LECTURES IN
THERMODYNAMICS
Claus Borgnakke
CHAPTER 7
For the 8th Edition of:
Fundamentals of Thermodynamics
Claus Borgnakke, Richard Sonntag
John Wiley & Sons, 2013
Chapter 7
•
•
•
•
•
•
The Entropy Equation for the Control Volume
The Steady State Process and The Transient Flow Process
The Steady State Single Flow Process and the Shaft Work
The Entropy Generation
Device Efficiency
General Analysis and Formulation
2
The Entropy Equation
Add flow terms to the Entropy Balance Equation
.
dSCV
.
.
.
Q
= ∑ misi – ∑ mese + ∑ T + Sgen
dt
CS
.
Sgen ≥ 0
B
Rate of change in Entropy = + in – out + gen
SCV = ∫ ρs dV = mCV savg
A
= mAsA + mBsB + mCsC + ….
If distributed heat transfer
.
.
.
δQ
Q
∑ T = ∫ T = ∫ (Q/A)local /TCS dA
CS
.
.
.
.
.
Sgen = ∫ ρsgen dV = Sgen A + Sgen B + Sgen C + …
3
The Entropy Equation Steady State
Steady State Process
No storage effect:
dmCV
dSCV
= 0,
=0
dt
dt
Entropy Eq.:
.
.
.
.
Q
∑ mese = ∑ misi + ∑ T + Sgen
CS
Rate out = Rate in + Heat transfer effect + Generation
.
If single flow divide by m:
q
se = si + ∑ T + sgen
CS
Single adiabatic flow:
se = si + sgen ≥ si
Following the flow s can only go up.
4
The Steady State Entropy Equation
Example 7.2
A reversible adiabatic steady flow of steam
enters a nozzle as shown in the figure. For the
given conditions determine the exit velocity.
Solution _______________________________________
C.V. Nozzle, we neglect potential energies, w = 0, q = 0 and sgen = 0.
.
.
.
.
.
Continuity Eq.:
0 = mi – me  mi = me = m
Energy Eq.:
2
2
0 = hi + ½ Vi – ( he + ½ Ve )
Entropy Eq.: 0 = si – se
State i, B.1.3: hi = 3051.2 kJ/kg, si = 7.1228 kJ/kg-K
State e, B.1.3: (P, s = si)  Te = 159.1oC, he = 2780.2 kJ/kg
From energy equation
30 × 30 (m/s)2
2
2
½ Ve = hi + ½ Vi – he = (3051.2 – 2780.2) kJ/kg + 2 × 1000 J/kJ = 271.5 kJ/kg
Ve = 2 × 1000 J/kJ × 271.5 kJ/kg = 737 m/s
Remember 1 kJ/kg = 1000 J/kg = 1000 (m/s)2, see Table A.1
5
The Steady State Entropy Equation
Example 7.2E
A reversible adiabatic steady flow of steam
enters a nozzle as shown in the figure. For the
given conditions determine the exit velocity.
Solution ________________________________________
C.V. Nozzle, we neglect potential energies, w = 0, q = 0 and sgen = 0.
.
.
.
.
.
Continuity Eq.:
0 = mi – me  mi = me = m
Energy Eq.:
2
2
0 = hi + ½ V i – ( he + ½ V e )
Entropy Eq.: 0 = si – se
State i, F.7.2: hi = 1279.1 Btu/lbm, si = 1.7085 Btu/lbm-R
State e, F.7.2: (P, s = si)  Te = 314.2 F, he = 1193.9 Btu/lbm
From energy equation
2
2
½ Ve = hi + ½ Vi – he = (1279.1 – 1193.9) Btu/lbm +
100 × 100 (ft/s)2
2 × 25 037 (ft/s)2/(Btu/lbm)
= 85.4 Btu/lbm
Ve =
2 × 25 037 (ft/s)2/(Btu/lbm) × 85.4 Btu/lbm = 2070 ft/s
2
Remember 1 Btu/lbm = 32.174 × 778.17 = 25037 (ft/s) , see A.1
6
The Entropy Equation, Compressor Example
Example 7.3
An inventor reports having a refrigeration compressor with inlet of -20oC saturated vapor R-134a
and an exit flow at 1 MPa, 40oC. The compressor is adiabatic, does that process violate the
second law?
Solution _______________________________________
C.V. Compressor. Single adiabatic steady flow, q = 0.
Entropy Eq.: se = si + sgen  sgen = se – si
Table B.5.1 and B.5.2:
si = 1.7395, se = 1.7148 kJ/kg-K
As se < si the entropy generation is negative. Impossible.
The red curve indicates the constant s process
7
The Entropy Equation, Air Compressor Example
Example 7.4
An air compressor in a gas station receives ambient air at 100 kPa, 290 K and compresses it to
1000 kPa in a reversible adiabatic process. Find the specific work and the exit temperature.
Solution _______________________________________
C.V. Compressor. Single reversible adiabatic steady flow, q = 0, sgen = 0.
Ideal gas, solve with constant specific heats, A.5: CP0 = 1.004 kJ/kg-K, k = 1.4
Energy Eq.:
0 = hi – he – wC
 wC = hi – he
Entropy Eq.: se = si
Property Eq.6.23:
Te = Ti (Pe/Pi)
Te = 290 K (
(k-1)/k
1000 0.2857
)
= 559.9 K
100
From energy equation
wC ≈ CP0 (Ti – Te)
= 1.004 kJ/kg-K (290 – 559.9) K
= –271 kJ/kg
8
The Entropy Equation, Air Compressor Example
Example 7.4 solved with air tables
An air compressor in a gas station receives ambient air at 100 kPa, 290 K and compresses it to
1000 kPa in a reversible adiabatic process. Find the specific work and the exit temperature.
Solution ________________________________________________________
C.V. Compressor. Single reversible adiabatic steady flow, q = 0, sgen = 0.
Ideal gas, solve with air tables, A.7.
Energy Eq.:
0 = hi – he – wC
o
 wC = hi – he
o
Entropy Eq.: se = si  sTe = sTi + R ln(Pe/Pi)
A.7:
o
sTe = 6.83521 + 0.287 ln(1000/100) = 7.49605
Te = 555.7 K,
he = 560.98 kJ/kg
From energy equation
wC = hi – he = 290.43 – 560.98 = -270.55 kJ/kg
9
The Entropy Equation, Air Compressor Example
Example 7.4E
An air compressor in a gas station receives ambient air at 14.7 lbf/in, 520 R and compresses it to
147 lbf/in in a reversible adiabatic process. Find the specific work and the exit temperature.
Solution ________________________________________________________
C.V. Compressor. Single reversible adiabatic steady flow, q = 0, sgen = 0.
Ideal gas, solve with constant specific heats, F.4: CP0 = 0.24 Btu/lbm-R, k = 1.4
Energy Eq.:
0 = hi – he – wC
 wC = hi – he
Entropy Eq.: se = si
Property Eq.6.23:
Te = Ti (Pe/Pi)
(k-1)/k
147 0.2857
Te = 520 R (
)
= 1003.9 R
14.7
From energy equation
wC ≈ CP0 (Ti – Te)
= 0.24 Btu/lbm-R (520 – 1003.9) R
= –116.1 Btu/lbm
10
The Entropy Equation, Air Compressor Example
Example 7.4E solved by air tables
An air compressor in a gas station receives ambient air at 14.7 lbf/in, 520 R and compresses it to
147 lbf/in in a reversible adiabatic process. Find the specific work and the exit temperature.
Solution ________________________________________________________
C.V. Compressor. Single reversible adiabatic steady flow, q = 0, sgen = 0.
Ideal gas, solve with air tables, F.5.
Energy Eq.:
0 = hi – he – wC
o
 wC = hi – he
o
Entropy Eq.: se = si  sTe = sTi + R ln(Pe/Pi)
F.5:
o
sTe = 1.63074 + (53.34/778) ln(147/14.7) = 1.78861
Te = 996.6 R,
he = 240.28 Btu/lbm
From energy equation
wC = hi – he = 124.383 – 240.28 = –115.9 Btu/lbm
11
The Entropy Equation, Multiple Flows
Example 7.5
A de-superheater injects liquid water into a flow of superheated steam. With 2 kg/s steam at 300
kPa, 200oC flowing in what mass flow rate of liquid water at 20oC should be added to have an
exit with saturated vapor 300 kPa? Find also the entropy generation for the process.
Solution ________________________________________________________
C.V. De-superheater. Steady adiabatic flow, 2 inlets and 1 exit, q = 0, w = 0.
.
.
.
.
.
.
Continuity Eq. 4.9: 0 = m1 + m2  m3
 m3 = m1 + m2
.
.
.
Energy Eq. 4.10:
0 = m1h1 + m2h2  m3h3
.
.
.
.
Entropy Eq. 7.2:
0 = m1s1 + m2s2  m3s3 + Sgen
Table B.1:
h1 = 2865.54, h2 = 83.94, h3 = 2725.3 all kJ/kg
s1 = 7.3115,
s2 = 0.2966, s3 = 6.9918 all kJ/kg-K
12
The Entropy Equation, Multiple Flows
Example 7.5 continued
.
.
Use the energy equation with m3 from continuity equation and solve for m2
.
.
.
.
0 = m1h1 + m2h2  (m1 + m2) h3
2865.54  2725.3
.
. h1  h3
m2 = m1
= 2 kg/s
= 0.1062 kg/s
h3  h2
2725.3  83.94
.
.
.
m3 = m1 + m2 = 2.1062 kg/s
From Entropy equation
.
.
.
.
Sgen = m3s3  m1s1  m2s2
kg kJ
= (2.1062 × 6.9918 – 2 × 7.3115 – 0.1062 × 0.2966) s kg-K
= 0.072 kW/K
13
The Entropy Equation, Transient Process
For the transient process we integrate the rate equations in time.
dmCV
.
.
= ∑ mi ∑ me
dt
dECV
.
.
.
.
=
∑
m
h

∑
m
h
+
Q
W
i tot i
e tot e
CV
CV
dt
dSCV
.
.
.
.
= ∑ mi si ∑ me se + QCV/T +Sgen CV
dt
To give
(m2 – m1)C.V. =  mi –  me
E2 – E1 = ∑ mi htot i ∑ me htot e + QCV WCV
t .
S2 – S1 = ∑ mi si ∑ me se +  (QCV/T) dt +1S2gen CV
0
Notice
S2 – S1 = m2s2 – m1s1
etc. assuming uniform states 1, 2
mi si , me se
mass flow averaged values for si, se
1 2
1 2
E2 – E1 = m2(u2 + 2V2 + gZ2) – m1(u1 + 2V1 + gZ1)
Σ mi htot i = Σ mi (hi + 12 V2i + gZi)
14
The Steady State Process Single Flow
Example 7.6
Assume an air tank has 40 L of 100 kPa air at ambient 17oC. The adiabatic reversible
compressor starts to fill the tank to 1000 kPa and shuts off. Find T2 and total work.
15
The Steady State Process Single Flow
Example 7.6 Solved by constant specific heat
Assume an air tank has 40 L of 100 kPa air at ambient 17oC. The adiabatic reversible
compressor starts to fill the tank to 1000 kPa and shuts off. Find T2 and total work.
Constant s, Eq.6.23:
(k-1)/k
T2 = T1 (P2/P1)
= 290 K × 10
0.2857
= 559.9 K
m1 = P1V1/RT1 = 100 × 0.04/(0.287 × 290) = 0.04806 kg
m2 = P2V2/RT2 = 1000 × 0.04/(0.287 × 559.9) = 0.2489 kg
mi = m2  m1 = 0.2009 kg,
also ui = u1
1W2 = mi hi + m1u1 – m2u2 = mi RTi + m1(u1 – ui) – m2 Cv (T2 – Ti)
= mi ×0.287 kJ/kg-K ×290 K – m2 ×0.717 kJ/kg-K ×(559.9 – 290) K
= –31.4 kJ
16
The Entropy Equation
17
The Entropy Equation, Steady State Single Flow
Single Steady State Flow Shaft Work.
Gibbs relation:
T ds = dh – v dP
Actual Process:
δq = T ds – T δsgen = dh – v dP – T δsgen
e
Integrate i to e:
e
q = he – hi – 
 v dP – 
 T δsgen
i
i
Substitute this into energy equation and solve for the work term.
1 2
1 2
w = (hi + 2Vi + gZi) – (he + 2Ve + gZe) + q
e
e
1 2
1 2
= (hi + 2Vi + gZi) – (he + 2Ve + gZe) + he – hi – 
 v dP – 
 T δsgen
i
i
Result is
e
e
2
2
1
w=–
 v dP + 2( Vi – Ve ) + g( Zi – Ze ) – 
 T δsgen
i
i
Last integral is positive (δsgen > 0, absolute T > 0) and represents a loss of potential work.
18
The Entropy Equation, Steady State Single Flow
e
Shaft work:
e
1
2
2
w=–
 v dP + 2( Vi – Ve ) + g( Zi – Ze ) – 
 T δsgen
i
i
Conclusions:
1. Max work out in a reversible process (last term zero).
2. If flow comes in with more leaves with less kinetic energy work can come out.
3. If flow comes in with more leaves with less potential energy work can come out.
4. The first term is the important one for turbines, compressors, pumps. dP gives
sign of work, v influences magnitude. See Figure for area.
5. If there is no shaft, w = 0, the 4 right-hand-side terms balance out to zero. A drop
in pressure (dP < 0) can generate kinetic energy (nozzle) or the inverse (diffuser).
Process shown
could be a
compression of a
gas, w is negative.
19
The Entropy Equation, Steady State Single Flow
e
Shaft work:
e
1
2
2
w=–
 v dP + 2( Vi – Ve ) + g( Zi – Ze ) – 
 T δsgen
i
i
If flow is incompressible (v = C)
e
1
2
2
w = – v(Pe – Pi) + 2( Vi – Ve ) + g( Zi – Ze ) – 
 T δsgen
i
If also reversible then extended Bernoulli equation
1
2
2
w = – v(Pe – Pi) + 2( Vi – Ve ) + g( Zi – Ze )
and if w = 0 it is the Bernoulli equation
1
2
1
2
v Pi + 2 Vi + g Zi = v Pe + 2 Ve + g Ze
Conclusions:
1. This explains the pump work, flow is liquid. First term is difference in flow work.
2. This explains the windmill work, flow is air (incompressible if V < 100 m/s = c/3)
3. Last equation explains static pressure in the ocean with depth (∆Pv = ∆gZ).
4. Last equation explains the height of a water fountain jet (KEin + 0 = 0 + PEout).
5. Last equation says Pv + ke + pe = constant along a flow line.
6. If loss term is kept last eq. would say Pv + ke + pe = constant – loss along flow
20
The Entropy Equation, Steady State Single Flow
Example 7.7
Find the specific work to pump water from 100 kPa, 30oC to 5 MPa.
Solution _______________________________________
C.V. Pump. Steady adiabatic flow, q = 0, ∆ke = ∆pe = 0.
Energy Eq.:
CATT3:
0 = hi – he – w
Entropy Eq.: 0 = si – se
hi = 125.8 kJ/kg, si = 0.4368 kJ/kg-K  he = 130.8 kJ/kg ,
w = hi – he = 125.8 – 130.8 = – 5.0 kJ/kg
Manual interpolation in Table B.1.1 & B.1.4 gives:
w = 125.77 – 131.48 = – 5.71 kJ/kg
Round-off differences becomes important for si which determines he.
Incompressible:
w = – v(Pe – Pi)
= – 0.001004 m3/kg (5000 – 100) kPa
= – 4.92 kJ/kg
Water pump.
Notice the slight diffuser exit,
it reduces velocity to recover
more pressure. Inlet is in the
center on the back side.
Electric motor towards
camera.
21
The Entropy Equation, Steady State Single Flow
Example 7.8
Consider a flow of liquid water at 300 kPa, 20oC in a hose. How high an exit velocity can an
ideal nozzle generate?
Solution ______________________________________________________
C.V. Nozzle. Steady adiabatic single flow, q = 0, w = 0, ∆Z = 0.
Bernoulli:
1
2
1
2
1
2
v Pi + 2 Vi + g Zi = v Pi + 0 = v Pe + 2 Ve + g Ze = v P0 + 2 Ve
1 2
3
2 Ve = v (Pi – P0) = 0.001002 m /kg (300 – 100) kPa = 0.2004 kJ/kg
Ve = 2 × 0.2004 kJ/kg × 1000 J/kJ = 20 m/s
Recall units: J/kg = (m/s)2, process did not have to be adiabatic for this result.
22
The Entropy Equation, Polytropic Flow Work
We already did the work for v = C (pump) now:
Shaft Work in Polytropic process:
n ≠ 1:
w
e
e
n
1/n  1/n
w=–
v
dP
=
–
C
P
dP
=
–


n – 1 (Peve – Pivi)
i
i
=–
n = 1:
Pvn = C
nR
n – 1 (Te – Ti) (if also ideal gas)
e
e
1
w=–
 P dP = – Pivi ln ( Pe / Pi )
 v dP = – C 
i
i
The first expression looks very much like the expression for
boundary work, that had a factor –1/(n – 1) in front.
These expressions give the indicated shaft work
e
w= –
 v dP
(see P – v diagram)
q
i
23
The Entropy Equation, Single Flow Concepts
24
The Entropy Equation, Multiple Flows
To illustrate multiple flows look at the mixing chamber.
The continuity, energy and entropy equations for this case become
.
.
.
Continuity Eq. 4.9:
0 = m1 + m 2  m3
Energy Eq. 4.10:
.
.
.
.
0 = m1h1 + m2h2  m3h3 + Q
.
.
.
.
.
0 = m1s1 + m2s2  m3s3 + Q/T + Sgen
.
Scale the equations with the exit mass flow rate m3
Entropy Eq. 7.2:
Flow fractions
. .
y = m1/ m3 ;
. .
m2/ m3 = 1 – y
The energy equation becomes
h3 = y h1 + (1 y) h2 + q῀
The entropy equation gives
s3 = y s1 + (1  y) s2 + q῀ /T + s῀gen
This is a mass flow rate average plus the heat transfer and entropy generation influence.
Tildes indicate a pseudo specific quantity (scaled to exit flow).
25
The Device Efficiency
Device Efficiency
Compare actual devices to ideal devices.
Many devices are adiabatic: Pumps, Turbines, Compressors, Nozzles, Diffusers
The ideal device is reversible adiabatic and therefore isentropic.
Isentropic Turbine Efficiency:
wT ac hi – he ac
ηturbine =
=
wT s h i – hes
Isentropic Compressor Efficiency:
wC s he ac – hi
ηcomp = w
= h –h
C ac
es
i
wC is measured positive in (non-standard sign)
2
Isentropic Nozzle Efficiency:
ηnoz =
Ve ac/2
2
Ve s/2
Use kinetic energy difference if inlet ke is significant.
These are very different from complete cycle efficiencies which expresses a conversion
efficiency from heat input to work output (heat engine) or work input to heat transfer
(refrigerator or HP)
26
The Device Efficiency, Example
Example 7.10
A turbine receives steam at 1 MPa, 300oC and it leaves at 15 kPa. The specific work output is
measured to be 600 kJ/kg. Find the isentropic efficiency.
Solution ______________________________________________________
C.V. Ideal Turbine. Given inlet state and exit pressure, single flow steady state device, sgen = 0.
Energy Eq.:
0 = hi – hes – wT s
Entropy Eq.: 0 = si – ses + 0
Table B.1.3:
hi = 3051.2 kJ/kg, si = 7.1228 kJ/kg-K
Table B.1.2:
15 kPa,
ses = 7.1228 kJ/kg-K < sg = 8.0084 kJ/kg-K so state is two-phase
ses = 7.1228 = 0.7548 + xes 7.2536

xes = 0.8779
hes = 225.9 + 0.8779 × 2373.1 = 2309.3 kJ/kg
From energy equation
wT s = hi – hes = 3051.2 – 2309.3 = 741.9 kJ/kg
and with wT s = 600 kJ/kg
wT ac 600
ηturbine = w
= 741.9 = 0.809
Ts
or 80.9%
27
The Device Efficiency, Example
Example 7.11
Air enters a gas turbine at 1600 K and exits at 100 kPa, 830 K. The gas turbine efficiency is
estimated to be 85%. What is the turbine inlet pressure?
Solution ______________________________________________________
C.V. Gas Turbine. Given Ti, Pe, Te ac and ηturbine. Unknown Pi.
Use air Tables A.7:
o
hi = 1757.3 kJ/kg, sTi = 8.6905 kJ/kg-K,
he ac = 855.3 kJ/kg
Actual Turbine:
wT ac = hi – he ac = 1757.3 – 855.3 = 902 kJ/kg
Ideal Turbine:
wT s = hi – hes ;
Efficiency:
wT s = wT ac / ηturbine = 902 /0.85 = 1061.2 kJ/kg = hi – hes
0 = si – ses
hes = hi – wT s = 1757.3 – 1061.2 = 696.1 kJ/kg
 A.7:
Ideal Turbine, Entropy:
Tes = 683.7 K,
o
o
sTes = 7.7148 kJ/kg-K
o
0 = sTi – sTes – R ln ( Pi / Pe )
0 = 8.6905 – 7.7148 – 0.287 ln(Pi / 100 kPa)

Pi = 2995 kPa
28
The Device Efficiency, Example
Example 7.12
Air enters an automotive supercharger at 100 kPa, 300 K and is
compressed to 150 kPa. The efficiency is 70%. Find the specific
work and the exit temperature?
Solution ____________________________________
C.V. Supercharger (compressor).
Given Ti, Pi, Pe and ηC s.
Ideal compressor:
Unknown Te.
wC s = hi – hes ;
(k-1)/k
 Tes = Ti ( Pe / Pi )
0 = si – ses
= 300 K (150/100)
0.2857
= 336.9 K
wC s = hi – hes = CP ( Ti – Tes ) = 1.004 (300 – 336.9) = –37.1 kJ/kg
Actual compressor:
wC ac = wC s / ηC s = –37.1/0.7 = –53 kJ/kg = CP ( Ti – Te ac )
–53 kJ/kg
Te ac = Ti – wC ac/ CP = 300 – 1.004 kJ/kg-K = 352.8 K
29
The General Control Volume Analysis
Formulation Steps.
1. Make a sketch of the physical system, show flows, fluxes, heat and work terms.
2. Choose a CV as the whole or a part of the system. Label all flows, fluxes crossing CS they will
appear in the equations. If they are unwanted choose a CV so they do not cross CS.
Label all parts with storage and note if they change state they will also appear in equations.
3. Write the 3 General Laws for the chosen CV: Continuity, Energy and Entropy equations.
Transfer terms leaving one CV to another CV appear with different signs on the RHS of the
corresponding equations. All storage terms show up on the LHS of the equations.
Write the most general form (rate form or time integrated form, same principle).
4. Write the specific, particular or special laws for this system or devices; often this becomes the
process equation like V = C, P = C, Q = 0 etc., often an approximation of reality. We make a math
model of the system.
5. Make a list of all states for different parts with storage like A1, A2 and the flows.
Note which 2 properties fixes a given state like (#1, #2) as in (PA1, TA1) and so on.
6. Tally up how many unknowns, you must have just as many equations to determine them.
The ideal gas law does not give a property, you get a 3rd property if you already have two properties.
30
Nozzles, Diffusers
A diffuser reduces velocity.
A jet engine nozzle.
A water nozzle
Diffuser on a Ford GT
Nozzles with lift fans for the F35B
31
Heat Exchangers
A plate heat exchanger. The black lines are seals (gaskets).
Alternate layers are used, one fluid is red the other fluid flow
is blue. Flow is between the two holes not covered by a gasket
around it.
A storage rack with the plates for
heat exchangers.
32
Heat Exchangers
© C. Borgnakke
A counter-flow heat exchanger
A nitrogen compressor with an intercooler driven by a hydraulic
motor. The intercooler sketch is on the left and the real thing is
the pipe shown on the right in the red box.
33
Turbines
Turbines: Wind
Steam
Steam turbine blades
Water
Turbo-charger automotive
Jet eng. Compressor Turbine Nozzle
34