Thermodynamic Processes & Isentropic...

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Thermodynamic Processes & Isentropic Efficiency Dr. Md. Zahurul Haq Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology (BUET) Dhaka-1000, Bangladesh [email protected] http://teacher.buet.ac.bd/zahurul/ ME 6101: Classical Thermodynamics c Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 1 / 25 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. For example, a centrifugal air compressor that operates with a constant mass rate of flow into and out it, constant properties at each point across the inlet and exit ducts, a constant rate of heat transfer to the surroundings, and a constant power input. At each point in the compressor the properties are constant with time. c Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 2 / 25 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. In a diffuser, the gas or liquid decelerates 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 W cv drops out. ΔPE 0 c Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 3 / 25 Steady-State, Steady Flow (SSSF) Processes [Moran Ex. 4.3]: Converging-diverging Steam Nozzle: Estimate A 2 . T145 Steady state dE cv /dt = 0 Z 2 = Z 1 W cv = 0 & Q = 0 m i = m e = 2 kg/s dE cv dt = Q + W cv + i m i h i + V 2 i 2 + gz i - e m e h e + V 2 e 2 + gz e 0 =(h 1 - h 2 )+ 1 2 (V 2 1 - V 2 2 ) h 2 = h 1 + 1 2 (V 2 1 - V 2 2 ) ρ 2 = ρ(Steam, P = P 2 , h = h 2 )= 6.143 kg/m 3 = m 2 = ρ 2 A 2 V 2 = A 2 = 4.896 × 10 -4 m 2 c Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 4 / 25

Transcript of Thermodynamic Processes & Isentropic...

Page 1: Thermodynamic Processes & Isentropic Efficiencyteacher.buet.ac.bd/zahurul/ME6101/ME6101_ThermoProcesses.pdf · Thermodynamic Processes & Isentropic Efficiency Dr. Md. Zahurul Haq

Thermodynamic Processes & Isentropic Efficiency

Dr. Md. Zahurul Haq

ProfessorDepartment of Mechanical Engineering

Bangladesh University of Engineering & Technology (BUET)Dhaka-1000, Bangladesh

[email protected]://teacher.buet.ac.bd/zahurul/

ME 6101: Classical Thermodynamics

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 1 / 25

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.

For example, a centrifugal air compressor that operates with a constant

mass rate of flow into and out it, constant properties at each point

across the inlet and exit ducts, a constant rate of heat transfer to the

surroundings, and a constant power input. At each point in the

compressor the properties are constant with time.

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 2 / 25

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.

In a diffuser, the gas or liquid decelerates 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

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 3 / 25

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

2

e

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 ⊳

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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.

There is usually no significant heat transfer with the surroundings,

and the change in potential energy from inlet to exit is negligible.c

Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 5 / 25

Steady-State, Steady Flow (SSSF) Processes

⊲ [Cengel Ex. 5.8]: R-134a enters the capillary tube of a refrigerator as

saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12 MPa.

Determine x2 and ∆T .

T148

⇒ Steady state ⇒ dEcv/dt = 0

⇒ Z2 = Z1 & V2 ≃ V1

⇒ _Q ≃ 0 & _Wcv = 0

h2∼= h1

⇒ h1 = enthapy(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 ⊳

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 6 / 25

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.

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 7 / 25

Steady-State, Steady Flow (SSSF) Processes

⊲ [Moran Ex 4.4]: Heat Transfer from Steam Turbine: Determine heat loss, Q .

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

2

e

2+ gze

)

− _Wcv = _Q + _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) ⊳

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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 T164c

Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 9 / 25

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

2

e

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 = 0.72 kg/s

=⇒ _Wcv = 119.4 kW (work input required) ⊳

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 10 / 25

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

2

e

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. 〉

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 11 / 25

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.

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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 ⊳

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 13 / 25

Isentropic (First Law) Efficiency

⊲ [Borgnakke Ex. 7.2]: Reversible adiabatic flow of steam 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⊳

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 14 / 25

Isentropic (First Law) Efficiency

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

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 15 / 25

Isentropic (First Law) Efficiency

⊲ [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 ) =

_qTb

+ _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 ⊳.

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Isentropic (First Law) Efficiency

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−h2h1−h2s

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 17 / 25

Isentropic (First Law) Efficiency

⊲ [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 ⊳

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 18 / 25

Isentropic (First Law) Efficiency

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−h1

h2−h1

c

Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 19 / 25

Isentropic (First Law) Efficiency

⊲ [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 ⊳

c

Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 20 / 25

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Isentropic (First Law) Efficiency

Steady-state Flow Process

First Law: CM system & reversible process:

⇒ δq = du − δw = d(h − Pv) + Pdv = dh − vdP

Second Law: δq = Tds ⇒ Tds = dh − vdP

⇒∫2

1 Tds =∫2

1(dh − vdP) = (h2 − h1) −∫2

1 vdP

First Law: CV system & reversible process:

⇒ 0 = q12 + w12 + (h1 − h2) + ∆(ke) + ∆(pe)

=⇒ −w12 = q12 + (h1 − h2) + 0 + 0 =∫2

1 Tds + (h1 − h2) = −∫2

1 vdP

wsf = w12 =∫2

1 vdP

W12 =

2

1V dP

T754

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Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 21 / 25

Isentropic (First Law) Efficiency

Pvn = constant

W12 =

2

1V dP

T754

wsf = w12 =∫2

1 vdP

−w12 :

{= n

1−n (P2v2 − P1v1) =nR(T2−T1)

1−n : n 6= 1

= RT ln

(

v2v1

)

= −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 Pv 1.3 =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)

1−n = 169.5 kJ/kg ⊳

⇒ q12 = cP (T2 −T1) + w12 = −37.5 kJ/kg ⊳

c

Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 22 / 25

Isentropic (First Law) Efficiency

Pvn = constant

W

W12 = −∫

2

1PdV

T753

wb = w12 = −∫2

1 Pdv

−w12 :

{= P2v2−P1v1

1−n =R(T2−T1)

1−n : n 6= 1

= RT ln

(

v2v1

)

= −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 Pv 1.3 =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)

1−n = 130.4 kJ/kg ⊳

⇒ q12 = cV (T2 −T1) + w12 = −32.2 kJ/kg ⊳

c

Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 23 / 25

Isentropic (First Law) Efficiency

⊲ [Cengel Ex.7.12]: Compression & Pumping Works

T191

⇒ wP =∫

2

1vdP ⋍ vf (P2 − P1) = 0.939 kJ/kg ⊳

⇒ wC = h2 − h1 = 518.6 kJ/kg ⊳

=⇒ wc >> wp

c

Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 24 / 25

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Isentropic (First Law) Efficiency

Isentropic Efficiencies of Some Devices

Device First-Law Efficiency Typical Range

Turbine ηt =

_wa

_ws= wa

ws= h2−h1

h2s−h180 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

≃V2

2

V22s

90%

Boiler

Heat-exchanger

Throttling-device

c

Dr. Md. Zahurul Haq (BUET) Thermodynamic Processes & Efficiency ME 6101 (2017) 25 / 25