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ENGR 2213 ENGR 2213 Thermodynamics Thermodynamics F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma

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ENGR 2213 Thermodynamics. F. C. Lai School of Aerospace and Mechanical Engineering University of Oklahoma. Reversible Steady-Flow Work. For a steady-flow device undergoing an internally reversible process,. ( δ q) rev - ( δ w) rev = dh + d(ke) + d(pe). - PowerPoint PPT Presentation

### Transcript of ENGR 2213 Thermodynamics

ENGR 2213 ThermodynamicsENGR 2213 Thermodynamics

F. C. LaiSchool of Aerospace and Mechanical EngineeringUniversity of Oklahoma

(δq)rev - (δw)rev = dh + d(ke) + d(pe)

(δq)rev = T ds = dh - v dp

For a steady-flow device undergoing an internally reversible process,

Neglect the changes in kinetic and potential energies,

(δq)rev - (δw)rev = dh

(δw)rev = - v dp 2rev 1w v dp

WorkWork

21w p dv reversible work in closed systems

2rev 1w v dp reversible work associated with

an internally reversible processan steady-flow device

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

WorkWork

2rev 1w v dp

To minimize the work input during a compression process► Keep the specific volume of the working fluid as small as possible.To maximize the work output during an expansion process► Keep the specific volume of the working fluid as large as possible.

WorkWork

Steam Power Plant► Pump, which handles liquid water that has a small specific volume, requires less work.

Gas Power Plant

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

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

WorkWork

1. To approach an internally reversible process as much as possible by minimizing the irreversibilities 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.

1n2

rev 1Cw dpp

2rev 1w v dp

1 11n n2C p dp

Polytropic Processes (pvn = constant)

1 n 12n n1

n C pn 1

1 1 2 2

n (p v p v )n 1

1 2rev

kR(T T )wk 1

Polytropic Processes (pvn = constant)

1 2rev

nR(T T )wn 1

n 1n

1 2

1

nRT p1n 1 p

Isentropic Processes (pvk = constant)k 1k

1 2

1

kRT p1k 1 p

2rev 1

Cw dpp

12

dpCp

Isothermal Processes (pv = constant)

1

2

pC lnp

1

2

pRTlnp

p

v

1. n = k

1

2. 1 < n < k

2

3. n = 1

3

W

Example 1Example 1

Air entering a compressor at p1 = 100 kPa and T1 = 20 ºC and exiting at p2 = 500 kPa. If the airundergoes a polytropic process with n = 1.3, determine the work and heat transfer per unit mass of flow rate.

Example 1 (continued)Example 1 (continued)n 1n

2 2

1 1

T pT p

1 2W nR (T T )m n 1

n 1n

22 1

1

pT Tp

1.3 11.3500293

100

= 425 K

1.3 0.287 293 4251.3 1

= - 164.15 kJ/kg

Polytropic processes

Example 1 (continued)Example 1 (continued)

2 22 1

2 1 2 1Q W V V(h h ) g(z z )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

Isentropic Efficiency for Turbines Isentropic Efficiency for Turbines

Isentropic efficiency for a turbine is defined as the ratio of the actual performance of a turbine to the performance that would be achieved by undergoing an isentropic process for the sameinlet state and the same exit pressure.

actualt

isentropic

WW

Isentropic EfficiencyIsentropic Efficiency

2 22 1

2 1 2 1Q W V V(h h ) g(z z )m m 2

1 2ss

W h hm

1 2W h hm

Turbines

h

s

1

2

2s

h1 – h2

h1 – h2s

1 2t

1 2s

h hh h

Example 2Example 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 is 74 kJ/kg and if the turbine operates adiabatically, determine the isentropic efficiency for the turbine.

actual actualt

isentropic 1 2s

W WW h h

Example 2 (continued)Example 2 (continued)

r2 2

1 r1

ppp p

1 2ss

W h hm

2r2 r1

1

pp pp

Table A-17 T1 = 390 Kpr1 = 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

actualt

isentropic

W 74W 105.6

= 0.7

Isentropic Efficiency for CompressorsIsentropic Efficiency for Compressors

Isentropic efficiency for a compressor is defined as the ratio of the performance of a compressor that would be achieved by undergoing an isentropic process to the actual performance for the same inlet state and the same exit pressure.

isentropicc

actual

WW

Isentropic EfficiencyIsentropic Efficiency

2 22 1

2 1 2 1Q W V V(h h ) g(z z )m m 2

2s 1s

W h hm

2 1W h hm

Compressors

h

s

2s

h1 – h2s

2s 1c

2 1

h hh h

h1 – h2

2

1

Example 3Example 3

isentropic 2s 1c

actual 2 1

W h hW h h

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.

2s 12 1

c

h hh h

Example 3 (continued)Example 3 (continued)

r2 2

1 r1

ppp p

2s 12 1

c

h hh h

2r2 r1

1

pp pp

Table A-17 T1 = 295 Kpr1 = 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/kg

493.0 295.17 295.17 536.4 kJ/kg0.82

Table A-17 h2 = 536.4 kJ/kg, T2 = 532 K

Example 4Example 4

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

Example 4 (continued)Example 4 (continued)

T

S

1 2

34

Given:p1 = p2 = 1.5 MPap3 = p4 = 100 kPax1 = 0.25W23 = 403.8 kJ/kg

Find:Q12 = ?Q34 = ?η = ?

Example 4 (continued)Example 4 (continued)

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/kgsf = 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/kgs2 = sg = 6.4448 kJ/kg K

Example 4 (continued)Example 4 (continued)

Q12 = m(h2 – h1)

Table A-5 p3 = p4 = 100 kPa, hf = 417.46 kJ/kg, hfg = 2258.0 kJ/kg, hg = 2675.5 kJ/kgsf = 1.3026 kJ/kg K, sfg = 6.0568 kJ/kg K, sg = 7.3594 kJ/kg K

= 0.5(2792.2 – 1331.67) = 730.27 kJQ34 = m(h4 – h3)s3 = s2 = 6.4448 kJ/kg K s4 = s1 = 3.3474 kJ/kg K

3 f3

g f

s s 6.4448 1.3026x 0.849s s 6.0568

Example 4 (continued)Example 4 (continued)

4 f4

g f

s s 3.3474 1.3026x 0.338s s 6.0568

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

34L

H 12

QQ 576.921 1 1 0.21Q Q 730.27