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EXAM 3 Solution. ES3001 Thermodynamics Problem 1 (50 pts) Air as an ideal gas [R = 0.287 kJ/(kgK)] contained within a piston-cylinder assembly is compressed between two specified states. Can this process occur adiabatically? If yes, determine the work in appropriate units for an adiabatic process between these states. If no, determine the direction of heat transfer. State 1: p1 = 0.1 MPa, T1 = 27 oC. State 2: p2 = 0.5 MPa, T2 = 207 oC

d Edt

= Q − W + m[(u1 −u2 )+(V1

2 −V22 )

2+ g(z1 − z2 )]

reduces to using Table A-22: Wm= [(u1 −u2 )]= 214.07−344.70 = −130.63

kJkg

So based on being adiabatic and 1st law gives the above work. Now, determine if the entropy production is appropriate.

ΔS = δQT

!

"#

$

%&b

∫ +σ

if adiabatic: ΔS =σ

s2 − s1 = s2o − s1

o − R ln( p2p1) =σ

s2o − s1

o − R ln( p2p1) = 2.1776−1.70203− .287ln(5) =σ

σ = .01366 kJkgK

≥0 Which does satisfy the 2nd law, so it can operate adiabatically.

2

inlet (state 1)

exit (state 2)

Problem. 2 (30 pts) Air at 400 kPa, 740 K enters a turbine operating at steady state and exits at 100 kPa, 550 K. Heat transfer from the turbine occurs at an average outer surface (boundary) temperature of 325 K at the rate of 30 kJ per kg of air flowing. Kinetic and potential energy effects are negligible. The air can be modeled as an ideal gas with Cp = 1.063 kJ/(kg K) and

0.287 kJ/(kg K)R = . 325 KbT =

a) (25 pts) Calculate the rate at which power is developed in kJ per kg of air flowing:

First Law reduces to: (h2 − h1) =

Qcvm−Wcv

m

cpkJkgK

(T2 −T1) =−30 kJ

kgm

−Wcv

m cpkJkgK

(T1 −T2 )−30 kJkgm

=Wcv

m

Wcv

m= cp

kJkgK

(T1 −T2 )−30 kJkgm

=1.063(740−550)−30 = 201.97−30 =171.97 kJkg

alternately, look up h2 and h1, table A-22

Wcv

m= (h1 − h2 )−

30 kJkgm

= (756.44−554.74)−30 = 201.7−30 =171.7 kJkg

b) (10 pts) Calculate the rate of entropy production within the turbine in kJ/K per kg of air flowing:

0 = QTb+ m(s1 − s2 ) + σ

σm= −

QmTb

+ (s2 − s1) = −QmTb

+[c p ln(T2

T1

)− R ln(p2

p1

)]

σm=

30 kJkg

m325K+[1.063 kJ

kgKln(550

740)− R ln(100

400)]= 0.09231

m+[−.3154+ .3979]= 0.1748 kJ

kgK

Turbine