Thermodynamic Processes • States of a thermodynamic system can be changed by interacting with its surrounding through work and heat. When this change occurs in a system, it is said that the system is undergoing a process. • A thermodynamic cycle is a sequence of different processes that begins and ends at the same thermodynamic state.

Power plant, CH. 1, Prepared by DR Assim AL Daraje Introduction to Cycles Analysis

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Different Processes •Some sample processes: Isothermal process: temperature is constant T=C Isobaric process: pressure is constant, P=C Isentropic process: entropy is constant, s=C Isochoric or Isometric Process: Constant-volume process, v=C Adiabatic process: no heat transfer, Q=0, where Q is constant

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1

2 1

1 2

Isothermal process: T=constantEnergy balance U=Q-W, for ideal gas U= H=0since both are functions of temperature only

Q=W, W= P

ln ln

Isobaric process:

mRT dVdV dV mRTV V

V PmRT mRTV P

∆ ∆ ∆

= =

= =

∫ ∫ ∫

→

→2

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2 1 2 1 2 1

2 2 1 1 2 1

P=constant

U=Q-W, W= PdV=P dV=P(V )

( ) ( ) ( )( ) ( )

V

Q U P V V U U P V VU PV U PV H H H

∆ −

= ∆ + − = − + −= + − + = − = ∆

∫ ∫

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Types of Cycles

1- Gas Power Systems – Brayton

2- Heat Engine (Steam plant) –Rankine

3- Internal Combustion Engines – Otto, Diesel, Stirling, Atckison.

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T-v Phase Diagram

Diagram courtesy of Jerry M. Seitzman, 2001. 5

@ Constant Pressure T > Tsat ⇒ “superheated vapor” T = Tsat ⇒ “two-phase liquid-vapor” T < Tsat ⇒ “compressed liquid”

T-v Phase Diagram

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P-v Phase Diagram

@ Constant Temperature P < Psat ⇒ “superheated vapor” P = Psat ⇒ “two-phase liquid-vapor” P > Psat ⇒ “compressed liquid” 7

Saturated Liquid Vapor Mixture (SLVM)

Quality is a function of the horizontal distances on P-v and T-v diagrams

gf

g

vapliq

vap

mmm

mmm

x+

=+

=

mass of vapor

mass of liquid

fg

fmix

fg

fmix

fg

fmix

fg

fmix

vvvv

ssss

hhhh

uuuux

−−

=−−

=−−

=−−

=

Quality:

( )( )

( )( )fgfmix

fgfmix

fgffgfmix

fgffgfmix

vvxvv

uuxuu

ssxsxsss

hhxhxhhh

−+=

−+=

−+=+=

−+=+=

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Property Evaluation Evaluate the remaining properties

– If “superheated vapor,” then go to superheated tables – If “compressed liquid” or “sub-cooled,” then go to

compressed liquid tables • If data does not exist, assume the following: v = vf h = hf u = uf s = sf where the saturated liquid property is evaluated at the

given temperature since pressure does not impact liquids that much

– If “liquid-vapor,” then continue using the “two-phase liquid-vapor” tables and find quality

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Example 1 Given: Steam at 2.0 kPa is saturated at 17.5 oC. In what state will the steam be at 40 oC if

the pressure is 2.0 kPa?

Analysis:

@ P = 2.0 kPa = 0.02 bar, and T = 40oC

Psat > P → “superheated vapor”

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Example - Cont’d Given: Steam initially at 1 MPa and 200 oC expands in a turbine to 40 oC and 83% quality.

What is the change in entropy?

Analysis:

s1 = 6.6940 kJ/kg-K (from superheated table for 1 MPa, 200 oC)

@ 40 oC, s2,f = 0.5725 kJ/kg-K; s2,g = 8.257 kJ/kg-K

s2 = s2,f +x2(s2,g-s2,f)

= 0.5725+0.83*(8.257- 0.5725)

s2 = 6.9506 kJ/kg-K

s2 - s1 = (6.9506 – 6.694) kJ/kg-K

s2 - s1 = 0.2566 kJ/kg-K

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Example 2 Given: A 3 kg mixture of water and water vapor at 70 oC is held at constant pressure while

heat is added. The enthalpy of the water increases by 50 kJ/kg. What is the change in entropy?

Analysis:

∆s = 0.1458 kJ/kg-K

∆s =δQrev

T∫ → ∆s =QTo

=50 kJ/kg

(70 + 273)K

integrate

Heat flow = enthalpy change since

Q = ∆H

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Example 3 Given: 3 kg of steam with a quality of 30% has a pressure of 12.056 bar. At that pressure, the

specific volume of a saturated fluid is vf = 1.5289 cm3/g. The specific volume of the vapor is vg = 14.1889 cm3/g. What is the specific volume of the steam?

Analysis:

vf = 1.5289 cm3/g

vg = 14.1889 cm3/

vmix = vf +x(vg-vf)

= 1.5289 +0.30*(14.1889 - 1.5289) cm3/g

vmix = 5.3269 cm3/g

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Problem 4 Given: 1 kg of steam is initially at 400 oC and 800 kPa. The steam expands adiabatically to

200 oC and 400 kPa in a closed process, performing 450 kJ of work, given the following information. Which law does this process violate: (zeroth law, first law, second law, first and second law)?

at 400 oC and 800 kPa at 200 oC and 400 kPa u = 2959.7 kJ/kg u = 2646.8 kJ/kg h = 3267.1 kJ/kg h = 2860.5 kJ/kg s = 7.5716 kJ/kg-K s = 7.1706 kJ/kg Analysis:

(1) Zeroth law not applicable, does not deal with thermal equilibrium

(2) Check first law

∆E = ∆U + ∆KE + ∆PE = Q − Wnegligible

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