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Thermodynamic Processes & Isentropic Efficiency
Dr. Md. Zahurul Haq
Professor
Department of Mechanical EngineeringBangladesh University of Engineering & Technology (BUET)
Dhaka-1000, Bangladesh
http://zahurul.buet.ac.bd/
ME 6101: Classical Thermodynamics
http://zahurul.buet.ac.bd/ME6101/
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 1 / 29
Overview
1 Steady-State, Steady Flow (SSSF) Processes
2 Isentropic (First Law) Efficiency
Nozzle
Diffuser
Turbine
Compressor & Pump
SSSF Process
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 2 / 29
Steady-State, Steady Flow (SSSF) Processes
Steady-State, Steady Flow (SSSF) Processes
Assumptions:
• Control volume does not move relative to the coordinate frame.
• State of the mass at each point in the control volume does not vary
with time.
• As for the mass that flows across the control surface, the mass flux
and the state of this mass at each discrete area of flow on the control
surface do not vary with time. The rates at which heat and work cross
the control surface remain constant.
⊲ Example: centrifugal compressor and turbines etc. at steady state.
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 3 / 29
Steady-State, Steady Flow (SSSF) Processes
Nozzles & Diffusers
T159
• A nozzle is a flow passage of varying cross-sectional area in which the
velocity of a gas or liquid increases in the direction of flow.
• A diffuser uses a reduction in the velocity of a fluid to cause an
increase in its pressure in the direction of flow.
• For a nozzle or diffuser, the only work is flow work at locations where
mass enters and exits the CV, so the term Wcv drops out.
• ∆PE ≈ 0© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 4 / 29
Steady-State, Steady Flow (SSSF) Processes
Moran Ex. 4.3: ⊲ Converging-diverging Steam Nozzle: Estimate A2.
T145
⇒ Steady state ⇒ dEcv/dt = 0
⇒ Z2 = Z1
⇒ Wcv = 0, Q = 0
⇒ mi = me = 2 kg/s
dEcv
dt= Q − Wcv +
∑
i
mi
(
hi +V
2
i
2+ gzi
)
−∑
e
me
(
he +V
2e
2+ gze
)
0 = (h1 − h2) +1
2(V2
1− V
2
2)
⇒ h2 = h1 +1
2(V2
1− V
2
2)
⇒ ρ2 = ρ(Steam,P = P2, h = h2) = 6.143 kg/m3
⇒ m2 = ρ2A2V2⇒ A2 = 4.896 × 10−4 m2 ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 5 / 29
Steady-State, Steady Flow (SSSF) Processes
Throttling Devices
A significant reduction in pressure can be achieved simply by introducing a
restriction into a line through which a gas or liquid flows.
T160
• For a control volume enclosing a throttling device, the only work is
flow work at locations where mass enters and exits the control volume,
so the term Wcv drops out.
• Q ≈ 0, and ∆PE ≈ 0.© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 6 / 29
Steady-State, Steady Flow (SSSF) Processes
Cengel Ex. 5.8: ⊲ R-134a enters the capillary tube as saturated liquid at
0.8 MPa and is throttled to 0.12 MPa. Determine x2 and ∆T .
T148
⇒ Steady state ⇒ dEcv/dt = 0
⇒ Z2 = Z1 & V2 ≃ V1
⇒ Q ≃ 0 & Wcv = 0
h2∼= h1
⇒ h1 = enthalpy(R134a,P1 = 0.8 MPa, x1 = 0.0)
⇒ h1 = h2 = hf , 0.12MPa + x2hfg , 0.12MPa⇒ x2 = 0.32 ⊳
⇒ T1 = T (R134a,P1 = 0.8 MPa, x1 = 0.0)⇒ T1 = 31.3oC
⇒ T2 = T (R134a,P2 = 0.12 MPa, sat.)⇒ T2 = −18.8oC
⇒ ∆T = −53.64oC ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 7 / 29
Steady-State, Steady Flow (SSSF) Processes
Turbines
A turbine is a device in which power is developed as a result of a gas or
liquid passing through a set of blades attached to a shaft free to rotate.
T161
Axial-flow steam or gas turbine.
T162
Hydraulic turbine installed in a dam.
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 8 / 29
Steady-State, Steady Flow (SSSF) Processes
Moran Ex 4.4: ⊲ Heat Transfer from Steam Turbine: Determine heat loss.
T146
⇒ Steady state ⇒ dEcv/dt = 0
⇒ Z2 = Z1
⇒ mi = me = (4600/3600) kg/s
dEcv
dt= Q − Wcv +
∑
i
mi
(
hi +V
2
i
2+ gzi
)
−∑
e
me
(
he +V
2e
2+ gze
)
0 = Q − Wcv + m[
(h1 − h2) +(
V2
1−V
2
2
2
)]
⇒ h1 = enthalpy(Steam,P = P1,T = T1)
⇒ h2 = enthalpy(Steam,P = P2, x = x2)
⇒ Qcv = −63.61 kW (heat loss) ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 9 / 29
Steady-State, Steady Flow (SSSF) Processes
Compressors & Pumps
Compressors and pumps are devices in which work is done on the substance
flowing through them in order to increase the pressure and/or elevation.
Compressor is used to compress a gas (vapour) and the term pump is
used when the substance is a liquid.
T163 T164
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 10 / 29
Steady-State, Steady Flow (SSSF) Processes
[Moran Ex. 4.5: ⊲ Air Compressor Power: Determine power required, Wcv .
T147
⇒ Steady state ⇒ dEcv/dt = 0
⇒ Z2 = Z1
⇒ Qcv = −180kJ/min = −3.0 kW
• m = ρAV
• ρ = P/RT
dEcv
dt= Q − Wcv +
∑
i
mi
(
hi +V
2
i
2+ gzi
)
−∑
e
me
(
he +V
2e
2+ gze
)
Wcv = Q + m[
(h1 − h2) +(
V2
1−V
2
2
2
)]
⇒ h1 − h2 = Cp(T1 − T2) = −160.8 kJ/kg
⇒ ρ1 = P1/RT1 = 1.20 m3/kg
⇒ m = ρ1A1V1 = 0.721 kg/s
⇒ Wcv = −119.4 kW (work input required) ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 11 / 29
Steady-State, Steady Flow (SSSF) Processes
Borgnakke Ex. 4.6: ⊲ A water pump is located 15 m down in a well, taking
water in at 10oC, 90 kPa at a rate of 1.5 kg/s. The exit line is a pipe of diameter
0.04 m that goes up to a receiver tank maintaining a gauge pressure of 400 kPa.
Assume that the process is adiabatic, with the same inlet and exit velocities, and
the water stays at 10oC. Find the required pump work.
T442
⇒ Steady state ⇒ dEcv/dt = 0
⇒ Q = 0, Vi = Ve , Ti = Te
⇒ Pi = 90 kPa, Pe = 501.325 kPa.
⇒ ze − zi = 15.0 m
⇒ m = 1.5 kg/s
dEcv
dt= Q − Wcv +
∑
i
mi
(
hi +V
2
i
2+ gzi
)
−∑
e
me
(
he +V
2e
2+ gze
)
Wcv = m[
(h1 − h2) +(
V2
1−V
2
2
2
)
+ g(z1 − z2)]
Wcv = −0.822 kW (work input required) ⊳ : 〈h1 − h2 =P1−P2
ρ, if ρ & T are constant. 〉
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 12 / 29
Steady-State, Steady Flow (SSSF) Processes
Heat Exchangers
T166
Common heat exchanger types. (a) Direct contact heat exchanger. (b)
Tube-within a-tube counterflow heat exchanger. (c) Tube-within-a-tube parallel
flow heat exchanger. (d) Cross-flow heat exchanger.
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 13 / 29
Steady-State, Steady Flow (SSSF) Processes
Cengel P5-85: ⊲ Condenser Cooling Water: Determine mwater .
T149
⇒ W = 0, Q = 0, ∆(KE ) = 0, ∆(PE ) = 0
⇒ m3 = (20000/3600) kg/s
⇒ m1 = m2 & m3 = m4
⇒ h1 = h(H2O, P = 100 kPa, T = 20)
⇒ h2 = h(H2O, P = 100 kPa, T = 30)
⇒ h3 = h(H2O, P = 20 kPa, x = 0.95)
⇒ h4 = h(H2O, P = 20 kPa, x = 0)
0 =∑
(mh)i −∑
(mh)e
⇒ (m3h3 + m1h1) = (m4h4 + m2h2)
⇒ mwater = m1 = 297.4 kg/s ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 14 / 29
Isentropic (First Law) Efficiency Nozzle
Isentropic Nozzle Efficiency
T181
• For nozzle: W = 0, ∆(PE ) = 0, V1 ∼ 0 ⇒ h1 = h2 +V 2
22
⇒ ηn ≡ Actual KE at nozzle exitIsentropic KE at nozzle exit
≡ V 22
V 22s
≈h1−h2h1−h2s
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 15 / 29
Isentropic (First Law) Efficiency Nozzle
Borgnakke Ex. 7.2: ⊲ Isentropic steam flow through nozzle, Ve = ?
T185
• Continuity equation: mi = me = m
• First law: 0 = 0 − 0 + (hi − he) +V 2
i −V 2
e
2+ 0
• Second law: si = se .
⇒ hi = h(steam, Pi = 1 MPa, Ti = 300oC ) = X
⇒ si = s(steam, Pi = 1 MPa, Ti = 300oC ) = X
⇒ se = si & Pe = 0.3 MPa: state ‘e’ defined.
⇒ he = h(Pe = 0.3 MPa, se = X) = X
⇒ V 2
e
2= (hi − he) +
V 2
i
2⇒ Ve = 736.7 m/s⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 16 / 29
Isentropic (First Law) Efficiency Nozzle
Moran Ex. 6.13: ⊲ Estimate nozzle efficiency.
T1083
• Continuity equation: mi = me = m
• First law: 0 = 0 − 0 + (hi − he) +V 2
i −V 2
e
2+ 0, valid for isentropic and
actual cases.
⇒ V 2
e
2= (hi − he) +
V 2
i
2⇒ Ve =
√
• For isentropic case, Ve,s =√
⇒ ηN =√
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 17 / 29
Isentropic (First Law) Efficiency Diffuser
Diffuser Isentropic Efficiency
T1084
• Diffuser isentropic efficiency, ηd ≡ Pout−Pin
Ps,out−Pin
• Coefficient of pressure recovery, Kp ≡ Pout−Pin(
ρinV2in
2
)
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 18 / 29
Isentropic (First Law) Efficiency Diffuser
T1085
1 P2 = ?
2 σcv = ?
3 Area ratio, A2/A1 = ?
4 Kp = ?
• SSSF, Q = 0, Wcv = 0, ss,2 = s1 : reversible diffuser
• 0 = 0 − 0 + (hi − he) +V 2i −V 2
e
2 + 0 → hs,2 = h1 +V 2
1 −V 22
2 : s = case
• State [s, 2] defined: → P2 = P1 + ηd(Ps,2 − P1) =√
(89.9 kPa)
• First law (actual case: h2 = h1 +V 2
1 −V 22
2 ⇒ state [2] defined.
• σcv = s2 − s1 =√
(10.3 J/kgK)
• m = V1A1v1
= V2A2v2
• AR = A2A1
= v2V1v1V2
=√
• Kp ≡ Pout−Pin(
ρinV2in
2
) =√
(0.496).
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 19 / 29
Isentropic (First Law) Efficiency Turbine
Isentropic Turbine Efficiency
T176
• wt = h1 − h2: for steady-state, adiabatic expansion.
• σcv
m= s2 − s1 ≧ 0: states with s2 < s1 is not attainable with adiabatic
expansion.
• wt |s = h1 − h2s : state ‘2s’ is for internally reversible expansion.
⇒ Isentropic turbine efficiency, ηt ≡ wt
wt |s= h1−h2
h1−h2s
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 20 / 29
Isentropic (First Law) Efficiency Turbine
Moran Ex. 6.6: ⊲ Entropy production in a steam turbine
T178
• Continuity equation: mi = me = m
• First law: 0 = q − w + (hi − he) +V 2
i −V 2
e
2+ 0
• Second law: (se − si ) =q
Tb+ σcv
m.
⇒ P1 = 30 bar, T1 = 400oC :⇒ h1 = X, s1 = X
⇒ T2 = 100oC , x2 = 1.0 :⇒ h2 = X, s2 = X
⇒ q = −23.92 kJ/kg⇒ σcv
m= 0.499 kJ/kg.K ⊳.
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 21 / 29
Isentropic (First Law) Efficiency Turbine
Cengel Ex.7.14: ⊲ Isentropic Efficiency of a Steam Turbine
T179
• (P1 = 3 MPa,T1 = 400oC ) :⇒ h1 = X, s1 = X
• (P2 = 50 kPa,T2 = 100oC ) :⇒ h2 = X
• (P2s = 50 kPa, s2s = s1) :⇒ h2s = X
⇒ ηt =h1−h2
h1−h2s= 66.7% ⊳
⇒ m = Powerh1−h2s
= 3.64 kg/s ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 22 / 29
Isentropic (First Law) Efficiency Compressor & Pump
Isentropic Compressor and Pump Efficiencies
T177
• −wc = (h2 − h1): for steady-state, adiabatic compression.
• σcv
m= s2 − s1 ≧ 0: states with s2 < s1 is not attainable with adiabatic
compression.
• −wc |s = (h2s − h1)
⇒ Isentropic compressor/pump efficiency, ηc ≡ −wc |s−wc
= h2s−h1h2−h1
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 23 / 29
Isentropic (First Law) Efficiency Compressor & Pump
Cengel Ex. 7.15: ⊲ Effect of efficiency on compressor power, if ηc = 80%
T180
• T2s = T1(P2/P1)(k−1)/k = 516.5 K
• ηc = h2s−h1
h2−h1
=cP(T2s−T1)
cP(T2−T1)⇒ T2 = 574.8 K
⇒ −Wc |s = m(h2s − h1) = mcP(T2s − T1) = 46.5 kW ⊳
⇒ Wc = −m(h2 − h1) = −mcP(T2 − T1) = −58.1 kW ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 24 / 29
Isentropic (First Law) Efficiency SSSF Process
Steady-state Flow Process
• First Law for SSSF: CV system & reversible process:
⇒ 0 = q12 − w12 + (h1 − h2) − ∆(ke) − ∆(pe)
• 2ndTds Equation: Tds = dh − vdP
• q12 =∫2
1 Tds = (h2 − h1) −∫2
1 vdP
• w12 = (h2 − h1) −∫2
1 vdP + (h1 − h2) − ∆(ke) − ∆(pe)
⇒ w12 = −∫2
1 vdP − ∆(ke) − ∆(pe) ≈ −∫2
1 vdP
wsf = w12 =∫2
1 vdP
T182 T183
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 25 / 29
Isentropic (First Law) Efficiency SSSF Process
Pvn = constant, wsf = w12 = −∫2
1 vdP
T182
w12 :
{= − n
n−1(P2v2 − P1v1) = −nR(T2−T1)
n−1 : n 6= 1
= −RT ln(
v1v2
)
= −RT ln(
P2P1
)
: n = 1
q12 = h2 − h1 + w12
Example: ⊲ A compressor operates at steady state with nitrogen entering at
100 kPa and 20oC and leaving at 500 kPa. During this compression process, the
relation between pressure and volume is Pv1.3 =constant.
• R = (8.314/28) = 0.2969 kJ/kg.K, cP = 1.0 kJ/kg.K
⇒ T2/T1 = (P2/P1)(n−1)/n⇒ T2 = 425 K.
⇒ w12 = −nR(T2−T1)
n−1= −169.5 kJ/kg ⊳
⇒ q12 = cP(T2 − T1) + w12 = −37.5 kJ/kg ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 26 / 29
Isentropic (First Law) Efficiency SSSF Process
Pvn = constant, wb = w12 =∫2
1 Pdv
T192
w12 :
{= −P2v2−P1v1
n−1 = −R(T2−T1)
n−1 : n 6= 1
= −RT ln(
v1v2
)
= −RT ln(
P2P1
)
: n = 1
q12 = u2 − u1 + w12
Example: ⊲ In a reversible process, nitrogen is compressed in a cylinder from
100 kPa and 20oC to 500 kPa. During this compression process, the relation
between pressure and volume is Pv1.3 =constant.
• R = (8.314/28) = 0.2969 kJ/kg.K, cv = R/(1.4 − 1) = 0.742 kJ/kg.K
⇒ T2/T1 = (P2/P1)(n−1)/n⇒ T2 = 425 K.
⇒ w12 = −R(T2−T1)
n−1= −130.4 kJ/kg ⊳
⇒ q12 = cV (T2 − T1) + w12 = −32.2 kJ/kg ⊳
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 27 / 29
Isentropic (First Law) Efficiency SSSF Process
Cengel Ex.7.12: ⊲ Compression & Pumping Works�
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T191
⇒ wP = −∫
2
1vdP ⋍ −vf (P2 − P1) =
−0.939 kJ/kg ⊳ as vf ≃ constant
⇒ wC = −(h2 − h1) = −518.6 kJ/kg ⊳
⇒ |wc | >> |wp |
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 28 / 29
Isentropic (First Law) Efficiency SSSF Process
Isentropic Efficiencies of Some Devices
Device First-Law Efficiency Typical Range
Turbine ηt ≡ wa
ws= wa
ws= h1−h2
h1−h2s80 to 90%
Compressor ηc ≡ ws
wa= ws
wa= h2s−h1
h2−h170 to 85%
Pump ηp ≡ ws
wa= ws
wa= h2s−h1
h2−h150 to 90%
Nozzle ηn ≡ ∆(ke)a∆(ke)s
= h1−h2h1−h2s
≃ V22
V22s
90%
© Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME6101 (2021) 29 / 29
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