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Entropy Balance For Combined System & Surrounding (Environment)

Environment State E1 State E2

State 1

State 2

Both system and environment conditions change during a process 

Qsys Wsys

State 1

State 2Qsys Wsys

State E1 State E2

Consider system only

Consider environment only

WenvQenv

(1)

(2)

Qenv = - Qsys

Wenv = - Wsys

REAL PROCESS 

Add (1) & (2) Qenv = -QsysSys+env Isolated system

(3)(4)

REVERSIBLE PROCESS 

(4)

INTERNALLY REVERSIBLE PROCESS 

(4)

(4)

Direction of a process is always in the direction that net entropy (sys + env) increases.

For any process

Carnot Power Cycle

TH=1000 K

TC=500 K

Q=1000 kJ

Q=1000 kJ

int. rev.

int. rev.

TH=1000 K

TC=500 K

QH=1000 kJ

QC=500 kJ

W=500 kJ

Carnot efficiency=1-Tc/TH=0.5

Example: Determine the total (system+surrounding) entropy change for 1000 kJ heat exchange between 2 reservoirs at TH=1000 K & TC=500 K. How it would chage if a Carnot engine used?

Entropy Balance for Hot Reservoir

Entropy Balance for Cold Reservoir

Entropy Change for Hot cold reservoirs similar to left

Entropy Change inside the device

=0 for a cycle

6.11 Isentropic ProcessClosed Systems Open Systems (CV) with 1i & 1e at steady-state

if Q=0if Q=0

if Q=0 and Reversible if Q=0 and

Reversible

Isentropic (s=constant)

process

Isentropic (s=constant)

process

STATE 1Mass=m

STATE 2Mass=m

Process over time

system state is constant over time

Entropy rate balance at any instant of time Entropy balance for the process over time

Reversible process in a closed system with Q=0

Rev. adiabatic process in an open system with 1i & 1e at SS

1 2i e

IDEALprocesses

Ssys=S1 Ssys=S2

6.12 Isentropic efficiency (Summary)

h2s

s1 s

h

h1 1

2s

h2 wt,rev2wt

h2s

s1 s

h

h1 1

2s

h2

wc,rev

2

wc

Turbine Pump or Compressor

Nozzle

In all of themOperating cost high or gain is less in real systems

If the inlet is the same, enthalpy (and T) at the exit is higher in real devices with 1i & 1e operating at

SS with Q=0

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SPECIAL CASE: isentropic process in an Ideal Gas with cp=constant

use use

For an isentropic process s2=s1 For an isentropic process s2=s1

Isentropic process in an ideal gas

For an ideal gas if

SPECIAL CASE: 1i1e SS Internally Reversible Process with Q

Entropy balance, for constant boundary temperature, Tb

Energy balance

if Work =0

=constant and Work=0

Use Tds relation

Bernoulli Equation

Reversible, incompressible flow without work

for KE=0 PE=0

Used in FM for pipe flow

z1z2

p1

p2

v1

v2

for a Reversible Tb=const. process

Pr. 6.143: Air modeled as an ideal gas enters a turbine operating at steady-state at 1040K, 278 kPa and exits at 120 kPa. The mass flow rate is 5.5 kg/s, and the power developed 1120 kW. Stray heat transfer and kinetic and potential energy effects are negligible. Determine (a) the temperature of the air at the turbine exit, in K, and (b) the isentropic turbine efficiency.

1

2

turbinep1=278 kPa T1=1040 K p2=120 kPa

1 inlet 1 exit at SS

Negligible KE & PE

T2= ? K

Energy balance for 1i & 1e at SS with KE=PE=0

TABLE A-22 @1040 Kh1 = 1091.85 kJ/kgso

1= 3.01260 kJ/kg/K

Search Table A-22 for h=888.21 kJ/kg 840

s1

s

T (K)

1040 1

2s860 2

A-22 @so=2.77148 kJ/kT2s ≈ 840 Kh2s ≈ 866.08 kJ/kg

Search A-22 for so

FIND:

ideal

Example: Air enters an insulated compressor operating at steady-state at 1 bar, 350 K with a mass flow rate of 1 kg/s and exits at 4 bar. The isentropic compressor efficiency is 82%. Determine the power input, in kW, and the rate of entropy production, in kW/K, using the ideal gas model with data from Table A-22. ENG. ASSUMPTIONS

- Steady State Operation- Negligible KE & PE- Insulated compressor‐ Air is an IG with cp=cp(T)

Data for airM=28.97kg/kmol R=0.287 kJ/kgK1

2

p2=4 barp1=1 bar T1=350 K

Insulated air compressorFIND

TABLE A-22 @350 Kh1 = 350.49 kJ/kgso

1= 1.85708 kJ/kg/K

521.04

s1 s

h (kJ/kg)

350.5 1

2s

558.48

wc,rev

2

wc

517.5

s1 s

T (K)

350.0 1

2s

553.6 2

TABLE A-22 searchfor so= 2.25495 kJ/kg/KT2s= 517.5 Kh2s = 521.04 kJ/kg

TABLE A-22 searchfor h= 558.48 kJ/kgT2= 553.6 Kso

2 = 2.32482 kJ/kg/K

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2‐phase mix.

Comp. liquid

Superheated vapor

T3=Tsat(0.08 bar)=41.51oCh3= hf(0.08bar)=173.88 kJ/kgs3= sf(0.08bar)=0.5926 kJ/kg/K

h1= 3425.1 kJ/kgs1=6.6622 kJ/kg/K

T2=Tsat(8 kPa)=41.51oCh2= 2336.7 kJ/kgs2= 7.4651 kJ/kg/K

h4= 188.86 kJ/kg (A-5)s4= 0.6061 kJ/kg/K(double interpolation)

Problem 6.165

FINDa) Net power output, b)Thermal efficiency, c) Isentropic turb. eficiency, td) Isentropic pump eficiency, pe) mass flow rate of cwf) Rate of entropy production for turbine, condenser, and pump, in kW/K, and rank them

h5 ≈ hf(20oC)=83.96 kJ/kgs5 ≈ sf(20oC)=0.2966 kJ/kg/K

h6 ≈ hf(35oC)=146.68 kJ/kgs6 ≈ sf(35oC)=0.5053 kJ/kg/K

1

2

3

4

5

6

1 2

3

4

5

6

Process

cycle

Energy rate balance for 1i1e CV with KE=PE=0, at steady-state

Turbine

Pump Steam Generator

Condenser, wf side

1-22-33-44-1

0 1186360-235747

-16330352750 0117003 117003

CV around cooling water side only

h (kJ/kg) s (kJ/kgK) phase1 3425.1 6.6622 SH Vap.2 2336.7 7.4651 x=0.9

3 173.88 0.5926 x=0.04 188.86 0.6061 C. Liq.

5 83.96 0.2966 Liquid6 146.68 0.5926 Liquid

3

2

5

6

Entropy rate balance for 1i1e at SS & Q=0Turbine (Q=01inlet & 1exit)

Pump (Q=0)1inlet & 1exit

Condenser at Steady‐State with 2 inlets and 2 exits

Most inefficiencies are in turbine

h

s

2

h1

h2

7.476.66

1

h2s 2s

wt wt,r

ev

p2s=p2=8 kPa and s2s=s1=6.6622 kJ/kg/K

0.6060.5926

h3 3

h4 4h4s 4s

exaggerated

wpp4s=p4=10 MPa and s4s=s3=0.5926 kJ/kg/KTable A5 @ p4 &s4s h4s=184.4 kJ/kg & T4s=41.9oC

wp,revcompare T4=43oC

0.7956.66222s 2084.0

0.59264s 184.40 C.Liq

<sf(10MPa)

Water

Air

21

34

CV for part (a)

CV for part (b)

T2=20oCp2=100 kPa

p3=p4=1-atmT3=275 K

p4=1-atmT4=290 K

x1=1p1=100 kPa

Air is an IG with

KE=0 & PE=0

(a) Energy rate balance for an open system at steady-state

STATE (1) x1=1 @ p1=100 kPa,A3T1=Tsat(100 kPa)=99.63oC

h1=hg=2675.5 kJ/kgs1=sg=7.3594 kJ/kg/K

STATE (2), p2=p1=100 kPa,T2=20oC<Tsat(100kPa)

compressed liquidSee A-5min. pressure listed is 25barassume incompressible liquidh2≈hf(20oC)=83.96 kJ/kg (A-2)s2≈sf(20oC)=0.2966 kJ/kg/K

(b) Use enlarged CV for Entropy rate balance

Tb=275 K

p=const.Air is an IG with cp=const

Pr. 6.110