SOLUTIONS: HOMEWORK #6

[Pages:22]AREN 2110 FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8

SOLUTIONS

SOLUTIONS: HOMEWORK #6

Chapter 5 Problems

5-45 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined.

Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible.

Properties The density and specific heat of the brass balls are given to be = 8522 kg/m3 and Cp = 0.385 kJ/kg.?C.

Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as

1Ei4n 2- E4o3ut =

1 4E2sys4t3em

Net energy transfer

Change in internal, kinetic,

by heat, work, and mass potential, etc. energies

-Qout = Uball = m(u2 - u1 )

Qout = mC(T1 - T2 )

The total amount of heat transfer from a ball is

Brass balls, 120?C

Water bath, 5?C

m = V = D3 = (8522 kg / m3 ) (0.05 m)3 = 0.558 kg

6

6

Qout = mC(T1 - T2 ) = (0.558 kg)(0.385 kJ / kg.? C)(120 - 74)? C = 9.88 kJ / ball

Then the rate of heat transfer from the balls to the water becomes Q&total = n&ballQball = (100 balls / min) ? (9.88 kJ / ball) = 988 kJ / min

Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature constant at 50?C since energy input must be equal to energy output for a system whose energy level remains constant. That is, Ein = Eout when Esystem = 0 .

5-58C It is mostly converted to internal energy as shown by a rise in the fluid temperature.

5-59C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by a decrease in the fluid temperature.

AREN 2110

SOLUTIONS

FALL 2006

HOMEWORK ASSIGNMENTS 6, 7 and 8

5-61 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exit temperature, and the exit area of the nozzle are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions.

Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The specific heat of air at the anticipated average temperature of 450 K is Cp = 1.02 kJ/kg.?C (Table A-2).

Analysis (a) There is only one inlet and one exit, and thus m&1 = m& 2 = m& . Using the ideal gas relation, the

specific volume and the mass flow rate of air are determined to be

v 1

=

RT1 P1

=

(0.287

kPa

m3/kg K)(473 300 kPa

K )

=

0.4525

m 3 /kg

m&

=

1 v1

A1V1

=

1 0.4525m 3 /kg

(0.008m2 )(30m/s)

= 0.5304

kg/s

(b) We take nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

1E&i4n 2- E4&o3ut

=

1 E&4sys4t4em2?04(s4te4a3dy)

= 0

Rate of net energy transfer Rate of change in internal, kinetic,

by heat, work, and mass

potential, etc. energies

E&in = E&out

m& (h1 + V12 / 2) = m& (h2 + V22 /2) (since Q& W& pe 0)

Substituting,

( ) 0

=

h2

-

h1

+

V22

- 2

V12

0 = C p,ave T2 - T1

+ V22 - V12 2

0

=

(1.02

kJ/kg

K)(T2

-

200o

C)

+

(180

m/s) 2

- 2

(30

m/s) 2

1 kJ/kg 1000 m 2 /s

2

It yields

T2 = 184.6?C

(c) The specific volume of air at the nozzle exit is

v 2

=

RT2 P2

=

(0.287

kPa m3/kg K)(184.6 + 273 K) 100 kPa

= 1.313 m3/kg

P1 = 300 kPa T1 = 200?C V1 = 30 m/s

AIR

P2 = 100 kPa V2 = 180 m/s

m& =

1 v2

A2 V2

0.5304kg/s

=

1

1.313m 3 /kg

A2 (180m/s)

A1 = 80 cm2

A2 = 0.00387 m2 = 38.7 cm2

5-77C Yes. Because energy (in the form of shaft work) is being added to the air.

AREN 2110 FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8

SOLUTIONS

5-79 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.

Properties From the steam tables (Tables A-4 through 6)

P1 T1

= =

10MPa 450o C

v1

=

0.02975m 3 /kg

h1 = 3240.9kJ/kg

and

P1 = 10 MPa T1 = 450?C V1 = 80 m/s

P 2 x2

= 10 kPa

= 0.92

h2

= hf

+ x2 h fg

= 191.83 + 0.92 ? 2392.8 = 2393.2kJ/kg

Analysis (a) The change in kinetic energy is determined from

ke =

V22

- V12 2

=

(50m/s)2

- (80m/s)2 2

1kJ/kg 1000m2 /s 2

= -1.95kJ/kg

?mS=T1E2AkMg/s ?

W

(b) There is only one inlet and one exit, and thus m&1 = m& 2 = m& . We take the turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

1E&i4n 2- E4&o3ut

=

1 E&4sys4t4em2?04(s4te4a3dy)

= 0

Rate of net energy transfer Rate of change in internal, kinetic,

by heat, work, and mass

potential, etc. energies

E&in = E&out m& (h1 + V12 / 2) = W&out + m& (h2 + V22 /2) (since Q& pe 0)

W& out

=

-m& h2

-

h1

+

V22

- 2

V12

Then the power output of the turbine is determined by substitution to be

W&out = -(12 kg/s)(2393.2 - 3240.9 - 1.95)kJ/kg = 10.2 MW

(c) The inlet area of the turbine is determined from the mass flow rate relation,

P2 = 10 kPa x2 = 0.92 V2 = 50 m/s

m& =

1 v1

A1V1

A1

=

m& v1 V1

=

(12 kg/s)(0.02975 m3/kg) 80 m/s

= 0.00446

m 2

AREN 2110 FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8

SOLUTIONS

5-90 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input required is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 Helium is an ideal gas with constant specific heats.

Properties The constant pressure specific heat of helium is Cp = 5.1926 kJ/kg?K (Table A-2a).

Analysis There is only one inlet and one exit, and thus m&1 = m& 2 = m& . We take the compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steadyflow system can be expressed in the rate form as

1E&i4n 2- E4&o3ut

=

1 E&4sys4t4em2?04(s4te4a3dy)

=0

Rate of net energy transfer Rate of change in internal, kinetic,

by heat, work, and mass

potential, etc. energies

E&in = E&out

P2 = 700 kPa T2 = 430 K

W&in + m& h1 = Q&out + m& h2 (since ke pe 0) W&in - Q&out = m& (h2 - h1 ) = m& Cp (T2 - T1 )

Thus,

W&in = Q& out + m& C p (T2 - T1 )

m? =9H0keg/mi ?

W

= (90/60kg/s)(20 kJ/kg) + (90/60kg/s)(5.1926kJ/kg K)(430 - 310)K = 965kW

P1 = 120 kPa T1 = 310 K

5-104 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of cold water is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat loss to the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There are no work interactions.

Properties Noting that T < Tsat @ 250 kPa = 127.44?C, the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus,

h1 hf @ 80?C = 334.91 kJ/kg h2 hf @ 20?C = 83.96 kJ/kg h3 hf @ 42?C = 175.92 kJ/kg

Analysis We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as

Mass balance: m& in - m& out = E&system?0 (steady) = 0 m& 1 + m& 2 = m& 3

Energy balance:

1E& i4n 2- E4&3out

=

1E&4sys4tem2?04(s4tea3dy)

=0

Rate of net energy transfer Rate of change in internal, kinetic,

by heat, work, and mass

potential, etc. energies

E& in = E& out

m& 1h1 + m& 2 h2 = m& 3h3 (since Q& = W& = ke pe 0)

mT? 11

= =

80?C 0.5 kg/s

Tm?22= 20?C

H2O (P = 250 kPa)

T3 = 42?C

AREN 2110 FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8

5-104 CONTINUED

Combining the two relations and solving for m& 2 gives

m&1h1 + m& 2h2 = (m&1 + m& 2 )h3

m& 2

=

h1 h3

- -

h3 h2

m&1

Substituting, the mass flow rate of cold water stream is determined to be

m& 2

=

(334.91 - 175.92)kJ/kg (175.92 - 83.96)kJ/kg

(0.5

kg/s )

=

0.864

kg/s

SOLUTIONS

5-106 Feedwater is heated in a chamber by mixing it with superheated steam. If the mixture is saturated liquid, the ratio of the mass flow rates of the feedwater and the superheated vapor is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.

Properties Noting that T < Tsat @ 800 kPa = 170.43?C, the cold water stream and the mixture exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus,

h1 hf @ 50?C = 209.33 kJ/kg h3 hf @ 800 kPa = 721.11 kJ/kg and

P2 T2

= =

800kPa 200o C

h2

=

2839.3kJ/kg

Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as

Mass balance:

m& in - m& out = m& system?0 (steady) = 0 m& in = m& out m&1 + m& 2 = m& 3

Energy balance:

1E&i4n 2- E4&o3ut

=

1E&4sys4t4em2?04(s4te4a3dy)

=0

Rate of net energy transfer Rate of change in internal, kinetic,

by heat, work, and mass

potential, etc. energies

E&in = E&out m&1h1 + m& 2h2 = m& 3h3

(since Q& W& ke pe 0)

Combining the two,

m&1h1 + m& 2h2 = (m&1 + m& 2 )h3

Dividing by m& 2 yields

yh1 + h2 = (y +1)h3

Solving for y:

y = h3 - h2 h1 - h3

where y = m&1 / m& 2 is the desired mass flow rate ratio. Substituting,

Tm?11= 50?C mT?22= 200?C

H2O (P = 800 kPa)

Sat. liquid

y = 721.11 - 2839.3 = 4.14 209.33 - 721.11

AREN 2110 FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8

SOLUTIONS

SOLUTIONS: HOMEWORK #7

Chapter 6 Problems

6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity.

6-3C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a room.

6-10C Heat engines are cyclic devices that receive heat from a source, convert some of it to work, and reject the rest to a sink.

6-12C No. Because 100% of the work can be converted to heat.

6-15C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics.

6-20 The power output and fuel consumption rate of a power plant are given. The overall efficiency is to be determined.

Assumptions The plant operates steadily.

Properties The heating value of coal is given to be 30,000 kJ/kg. Analysis The rate of energy supply (in chemical form) to this power plant is

Q& H = m& coal ucoal = (60,000 kg/h)(30,000 kJ/kg) = 1.8?109 kJ/h = 500 MW

60 t/h coal

Then the overall efficiency of the plant becomes

overall

=

W& net,out Q& H

150 MW =

500 MW

= 0.300 = 30.0%

Furnace HE sink

6-40C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium.

6-47C No. The refrigerator captures energy from a cold medium and carries it to a warm medium. It does not create it.

AREN 2110

SOLUTIONS

FALL 2006

HOMEWORK ASSIGNMENTS 6, 7 and 8

6-54 The COP and the power consumption of a refrigerator are given. The time it will take to cool 5

watermelons is to be determined.

Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator through its walls, door, etc. is negligible. 3 The watermelons are the only items in the refrigerator to be cooled.

Properties The specific heat of watermelons is given to be C = 4.2 kJ/kg.?C.

Analysis The total amount of heat that needs to be removed from the watermelons is

( ) QL ( ) = mCT watermelons = 5? (10kg) 4.2kJ/kgo C (20 - 8)o C = 2520kJ

Kitchen air

The rate at which this refrigerator removes heat is

( ) Q& L = ( ) COPR W&net,in = (2.5)(0.45kW) = 1.125kW

450 W

R

COP = 2.5

That is, this refrigerator can remove 1.125 kJ of heat per second. Thus the time required to remove 2520 kJ of heat is

cool space

t =

QL Q& L

= 2520kJ 1.125kJ/s

= 2240s = 37.3min

This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air, which will increase the work load. Thus, in reality, it will take longer to cool the watermelons.

6-81 The sink temperature of a Carnot heat engine and the rates of heat supply and heat rejection are given. The source temperature and the thermal efficiency of the engine are to be determined.

Assumptions The Carnot heat engine operates steadily.

Analysis

(a)

For reversible cyclic devices we have

Q H QL

rev

=

TH TL

Thus the temperature of the source TH must be

T H

=

QH QL

rev TL

=

650 200

kJ kJ

(290

K

)

=

942.5

K

(b) The thermal efficiency of a Carnot heat engine depends on the source and the sink temperatures only, and is determined from

source

650 kJ

HE

200 kJ

17?C

th,C

= 1 - TL TH

= 1 - 290 K = 0.69 or 69% 942.5 K

AREN 2110

SOLUTIONS

FALL 2006

HOMEWORK ASSIGNMENTS 6, 7 and 8

6-95 The refrigerated space and the environment temperatures for a refrigerator and the rate of heat

removal from the refrigerated space are given. The minimum power input required is to be determined.

Assumptions The refrigerator operates steadily.

Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversible manner. The coefficient of performance of a reversible refrigerator depends on the temperature limits in the cycle only, and is determined from

COPR,rev

=

(TH

1

/ TL )-1

=

1

(25 + 273K)/(- 8 + 273K)-1

= 8.03

25?C

The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator,

W&net, in, min

=

Q& L COPR, max

=

300 kJ / min 8.03

= 37.36 kJ / min = 0.623 kW

R

300 kJ/min

-8?C

6-103 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. It is to be determined if this heat pump can do the job.

Assumptions The heat pump operates steadily.

Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from

COPHP,rev

=

1

1- (TL / TH

)

=

1

1- (2 + 273K)/(22 + 273K)

= 14.75

The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to be

W& net,in,min

= Q& H COPHP

=

110,000kJ/h 14.75

1h 3600s

=

2.07kW

House 22?C

110,000 kJ/h

HP

8 kW

This heat pump is powerful enough since 8 kW > 2.07 kW.

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