EXAMPLE PROBLEMS

FOR MIDTERM I

PROBLEM 1

For an air standard ideal Otto cycle the pressure and temperature at the beginning of compression are 90 kPa and 300 K respectively. The temperature at the end of heat input at constant volume is 2300 K. The compression ratio is 9.5. (R = 0.287 kJ/kg-K and Cp = 1.005 kJ/kg-K for the whole cycle)

Calculate;a) The work done during compression.b) The work done during expansionc) The mean effective pressure

1

2

3

4

P

v

GIVENP1 = 90 kPaT1 = 300 KR = 0.287 kJ/kg-KCp = 1.005 kJ/kg-KR = 9.5T3 = 2300 K

REQUIREDWc = ?

SOLUTIONThe logic of the solution will be from TOP to BOTTOMThe solution will be from BOTTOM to TOP

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21

11

s

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Cv = 0.718 kJ/kg-Kv1 = 0.956667 m3/kgv2 = 0.1007 m3/kgvs = 0.8560 m3/kg

k = 1.4T2 = 738 KT4 = 935 K

Wcomp = -314 kJ/kgWexp = 980 kJ/kgWnet = 666 kJ/kgPe = 778 kJ/kg

2C1C1

6.22x2C0.22x1C47.186x2.035.155x8.0C

Kkmol/kJ6.22573x10x99.332.20)N(C

Kkmol/kJ0.22573x10x7.573.18)O(C

Kkmol/kJ47.186573x39.037)OHHC(C

Kkmol/kJ35.155573x45.05.102)HC(CK573At)b

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Problem 3The temperature and pressure of the reactants at the

beginning of compression in an otto cycle is 25 C and 90 kPa respectively. The temperature and pressure of the reactants at the end of compression is 320 C and 1500 kPa. The fuel used is gasoline (C7.76H13.1) and the excess air coefficient is 0.9. Find the heat lost per kg of reactants, during compression.

a) The mean molar specific heat of the reactants during compression.

b) The mean specific heat of the reactants per kg of mixture, during compression

c) The change of the internal energy of the reactants during compression.

d) The work done on the reactants during compressione) The heat lost by the reactants during compression.

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1

2

3

4

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Problem 1 ( 15 points ) An air-standard dual cycle (isentropic compression and expansion of only air as working gas) has a compression ratio of 15 and a maximum pressure of 10 MPa . One fifth of the heat addition takes place at constant volume and the rest at constant pressure. If the temperature and pressure at the beginning of compression are 320 K and 0.1 MPa respectively, determine the cycle thermal efficiency (Rair = 0.287 kJ/kg-K and isentropic gas constant k = 1.4).

GIVEN

Air-standard dual cycle

r = 15

Pmax = 10 Mpa

Qv = 0,2 Qin

Qp = 0,8 Qin

P1 = 0,1 MPa

T1 = 320 K

REQUIRED

th = ?

1

2

3 4

5

44

4

3T

C

Q4T

)3T4T(CQ

)2T3T(CQ1k

RC

C

R1k

RCC

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Q

Q1

p

p

pp

vv

v

v

vp

pvin

in

outTH

SOLUTION

v1 = 0,9184 m3/kg

v2 = 0,061227 m3/kg

T2 = 945 K

P2 = 4431 kPa

T3 = 1707 K

Qv = 546 kJ/kg

Qp = 2185 kJ/kg

T4 = 3882 K

v4 = 0,1114 m3/kg

= 8,24

T5 = 1670 K

Qout = 1356 kJ/kg

th = 0,50