TD solution ch3 - 교과서 솔루션 PDF

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3-10 3-27 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, the volume change of the cylinder, and the enthalpy change are to be determined. Analysis (a) The final pressure is equal to the initial pressure, which is determined from

P2

P1

Patm 

mp g

SD /4 2

88 kPa 

(12 kg)(9.81 m/s 2 ) § 1 kN ¨ 2 ¨ S (0.25 m) /4 © 1000 kg.m/s2

· ¸ 90.4kPa ¸ ¹

(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10qC and at the final state of 90.4 kPa and 15qC are (from EES)

v1 = 0.2302 m3/kg 3

v 2 = 0.2544 m /kg

h1 = 247.77 kJ/kg h2 = 268.18 kJ/kg

The initial and the final volumes and the volume change are

V1 mv 1

(0.85 kg)(0.2302 m3 /kg) 0.1957 m 3

V 2 mv 2

(0.85kg)(0.2544 m3 /kg) 0.2162 m 3

'V

R-134a 0.85 kg -10qC

Q

0.2162  0.1957 0.0205m3

(c) The total enthalpy change is determined from

'H

m (h 2  h1 ) (0.85kg)(268.18 247.77)kJ/kg 17.4kJ/kg

3-28 The temperature of R-134a at a specified state is to be determined. Analysis Since the specified specific volume is higher thanv g for 800 kPa, this is a superheated vapor state. From R-134a tables,

P

v

½ ¾T 3 0.04471m /kg ¿

600 kPa

80qC (Table A - 13)

3-29 A rigid container that is filled with R-134a is heated. The final temperature and initial pressure are to be determined. Analysis This is a constant volume process. The specific volume is

v1 v 2

V m

1.348 m3 10 kg

R-134a -40°C 10 kg 1.348 m3

3

0.1348 m /kg

The initial state is determined to be a mixture, and thus the pressure is the saturation pressure at the given temperature

P1

Psat @ -40qC

51.25kPa (Table A -11)

The final state is superheated vapor and the temperature is determined by interpolation to be

P2

v2

½ ¾ T2 3 0.1348 m /kg ¿

200 kPa

66.3qC (Table A - 13)

P

2

1

v

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3-13 3-34 A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be determined. Analysis The initial specific volume is

v1

V1 m

0.1546 m3 1 kg

0.1546 m 3/kg

H 2O 350°C 1 kg 0.1546 m3

This is a constant-pressure process. The initial state is determined to be superheated vapor and thus the pressure is determined to be

T1

v1

350qC

½ ¾ P1 0.1546 m /kg ¿ 3

P2

1.8 MPa (Table A - 6) P

The saturation temperature at 1.8 MPa is 207.11°C. Since the final temperature is less than this temperature, the final state is compressed liquid. Using the incompressible liquid approximation,

v2 v

f @100qC

2

1

0.001043 m3 /kg (Table A - 4)

v The final volume is then

V 2 mv 2

(1 kg)(0.001043 m3 /kg) 0.001043m3

3-35 The volume of a container that contains water at a specified state is to be determined. Analysis The specific volume is determined from steam tables by interpolation to be

P 100 kPa ½ ¾ v T 150 qC ¿

1.9367 m3 /kg (Table A - 6)

The volume of the container is then

V

mv

(3 kg)(1.9367 m3 /kg) 5.81m3

Water 3 kg 100 kPa 150q C

3-36 Water is boiled at sea level (1 atm pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of the heat generated to the water. The rate of evaporation of water is to be determined. Properties The properties of water at 1 atm and thus at the saturation temperature of 100qC are hfg = 2256.4 kJ/kg (Table A4). Analysis The net rate of heat transfer to the water is

Q

0.60 u 3 kW 1.8 kW

Noting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, the rate of evaporation of water is determined to be  Q 1.8 kJ/s 3 m evapo ration = 0.80 u10 kg/s 2.872kg/h h fg 2256.4 kJ/kg

H2O 100qC

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3-16 3-40 A piston-cylinder device fitted with stops contains water at a specified state. Now the water is cooled until a final pressure. The process is to be indicated on the T-v diagram and the change in internal energy is to be determined. Analysis The process is shown on T-v diagram. The internal energy at the initial state is

P1 T1

200 kPa½ ¾ u1 300qC ¿

2808.8 kJ/kg (T able A - 6) Water 200 kPa 300qC

State 2 is saturated vapor at the initial pressure. Then,

P2 x2

½ ¾ v2 1 (sat. vapor) ¿ 200 kPa

3

0.8858 m /kg (Table A - 5)

Process 2-3 is a constant-volume process. Thus,

P3

v3

100 kPa ½ 1508.6 kJ/kg (Table A - 5) ¾ u v 2 0.8858 m 3 /kg ¿ 3

200 kPa

T

1

300qC

100 kPa 2

The overall change in internal energy is

' u u1  u3

Q

3

2808.8  1508.6 1300kJ/kg

v

3-41 Saturated steam at Tsat = 40qC condenses on the outer surface of a cooling tube at a rate of 130 kg/h. The rate of heat transfer from the steam to the cooling water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The condensate leaves the condenser as a saturated liquid at 30qC. Properties The properties of water at the saturation temperature of 40qC are hfg = 2406.0 kJ/kg (Table A-4). Analysis Noting that 2406.0 kJ of heat is released as 1 kg of saturated vapor at 40qC condenses, the rate of heat transfer from the steam to the cooling water in the tube is determined directly from

Q m evaph

fg

40qC

L = 35 m

D = 3 cm

(130 kg/h)(2406.0 kJ/kg) 312,780 kJ/h = 86.9kW

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3-23 3-50 A rigid tank contains saturated liquid-vapor mixture of R-134a. The quality and total mass of the refrigerant are to be determined. Analysis At 400 kPa, v f = 0.0007905 m3/kg andvg = 0.051266 m3/kg (Table A-12). The volume occupied by the liquid and the vapor phases are

Vf

0.2 u 0.14 0.028 m3 and V g

0.8 u0.14 0.112 m3

R-134a 0.14 m3 400 kPa

Thus the mass of each phase is

mf mg

Vf

0.028 m 3

vf

0.0007905m /kg

35.42 kg

3

Vg

0.112 m

vg

3

0.051266 m 3/kg

2.185 kg

Then the total mass and the quality of the refrigerant are mt = mf + mg = 35.42 + 2.185 = 37.61 kg

x

mg

2.185 kg

mt

37.61kg

0.05810

3-51 Superheated water vapor cools at constant volume until the temperature drops to 120°C. At the final state, the pressure, the quality, and the enthalpy are to be determined. Analysis This is a constant volume process (v = V/m = constant), and the initial specific volume is determined to be

P1 T1

1.4 MPa $ 250 C

½ ¾v 1 ¿

0.16356 m 3 /kg

(Table A-6) H 2O 1.4 MPa 250 qC

At 120°C, v f = 0.001060 m3/kg andv g = 0.89133 m3/kg. Thus at the final state, the tank will contain saturated liquid-vapor mixture since vf < v < vg , and the final pressure must be the saturation pressure at the final temperature,

P

P sat@120$C

198.67 kPa

T

1

(b) The quality at the final state is determined from

x2

v2  v f v fg

0.16356 0.001060 0.1825 0.89133 0.001060

(c) The enthalpy at the final state is determined from

h h f  xh fg

503.81 0.1825u 2202.1 905.7 kJ/kg

2

v

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3-28 3-59 A piston-cylinder device that is initially filled with water is heated at constant pressure until all the liquid has v diagram vaporized. The mass of water, the final temperature, and the total enthalpy change are to be determined, and the Tis to be drawn. Analysis Initially the cylinder contains compressed liquid (since P > P sat@40°C) that can be approximated as a saturated liquid at the specified temperature (Table A-4),

v 1 # v f@40qC

0.001008 m 3 /kg

h 1 # h f@40qC

167.53 kJ/kg H2 O 40qC 200 kPa

(a) The mass is determined from

m

V1 v1

0.050 m 3 3

49.61 kg

0.001008 m /kg

(b) At the final state, the cylinder contains saturated vapor and thus the final temperature must be the saturation temperature at the final pressure,

T

Tsat@200 kPa

120.21qC

T

2 1

(c) The final enthalpy is h2 = hg @ 200 kPa = 2706.3 kJ/kg. Thus,

'H

m( h2  h1 ) (49.61kg)(2706.3 167.53)kJ/kg 125,950 kJ

v

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3-32 3-69 An automobile tire is under inflated with air. The amount of air that needs to be added to the tire to raise its pressure to the recommended value is to be determined. Assumptions 1 At specified conditions, air behaves as an ideal gas. 2 The volume of the tire remains constant. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis The initial and final absolute pressures in the tire are P1 = Pg1 + Patm = 135 + 100 = 235 kPa

Tire 0.015 m3 32q C 135 kPa (gage)

P2 = Pg2 + Patm = 225 + 100 = 325 kPa Treating air as an ideal gas, the initial mass in the tire is

m1

P1V

(235 kPa)(0.015 m 3 )

RT1

(0.287 kPa  m3 /kg  K)(305 K)

0.04027 kg

Noting that the temperature and the volume of the tire remain constant, the final mass in the tire becomes

m2

P2V RT2

(325 kPa)(0.015 m3 ) (0.287 kPa  m3 /kg  K)(305 K)

0.05569 kg

Thus the amount of air that needs to be added is

'm m2 m1

0.05569  0.04027 0.0154kg

3-70 Two rigid tanks connected by a valve to each other contain air at specified conditions. The volume of the second tank and the final equilibrium pressure when the valve is opened are to be determined. Assumptions At specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1). Analysis Let's call the first and the second tanks A and B. Treating air as an ideal gas, the volume of the second tank and the mass of air in the first tank are determined to be

VB

§ m1 RT1 · ¨¨ ¸¸ © P1 ¹

§PV · m A ¨¨ 1 ¸¸ © RT 1 ¹A

B

(3 kg)(0.287 kPa  m 3 /kg K)(308 K) 200 kPa

1.326 m3

3

(0.287 kPa  m /kg  K)(283 K)

4.309 kg

V A  VB

m mA  mB

V = 1 m3 T = 10 qC P = 35 0 kPa

Thus,

V

B

Air

3

(350 kPa)(1.0 m )

A

u

Air m = 3 kg T = 35qC P = 200 kPa

1.0 1.326 2.326 m3 4.309  3 7.309 kg

Then the final equilibrium pressure becomes

P2

mRT 2

V

(7.309 kg)(0.287 kPa m3 /kg K)(293 K) 2.326 m

3

264 kPa

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3-34 Compressibility Factor

3-74C All gases have the same compressibility factor Z at the same reduced temperature and pressure.

3-75C Reduced pressure is the pressure normalized with respect to the critical pressure; and reduced temperature is the temperature normalized with respect to the critical temperature.

3-76 The specific volume of steam is to be determined using the ideal gas relation, the compressibility chart, and the steam tables. The errors involved in the first two approaches are also to be determined. Properties The gas constant, the critical pressure, and the critical temperature of water are, from Table A-1, R = 0.4615 kPa·m3/kg·K,

Tcr = 647.1 K,

Pcr = 22.06 MPa

Analysis (a) From the ideal gas equation of state,

v

RT P

(0.4615 kPa m 3/kg K)(623.15 K) 15,000 kPa

0.01917 m 3/kg (67.0% error)

(b) From the compressibility chart (Fig. A-15),

PR TR

P Pcr T Tcr

½ 0.453 ° 22.06 MPa ° ¾ Z 673 K ° 1.04 °¿ 647.1 K 10 MPa

H2 O

0.65

15 MPa 350qC

Thus,

v

(0.65)(0.01917 m 3/kg)

Zv ideal

0.01246m3 /kg (8.5% error)

(c) From the superheated steam table (Table A-6),

`

P 15 MPa v T 350qC

0.01148 m3 /kg

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3-43 Other Equations of State

3-87C The constant a represents the increase in pressure as a result of intermolecular forces; the constant b represents the volume occupied by the molecules. They are determined from the requirement that the critical isotherm has an inflection point at the critical point.

3-88 The pressure of nitrogen in a tank at a specified state is to be determined using the ideal gas, van der Waals, and Beattie-Bridgeman equations. The error involved in each case is to be determined. Properties The gas constant, molar mass, critical pressure, and critical temperature of nitrogen are (Table A-1) R = 0.2968 kPa·m3/kg·K,

M = 28.013 kg/kmol,

Tcr = 126.2 K,

Pcr = 3.39 MPa

Analysis The specific volume of nitrogen is

3.27 m3 100 kg

V

v

m

0.0327 m 3/kg

N2 3

0.0327 m /kg 175 K

(a) From the ideal gas equation of state,

P

RT

(0.2968 kPa  m3 /kg  K)(175 K)

v

0.0327 m /kg

3

1588 kPa (5.5% error)

(b) The van der Waals constants for nitrogen are determined from

a

2 27R2 T cr 64 Pcr

b

RTcr 8Pcr

P

RT a  2 v b v

(27)(0.2968 kPa m3 / kg K)2 (126.2 K)2 (64)(3390 kPa)

6

2

0.175 m  kPa / kg

(0.2968 kPa m3 / kg K)(126.2 K) 3 0.00138 m / kg 8 u3390 kPa

Then,

0.2968u 175 0.175  0.0327 0.00138 (0.0327) 2

1495 kPa (0.7% error)

(c) The constants in the Beattie-Bridgeman equation are

A B c since v

0.02617· § § a· Ao ¨ 1 ¸ 136.2315¨1  ¸ 132.339 v 0.9160 ¹ © ¹ © §  0.00691· § b· Bo ¨ 1 ¸ 0.05046¨1 ¸ 0.05084 v 0.9160 ¹ © ¹ © 4

Mv P

3

3

4.2u10 m  K /kmol

(28.013 kg/kmol)(0.0327 m 3/kg) 0.9160 m 3/kmol . Substituting, A Ru T § c · ¨1  ¸v  B   2 v 2 © vT 3 ¹ v

132.339 4.2 u10 4 · 8.314 u175 § ¨1  ¸0.9160  0.05084  2 ¨ 0.91602 (0.9160) © 0.9160u 1753 ¸¹ 1504 kPa (0.07% error)

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3-57 3-107 A piston-cylinder device that is filled with water is cooled. The final pressure and volume of the water are to be determined. Analysis The initial specific volume is

v1

0.2825 m3 1 kg

V1 m

0.2825 m 3/kg

H2O 350°C 1 kg 0.2825 m3

This is a constant-pressure process. The initial state is determined to be superheated vapor and thus the pressure is determined to be

T1 350 qC

v1

½ ¾ P1 0.2825 m /kg ¿

P2

3

1.0 MPa (Table A - 6)

P

The saturation temperature at 1 MPa is 179.88°C. Since the final temperature is less than this temperature, the final state is compressed liquid. Using the incompressible liquid approximation,

2

1

0.001012 m3 /kg (Table A - 4)

v 2 v f @ 50qC

v

The final volume is then

V 2 mv 2

(1 kg)(0.001012 ft 3/lbm) 0.001012m3

3-108 Ethane is heated at constant pressure. The final temperature is to be determined using ideal gas equation and the compressibility charts. Properties The gas constant, the critical pressure, and the critical temperature of ethane are, from Table A-1, R = 0.2765 kPa·m3/kg·K,

Tcr = 305.5 K,

Pcr = 4.48 MPa

Analysis From the ideal gas equation,

T2

T1

v2 v1

(373 K)(1.6) 596.8K

From the compressibility chart at the initial state (Fig. A-15),

TR1 PR1

T1 T cr P1 Pcr

373 K 305.5 K

½ ° ° ¾ Z1 2.232 ° °¿

Ethane 10 MPa 100qC

1.221

10 MPa 4.48 MPa

0.61, v R1

0.35

Q

At the final state,

½ ¾ Z2 1.6(0.35) 0.56 ¿

2.232

PR 2

PR 1

v R2

1.6v R1

0.83

Thus,

T2

P2 v 2 Z2 R

P2 v R2T cr Z 2 Pcr

10,000kPa (0.56)(305.5 K) 4480 kPa 0.83

460 K

Of these two results, the accuracy of the second result is limited by the accuracy with which the charts may be read. Accepting the error associated with reading charts, the second temperature is the more accurate.

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