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Prob:3:The Ideal Brayton cycle

Given

P1 = P4 = 100 kPa

P2 = P3 = 1100 kPa

T1 = 20 C = 293 K

T2 = 1200 C = 1473 K

For

Argon

k = 1.667

R = 0.208 kJ/kg/K

P1v1 = RT1

100*v1 = 0.208*293

v1 = 0.6094 m 3/kg

1 to 2 is Isentropic compression

T2 = 764.8 K

P2*v2 = RT2

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1100*v2 = 0.208*764.8

v2 = 0.1446 m 3/kg

2 to 3 constant pressure heat addition

P3 = 1100 kPa

T3 = 1473 K

v3/T3 = v2/T2

v3 = v2*(T3/T2)

v3 = 0.1446*(1473/764.8)

v3 = 0.2785 m 3/kg

3-4 is isentropic expansion

T4 = 564.3 K

Work Produced

W = Cp*(T3-T4)

W = 0.520*(1473-564.3)

Wp = 472.52 kJ/kg

Work Consumed

Wc = 0.520*(764.8-293)

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Wc = 245.34 kJ/kg

Net Work = Wp-Wc

Net work = 227 kJ/kg

Prob:6.

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P:11

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Problem 2:

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Prob: Problem 1:The inlet condtions are given as

P1 = 105 kPa

T1 = 20 C = 293 K

Pv = RT

P1*v1 = RT1

105*v1 = 0.287*293

v1 =0.801

Also

Compression ratio = 8.5

r = P2/P1 = 8.5

Since

1-2 is isentropic process we can use

Therefore

T2 = 540 K

P2*v2 = RT2

8.5*105*v2 = 0.287*540

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v2 = 0.174

2-3 is Constant Volume Process

1500 = Cv(T3-T2)

1500 = 0.718*(T3-540)

T3 = 2629.14 K

Also We have

P3/T3 = P2/T2

P3 = P2*(T3/T2)

P3 = 8.5*105*(2629.14/540)

P3 = 4345.38 kPa

v3 = 0.174 m 3/kg

3-4 is isentropic expansion process

v4 = v1 = 0.801

P4 = 4345.38*(0.174/0.801)1.4

P4 = 511.07 kPa

P4*v4 = R*T4

511.07*0.801 = 0.287*T4

T4 = 1426.37 K

Net Work = Cv(T3-T4)

W = 0.718*(2629.14-1426.37)

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W = 860 kJ/kg

Problem:9

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Problem5:

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Problem 4:

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Problem 8:

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Problem 7:

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Problem 12: