Lecture 95

8/18/2019 Lecture 95

1/25

ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics

F. C. Lai

School of Aerospace and MechanicalEngineering

University of Oklahoma

8/18/2019 Lecture 95

2/25

Reversible Steady-Flow Work Reversible Steady-Flow Work

(δq)rev - (δw)rev= dh + d(ke) + d(pe)

(δq)rev = T ds = dh - v dp

For a steady-flow device undergoing an internallyreversible process,

Neglect the changes in kinetic and potential energies,

(δq)rev - (δw)

rev= dh

(δw)rev= - v dp 2rev 1

w v dp= −∫

8/18/2019 Lecture 95

3/25

Work Work

2

1w p dv= ∫ reversible work in closed systems

2rev 1w v dp= −∫ reversible work associated withan internally reversible processan steady-flow device

► The larger the specific volume, the larger thereversible work produced or consumed by thesteady-flow device.

8/18/2019 Lecture 95

4/25

Work Work

2rev 1

w v dp= −∫

To minimize the work input during a compressionprocess

► Keep the specific volume of the working fluid as small as possible.

To maximize the work output during an expansionprocess

► Keep the specific volume of the working fluid

as large as possible.

8/18/2019 Lecture 95

5/25

Work Work

Steam Power Plant

► Pump, which handles liquid water that has asmall specific volume, requires less work.

Gas Power Plant

Why does a steam power plant usually have abetter efficiency than a gas power plant?

► Compressor, which handles air that has alarge specific volume, requires more work.

8/18/2019 Lecture 95

6/25

Work Work

1. To approach an internally reversible process as

much as possible by minimizing theirreversibilities such as friction, turbulence, and

non-quasi-equilibrium compression.

Minimizing the Compressor Work

2.To keep the specific volume of the gas as small as possible by maintaining the gas temperature as low as possible during the compression process. This requires that the gas be cooled

as it is compressed.

8/18/2019 Lecture 95

7/25

Steady-Flow Work Steady-Flow Work

1

n2rev 1

Cw dp

p

= −∫ ÷

2rev 1

w v dp= −∫

1 11n n2

C p dp−

= − ∫

Polytropic Processes (pvn = constant)

1 n 1

2n n1

nC p

n 1

−

= −− 1 1 2 2

n(p v p v )

n 1= −

−

8/18/2019 Lecture 95

8/25


1 2rev

kR(T T )w

k 1

−=

−

Polytropic Processes (pvn = constant)

1 2rev nR(T T )w

n 1−= −

n 1

n

1 2

1

nRT p1n 1 p

−

= − ÷ −

Isentropic Processes (pvk

= constant)k 1k

1 2

1

kRT p1

k 1 p

− = − ÷ −

8/18/2019 Lecture 95

9/25


2

rev 1

C

w dpp= −∫ 1

2

dp

C p= − ∫

Isothermal Processes (pv = constant)

1

2

pC ln

p

= ÷

1

2

pRTln

p

= ÷

p

v

1. n = k

1

2. 1 < n < k

2

3. n = 1

3

W

8/18/2019 Lecture 95

10/25

Eam!le 1Eam!le 1

Air entering a compressor at p1 = 100 kPa and

T1 = 20 ºC and exiting at p2 = 500 kPa. If the air

undergoes a polytropic process with n = 1.3,determine the work and heat transfer per unitmass of flow rate.

8/18/2019 Lecture 95

11/25

Eam!le 1 "contin#ed$Eam!le 1 "contin#ed$

n 1

n2 2

1 1

T p

T p

−

= ÷

1 2

W nR

(T T )m n 1= −−

&

&

n 1

n2

2 11

pT T

p

−

= ÷

1.3 11.3500293

100

− = ÷

= 425 K

( ) ( )1.3

0.287 293 4251.3 1

= −−

= - 164.15 kJ/kg

Polytropic processes

8/18/2019 Lecture 95

12/25


2 22 1

2 1 2 1

Q W V V(h h ) g( )

m m 2

−= + − + + −

& &

& &

= - 164.15 + (426.35 – 293.17)

Table A-17, T1 = 293 K, h1 = 293.17 kJ/kg

T2 = 425 K, h2 = 426.35 kJ/kg

= - 30.97 kJ/kg

8/18/2019 Lecture 95

13/25

%sentro!ic E&&iciency &or T#rbines%sentro!ic E&&iciency &or T#rbines

Isentropic efficiency for a turbine is defined asthe ratio of the actual performance of a turbine

to the performance that would be achieved byundergoing an isentropic process for the sameinlet state and the same exit pressure.

!"#$!l#

%&en#r'p%"

W

Wη =

8/18/2019 Lecture 95

14/25

%sentro!ic E&&iciency%sentro!ic E&&iciency

2 22 1

2 1 2 1

Q W V V

(h h ) g( )m m 2

−

= + − + + −

& &

& &

1 2&

&

W h hm

= − ÷

&

&

1 2

Wh h

m= −

&

&

Turbines

h

s

1

2

2s

h1 – h2

h1 – h2s

1 2#

1 2&

h h

h h

−η =

−

8/18/2019 Lecture 95

15/25

Eam!le 2Eam!le 2

Air enters a turbine at p1 = 300 kPa and

T1 = 390 K and exits at p2 = 100 kPa. Given

that the actual work output from the turbine is74 kJ/kg and if the turbine operates adiabatically,determine the isentropic efficiency for the turbine.

!"#$!l !"#$!l#

%&en#r'p%" 1 2&

W WW h h

η = =−

8/18/2019 Lecture 95

16/25


r 2 2

1 r1

pp

p p=

1 2&

&

Wh h

m

= − ÷

&

&

2r2 r1

1

pp pp

=

Table A-17 T1 = 390 K

pr1 = 3.481, h1 = 390.88 kJ/kg

= 3.481 (100/300)= 1.1603

Table A-17 pr2 = 1.1603,

h2s = 285.27 kJ/kg= 390.88 – 285.27 = 105.6 kJ/kg

!"#$!l#

%&en#r'p%"

W 74

W 105.(

η = = = 0.7

8/18/2019 Lecture 95

17/25

%sentro!ic E&&iciency &or 'om!ressors%sentro!ic E&&iciency &or 'om!ressors

Isentropic efficiency for a compressor is definedas the ratio of the performance of a compressor

that would be achieved by undergoing anisentropic process to the actual performance forthe same inlet state and the same exit pressure.

%&en#r'p%""

!"#$!l

W

Wη =

8/18/2019 Lecture 95

18/25

%sentro!ic E&&iciency%sentro!ic E&&iciency

2 22 1

2 1 2 1

Q W V V

(h h ) g( )m m 2

−

= + − + + −

& &

& &

2& 1

&

W h hm

= − ÷

&&

2 1

Wh h

m= −

&

&

Compressors

h

s

2s

h1 – h2s

2& 1"

2 1

h h

h h

−η =

−

h1 – h2

2

1

8/18/2019 Lecture 95

19/25

Eam!le 3Eam!le 3

%&en#r'p%" 2& 1"

!"#$!l 2 1

W h h

W h h

−η = =

−

Air enters an insulated compressor at p1 = 95 kPa

and T1 = 22 ºC. Given that p2 /p1 = 6 and ηc = 0.82,

determine the exit temperature for the air.

2& 12 1

"

h hh h

−= +

η

8/18/2019 Lecture 95

20/25


r 2 2

1 r1

pp

p p=

2& 12 1

"

h hh h

−= +

η

2r2 r1

1

pp pp

=

Table A-17 T1 = 295 K

pr1 = 1.3068, h1 = 295.17 kJ/kg

= 1.3068 (6)= 7.841

Table A-17 pr2 = 7.841,

T2s = 490.29 K h2s = 493.0 kJ/kg493.0 295.17

295.17 53.4 k * kg0.82

−= + =

Table A-17 h2 = 536.4 kJ/kg,

T2 = 532 K

8/18/2019 Lecture 95

21/25

Eam!le (Eam!le (

0.5 kilogram of water executes a Carnot powercycle. During the isothermal expansion, the water

is heated until it is a saturated vapor from an initialstate where the pressure is 1.5 MPa and the qualityis 25%. The vapor then expands adiabatically topressure of 100 kPa. Find

(a) the heat addition and rejection from this cycle.(b) the cycle efficiency.

8/18/2019 Lecture 95

22/25

Eam!le ( "contin#ed$Eam!le ( "contin#ed$

T

S

1 2

34

Given:p1 = p2 = 1.5 MPa

p3 = p4 = 100 kPax1 = 0.25

W23 = 403.8 kJ/kg

Find:Q12 = ?

Q34 = ?

η = ?

8/18/2019 Lecture 95

23/25


Q12 = m(u2 – u1) + mp(v2 – v1)

Table A-5 p1 = p2 = 1.5 MPa,

hf = 844.84 kJ/kg, hfg = 1947.3 kJ/kg, hg = 2792.2 kJ/kg

sf = 2.315 kJ/kg K, sfg = 4.1298 kJ/kg K, sg = 6.4448 kJ/kg K

h1 = hf + x1 hfg

= m(h2 – h1)

s1 = sf + x1 sfg = 844.84 + 0.25(1947.3) = 1331.67 kJ/kg= 2.315 + 0.25(4.1298) = 3.3474 kJ/kg K

h2 = hg = 2792.2 kJ/kg

s2 = sg = 6.4448 kJ/kg K

8/18/2019 Lecture 95

24/25

8/18/2019 Lecture 95

25/25


4 + 4

g +

& & 3.3474 1.302, 0.338

& & .058

− −= = =

−

Q34 = m(h4 – h3)

h3 = hf + x3 hfg = 417.46 + 0.849(2258) = 2334.5 kJ/kgh4 = hf + x4 hfg = 417.46 + 0.338(2258) = 1180.66 kJ/kg

= 0.5(1180.66 – 2334.5) = -576.92 kJ

34-

12

QQ 57.921 1 1 0.21

Q Q 730.27η = − = − = − =

Lecture 95

Documents

Transcript of Lecture 95