efficiency = 1 - (1/r) ^{- 1}

efficiency = 1- (1/8) ^0.4

efficiency = 0.565

W = 0.565 x 750

work output W = 392.4 KJ/Kg

c) The thermal efficiency = 0.565 x100

_th = 56.5 %

d) the mean effective pressure for cycle

MEP = P₁r (r ^-1- 1)(r_p -1)/ ((r-1) x ( - 1))

where r_p is pressure ratio = T₃/T₂ = 1.9

putting all the values and, we get

mean effective pressure

MEP = 495 Kpa