Tutorial 3 thermo

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Thermodynamics (ME2121)

Tutorial 3

Lecturers: Prof. A.S.Mujumdar

Dr C. Yap

Tutor: Wang Shijun

email:[email protected]

Department of Mechanical Engineering

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Summary of Chapters 5&6

* The second law of Thermodynamics (P253, P262)

Kelvin-Planck Statement :

It is impossible for any device that operates on a

cycle to receive heat from a single reservoir and produce a net amount of work.

Clausius Statement:

It is impossible to construct a device that operates in a

cycle and produces no effect other than the transfer of heat from a

lower-temperature body to a higher-lower-temperature body.

* Principles of heat engines, Refrigerators and heat pumps (P257-260)

Heat engine:

Refrigerator:

Heat pump:

in out net th

Q

W

_.

=

η

L H L L H L in net L R T T T Q Q Q W Q COP − = − = = . L H H L H H in net H HP

T

Q

W

Q

COP

−

=

−

=

. 2

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3

Summary of Chapters 5&6

* Reversible and irreversible processes (P265-269)

Reversible process is a process that can be reversed without leaving

any trace on the surroundings

Typical irreversibilities:

Friction

Unrestrained expansion of a gas

Heat transfer through a finite difference

Mixing

Chemical reactions, etc.

* The Carnot cycle and its principles

Reversible isothermal expansion

Reversible adiabatic expansion (Isentropic)

Reversible isothermal compression

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4

Summary of Chapters 5&6

* Definition of Entropy

a measure of molecular disorder or

randomness of a system and it can be created but cannot be destroyed.

T rev

Q dS int, ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = δ

* Increases of entropy principle: The entropy of an isolated system during a

process always increases

or, in the limiting case of a reversible process, remains

constant

* property diagrams involving entropy (h-s, t-s) (p314), The Tds relations (p320)

Tds = du + Pdv (Gibs equation)

Tds = dh –vdP

* Entropy change of liquids and solids (small temperature change)

0 ≥ gen S 1 2 1 2 ln T T C s s T CdT T du ds= = ⇒ − = _av

* The entropy change of ideal gases (small temperature change)

1 2 1 2 , 1 2 1 2 1 2 , 1 2 ln ln ln ln P P R T T C s s P dP R T dT C P T vdv T dh ds v v R T T C s s v dv R T dT C T Pdv T du ds av p p av v v − = − ⇒ + = + = + = − ⇒ + = + =

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5

Summary of Chapters 5&6

* Reversible steady-flow work

v should be smaller for work-consuming devices---pumps, compressors, etc.

v should be greater for work-producing devices---steam turbines, etc.

* Calculation of the compressor work (P337)

* Isentropic processes (internally irreversible, adiabatic, const Cp and Cv) (p313)

Isentropic process

Polytropic process

Isothermal process

Pv = const

* Isentropic (or adiabatic) efficiencies of steady-flow devices (turbines,

compressors and pumps, nozzles)

* Entropy balance (P347)

{

S

2

S

1

S

_in

−

_out

+

_gen

=

∆

_system

=

−

3

2

1

4

3

42

1

Net entropy transfer by heat and mass

Change in entropy Entropy Generation

pe

ke

vdp

w

_rev_,_in

=

∫

+

∆

+

∆

const

Pv

n

=

const

Pv

k

=

The entropy change of a system during a

process is equal to the net entropy

transfer through the system and the

entropy generated within the system.

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6

Summary of Chapters 5&6

*Entropy transfer by Heat transfer

* Entropy change for a closed system and its surroundings

Since any closed system and its surroundings can be treated as an

adiabatic system in which the entropy generation equals the entropy

change of the adiabatic system

sur sys gen

S

=

∑

∆

=

∆

+

∆

∫

≅

∑

= = = 2 1 ) ( k k heat heat T Q T Q S const T T Q S

*Entropy transfer by Mass flow

∫

• ∆ • = = A mass t mass mass mass VdA s dt S ms S

S

ρ

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Summary of Chapters 5&6

* Entropy balance for a closed system

* Entropy change for a general steady flow process

Typical processes operated steadily in such devices as turbines,

compressors, nozzles, diffusers, heat exchangers, pipes, ducts, etc.

sys gen k k S S T Q ₊ ₌ _∆

∑

* Entropy balance for a control volume

1 2 S S S S S S S s m s m T Q initial final sys sys gen e e i i k k − = − = ∆ ∆ = + − +

∑

) 0 (sin ∆ = − − = ⋅ ⋅ ⋅ ⋅ ⋅

∑

sys k k i i e e gen ce S T Q s m s m S 7

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8

Problem 1

Problem C1 (Problem 5-107)

A Carnot heat pump is to be used to heat a house and maintain it at 20o_{C in winter. On a day}

when the average outdoor temperature remains at about 2o_{C, the house is estimated to lose}

heat at a rate of 82,200kJ/h. If the heat pump consumes 8kW of power while operating,

determine (a) how long the heat pump ran on that day; (b) the total heating costs, assuming an average price of 8.5cents/kWh for electricity; and (c) the heating cost for the same day if resistance heating is used instead of a heat pump

Solution

(a) The coefficient of performance of this Carnot heat pump depends on the temperature limits in the cycle only, and is determined from

3 . 16 ) 273 20 /( ) 273 2 ( 1 1 ) / ( 1 1 , = ₋ = ₋ ₊ ₊ = H L rev HP T T COP

The amount of heat the house lost that day is

kJ h h kJ day Q Q_H = &_H(1 ) = (82000 / )(24 ) =1.968×106

Then the required work input to this Carnot heat pump is determined from the definition of the coefficient of performance to be

kJ kJ COP Q W HP H in net 120736 3 . 16 10 968 . 1 6 , = × = =

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9

Problem 1

A Carnot heat pump is to be used to heat a house and maintain it at 20o_{C in winter. On a day}

when the average outdoor temperature remains at about 2o_{C, the house is estimated to lose}

heat at a rate of 82,200kJ/h. If the heat pump consumes 8kW of power while operating,

determine (a) how long the heat pump ran on that day; (b) the total heating costs, assuming an average price of 8.5cents/kWh for electricity; and (c) the heating cost for the same day if resistance heating is used instead of a heat pump

Solution

Thus the length of time the heat pump ran on that day is

h s s kJ kJ W W t in net in net ₁₅₀₉₂ ₄_.₁₉ / 8 120736 , , ₌ ₌ ₌ = ∆ &

(b) The total heating cost on that day is

85 . 2 $ ) / $ 085 . 0 )( 19 . 4 )( 8 ( ) )( ( _, ×∆ = = = × =W price W t price kW h kWh

Cost &_net_in

(c) If resistance heating were used, the entire heating load on that day would have to be met by electrical energy. Therefore, the heating system would consume 1968000kJ of thermal energy supplied by electricity that would cost

47 . 46 $ ) / $ 085 . 0 )( 3600 1 )( 10 968 . 1 ( cos ₌ _× ₌ _× 6 ₌ − kWh kJ kWh price Q t New _H

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Problem 2

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.96kg of refrigrant-134a as the working fluid. It is known that the maximum absolute temperature is 1.2 times the minimum absolute temperature, and the net work input to the cycle is 22kJ. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the minimum pressure in the cycle.

Solution

Assumptions: The refrigeration cycle is said to operate on

the closed Carnot cycle, which is totally reversible. The coefficient of performance of the cycle is

0 . 5 1 2 . 1 1 1 / 1 ₌ − = − = L H R T T COP Also, kJ kJ W COP Q W Q COP _L _R _in in L R = → = × =(5)(22 ) =110 Thus ) ( 132 22 110 kJ W Q Q_H = _L + = + = ₁₀

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11

Problem 2

Solution

and

Since the enthalpy of vaporization h_fg at a given T or P represents the amount of heat transfer per unit mass as substance is converted from saturated liquid to saturated vapor at that T or P. Therefore, T_H is the temperature that corresponds to the h_fg value of 137.5kJ/kg, and is

determined from the Table A-11 to be

K

C

T

_H

≅

61

o

=

334 kg

kJ

kg

kJ

m

Q

h

mh

h

m

Q

h T fg T fg T f g h H H H

/

5 .

137

96 .

0

132 )

(

@ @ @

=

−

=

(12)

12

Problem 2

Solution

C

K

T

_L H

278 .

3

5 .

3

o

2 .

1

334

2 .

1 =

=

≅

=

Therefore,

MPa

P

C sat@5.3o

0 .

354

min

=

Then,

Question: could we do calculation as follows and then

read off pressure from Table A-12?

kg

kJ

kg

kJ

m

Q

h

mh

h

m

Q

L T fg T fg T f g L L L L

/

?

96 .

0

110 )

(

@ @ @

=

−

=

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13

Problem 3

A Carnot heat engine receives heat at 750K and rejects the waste heat to the environment at 300K. the entire work output of the heat engine is used to drive a Carnot refrigerator that removes heat from the cooled space at –15o_{C at a rate of 400kJ/min and rejects it to the same}

environment at 300K. Determine (a) the rate of heat supplied to the heat engineand (b) the total rate of heat rejection to environment.

Solution

(a) The coefficient of performance of the Carnot refrigerator is

14 . 6 1 ) 258 /( ) 300 ( 1 1 / 1 , = ₋ = ₋ = K K T T COP L H C R

Then power input to the refrigerator becomes

min / 1 . 65 14 . 6 min / 400 , , kJ kJ COP Q W C R L in net = = = &

which is equal to the power output of the heat engine, W&_net_,_out

The thermal efficiency of the Carnot engine is determined from

6 . 0 750 300 1 1 , = − = − = K K T T H L C th η

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Problem 3

Solution

Then the rate of heat input to this engine is determined from the definition of the thermal efficiency to be

min

/

5 .

108

60 .

0 min

/

1 .

65

, , ,

kJ

W

Q

HE th out net HE H

=

η

=

&

(b) The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine ( QL,HE ) and the heat discarded by the refrigerator ( Q&H,R ),

min)

/

(

4 .

43

1 .

65

5 .

108

, , ,

Q

W

kJ

Q

&

_L _HE

=

&

_H _HE

−

&

_net_out

=

−

=

min)

/

(

1 .

465

1 .

65

400

, , ,

Q

W

kJ

Q

&

_H _R

=

&

_L _R

+

&

_net_in

=

+

=

min)

/

(

5 .

508

1 .

465

4 .

43

, ,

Q

kJ

Q

&

_Ambient

=

&

_L _HE

+

&

_H _R

=

+

=

Questions: May we take the HE and Refrigerator as a combined system to calculate the

total heat rejection, which is equal to the sum of Q_H,HEand Q_L,R? Does the work

interaction between HE and R affect the calculation in this case? Does it violate the first and second laws of Thermodynamics?

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Problem 4

The kitchen, bath and other ventilation fans in a house should be used sparingly since these fans can discharge a houseful of warmed or cooled air in just one hour. Consider a 200m2 _house

whose ceiling height is 2.8m. The house is heated by a 96 percent efficient gas heater and is maintained at 22o_{C and 92kPa. If the unit cost of natural gas is $0.60/therm (1therm=105500kJ),}

determine the cost of energy “vented out” by the fans in 1hr. Assume the average outdoor temperature during the heating season to be 5o_C.

Solution: Assumptions:

(1) Steady operating conditions exist

(2) The house is maintained at 22o_{C and 92kPa at all times}

(3) The infiltrating air is heated to 22o_{C before it is vented out}

(4) Air is an ideal gas with constant specific heats at room temperature (5) The volume occupied by the people, furniture, etc., is negligible

Properties:

The gas constant of air is R = 0.287kPa.m3 /kg.K

The specific heat of air at room temperature is

C

_p

=

1 .

0 kJ

/

kg

.

K

(Table A-2a) (Table A-1) A = 220 m2 H = 2.8 m P = 92 kPa T = 22 o_C T = 5 o_C

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Problem 4

Solution:

The density of air at the indoor conditions of 92kPa and 22o_{C is}

3 3 0 0 0 1.087 / ) 273 22 )( . / . 287 . 0 ( 92 m kg K kg m kPa kPa RT P = + = = ρ

Noting that the interior volume of the house is ₂₀₀ ₂_.₈ ₅₆₀ 3

m

= ×

the mass flow rate of air vented out becomes

s kg h kg hr m m kg t V

m&_air = &

ρ

_air / = (1.087 / 3)(560 3)/(1 ) = 608.7 / = 0.169 /

Noting that the indoor air vented out at 22o_{C is replaced by infiltrating outdoor}

air at 5o_{C, this corresponds to energy loss at a rate of}

) (

,fan air indoors outdoors air P indoors outdoors

loss m h h m C T T

Q& = & − = & −

A = 220 m2 H = 2.8 m P = 92 kPa T = 22 o_C

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Problem 4

Solution: kW s kJ C C K kg kJ s kg/ )(1.0 / . )(22o 5o ) 2.874 / 2.874 169 . 0 ( − = = =

Then the amount and cost of the heat “vented out” per hour becomes

96 . 0 / ) 1 )( 874 . 2 ( / , t kW h Q Loss Energy

Fuel− − = &_loss _fan∆ η_furnace =

kWh 3 = ) cos )(

(Fuel Energy Loss unit t of energy

loss Money − = − − − − − A = 220 m2 H = 2.8 m P = 92 kPa T = 22 o_C T = 5 o_C hr kWh therm therm hr kWh ) $0.614/ 3 . 29 1 )( / 60 . 0 )($ / 3 ( = =

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Problem 5

An ordinary egg can be approximated as a 5.5 cm diameter sphere. The egg is initially at a uniform temperature of 8 o_{C and is dropped into boiling water at 97}o_{C. Taking the properties of the egg to be}

ρ = 1020 kg/m3 _{and Cp = 3.32 kJ/(kg}o_{C), determine how much heat is transferred to the egg by the}

time the average temperature of the egg rises to 70o_{C and the amount of entropy generation}

associated with this heat transfer process.

Solution:

Assumptions

1. The egg is spherical in shape with a radius of 5.5cm. 2. The thermal properties of the egg are constant.

3. Energy absorption or release associated with any chemical and/or phase changes within the egg is negligible.

4. There are no changes in kinetic and potential energies.

Properties

Egg 8o_C Boiling

Water

The density and specific heat of the egg are given to be

3

/

1020kg m

=

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Problem 5

An ordinary egg can be approximated as a 5.5 cm diameter sphere. The egg is initially at a uniform temperature of 8 o_{C and is dropped into boiling water at 97}o_{C. Taking the properties of the egg to be}

ρ = 1020 kg/m3 _{and Cp = 3.32 kJ/(kg}o_{C), determine how much heat is transferred to the egg by the}

Egg 8o_C Boiling Water

3

2

1

43

42

1

in

E

out

E

system

E

−

=

∆

Change in internal, kinetic, Potential, etc. energies Net energy transfer

By heat, work, and mass

Analysis

We take the egg as the system. This is a closed system since no mass enters or leaves the egg. The energy balance for this closed system can be expressed as Energy balance:

(

u2 u1

)

mC

(

T2 T1

)

m U

Q_in = ∆ _egg = − = _V −

Then the mass of the egg and the amount of heat transfer become

(

kg m

) (

m

)

kg D V m 0.0889 6 055 . 0 / 1020 6 3 3 3 = = = = ρ ρπ π

(

u u

)

mC

(

T T

) (

kg

)

(

kJ kg C

)

C m U Q_in =∆ _egg = ₂ − ₁ = _p ₂ − ₁ = 0.0899 3.32 / ⋅o (70−8)o

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Problem 5

An ordinary egg can be approximated as a 5.5 cm diameter sphere. The egg is initially at a uniform temperature of 8 o_{C and is dropped into boiling water at 97}o_{C. Taking the properties of the egg to be ρ}

= 1020 kg/m3 _{and Cp = 3.32 kJ/(kg}o_{C), determine how much heat is transferred to the egg by the}

Solution:

Egg 8o_C Boiling

Water

We again take a single egg as the system. The entropy generated during this process can be determined by applying an entropy balance on an extended system that includes the egg and its immediate surroundings so that the boundary temperature of the extended system is at 97 o_{C at all times:}

{

1

2

3

4

3

42

1

in

S

out

S

gen

S

system

S

−

+

=

∆

Net entropy transfer

by heat and mass EntropyGeneration Changein entropy

system b in gen system gen b in S T Q S S S T Q ∆ + − = → ∆ = + where

(

)

(

kg

)(

kJ kg K

)

kJ K T T mC s s m S_system _av 0.0588 / 273 8 273 70 ln / 32 . 3 0889 . 0 ln 1 2 1 2 ₊ = + ⋅ = = − = ∆

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Problem 5

An ordinary egg can be approximated as a 5.5 cm diameter sphere. The egg is initially at a uniform temperature of 8 o_{C and is dropped into boiling water at 97}o_{C. Taking the properties}

of the egg to be ρ = 1020 kg/m3 _{and Cp = 3.32 kJ/(kg}o_{C), determine how much heat is}

transferred to the egg by the time the average temperature of the egg rises to 70o_{C and the}

amount of entropy generation associated with this heat transfer process.

Solution: Egg 8o_C Boiling Water Substituting, K kJ K kJ K kJ S T Q S _system b in gen 0.0588 / 0.00961 / 370 3 . 18 = + − = ∆ + − = (per egg)

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22

Problem 6

A 0.4m3 _{rigid tank is filled with saturated liquid water at 200}o_{C. A valve at the bottom of the tank}

is now opened, and one-half of the total mass is withdrawn from the tank in liquid form. Heat is transferred to water from a source at 250 o_{C so that the temperature in the tank remains constant.}

Determine (a) the amount of heat transfer and (b) the total entropy generation for this process.

Solution:

Assumptions

1. This is an unsteady process since the conditions within the device are

changing during the process, but it can be analyzed as uniform-flow process since the state of fluid leaving the device remains constant.

2. Kinetic and potential energies are negligible. 3. There are no work interactions involved.

4. The direction of heat transfer is to the tank (will be verified). 5. No water evaporates during discharging.

m_e Q H₂O V=0.4m3 200o_C T = const. Analysis

(a) We take the tank as the system, which is a control volume since mass crosses the boundary. Noting that the microscopic energies of flowing and nonflowing fluids are represented by enthalpy h and internal energy u, respectively, the mass and energy balances for this uniform-flow system can be expressed as

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23

Problem 6

Solution: ⎪⎩ ⎪ ⎨ ⎧ ⋅ = = = = ⇒ ⎭ ⎬ ⎫ = K kg kJ s s kg kJ h h liquid sat C T C f e C f e o e o o / 3309 . 2 / 45 . 852 . 200 200 @ 200 @ me Q H2O V=0.4m3 200o_C T = const.

Mass balance: min −mout = ∆msystem → me = m1 −m2

3

2

1

43

42

1

in

E

out

E

system

E

−

=

∆

Change in internal, kinetic, Potential, etc. energies Net energy transfer

By heat, work, and mass Energy balance:

) 0 ( 1 1 2 2 − ≅ ≅ ≅ + = m h m u m u SinceW ke pe Q_in _e _e Properties ⎪ ⎪ ⎩ ⎪⎪ ⎨ ⎧ ⋅ = = = = = = ⇒ ⎭ ⎬ ⎫ = K kg kJ s s kg kJ u u kg m liquid sat C T C f C f C f o o o o / 3309 . 2 / 65 . 850 / 001157 . 0 . 200 200 @ 1 200 @ 2 3 200 @ 1 1 ν ν

(24)

Problem 6

Solution: m_e Q H2O V=0.4m3 200o_C T = const.

The initial and the final masses in the tank are

(

kg

)

kg me m m kg kg m m V m = = = = = = = 86 . 172 72 . 345 2 1 2 1 72 . 345 / 001157 . 0 4 . 0 1 2 3 3 1 1 _ν

Now we determine the final internal energy and entropy,

00917 . 0 001157 . 0 12736 . 0 001157 . 0 002314 . 0 / 002314 . 0 86 . 172 4 . 0 2 2 3 3 2 2 = − − = − = = = = fg f x kg m kg m m V ν ν ν ν Problem C6 (Problem 6-133)

(25)

Problem 6

Solution:

(

)(

)

(

)(

)

⎪⎩ ⎪ ⎨ ⎧ ⋅ = + = + = = + = + = ⇒ ⎭ ⎬ ⎫ = = K kg kJ s x s s kg kJ u x u u x C T fg f fg f o / 3685 . 2 1014 . 1 00917 . 0 3309 . 2 / 65 . 866 7 . 1744 00917 . 0 65 . 850 00917 . 0 200 2 2 2 2 2 2

The heat transfer during this process is determined by substituting these values into the energy balance equation,

(

)(

) (

)(

) (

)(

)

kJ kg kJ kg kg kJ kg kJ kg kJ kg u m u m h m Q_in _e _e 3077 / 65 . 850 72 . 345 / 65 . 866 / 96 . 172 / 4 . 852 86 . 172 1 1 2 2 = − + = − + = Problem C6 (Problem 6-133)

(b) The total entropy generation is determined by considering a combined system that includes the tank and the heat source. Noting that no heat crosses the boundaries of this combined system and no mass enters, the entropy balance for it can be expressed as

(26)

Problem 6

Solution: me Q H₂O V=0.4m3 200o_C T = const.

{

1

2

3

4

3

42

1

in

S

out

S

gen

S

system

S

−

+

=

∆

Net entropy transfer by heat and mass

Change in entropy Entropy Generation source k gen e e

s

S

m

+

=

∆

+

∆

−

_tan

Therefore, the total entropy generated during the process is

(

)

source source e e source k e e gen T Q s m s m s m S S s m S = +∆ _tan +∆ = + ₂ ₂ − ₁ ₁ −

(

)(

) (

)(

)

(

)(

)

K kJ K kJ K kg kJ kg K kg kJ kg K kg kJ kg / 616 . 0 523 3077 / 3309 . 2 72 . 345 / 3685 . 2 86 . 172 / 3309 . 2 86 . 172 = − ⋅ − ⋅ + ⋅ = Problem C6 (Problem 6-133)

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