1

Atmospheric pressures at the top and at the bottom of the building are

kPamN

kPasmkg

Nmsmmkg

hgP

kPamN

kPasmkg

Nmsmmkg

hgP

bottombottom

toptop

70.100/1000

1/1

1)755.0)(/807.9)(/600,13(

)(

36.97/1000

1/1

1)730.0)(/807.9)(/600,13(

)(

2223

2223

=

⋅=

=

=

⋅=

=

ρ

ρ

Taking an air column between the top and the bottom of the building and writing a force balance per unit base area, we obtain

kPamN

kPasmkg

Nhsmmkg

PPgh

PPAW

topbottomair

topbottomair

)36.9770.100(/1000

1/1

1))(/807.9)(/18.1(

)(

/

2223 −=

⋅

−=

−=

ρ

It yields h = 288.6 m

which is also the height of the building.

2

Drawing the free body diagram of the piston and balancing the vertical forces yield PA P A W Fatm spring= + + Thus,

kPa4123./1000

11035

60)/807.9)(4()95( 224

2

=

×+

+=

++=

− mNkPa

mNsmkg

kPa

AFmg

PP springatm

The gage pressure is related to the vertical distance h between the two fluid levels by

P gh hP

ggagegage= =ρρ

(a) For mercury,

m600.1

/1000)/807.9)(/13600(

80 2

23=

⋅==

kPasmkg

smmkgkPa

gP

hHg

gage

ρ

(b) For water,

m8.16=

⋅==

kPasmkg

smmkgkPa

gP

hOH

gage

1/1000

)/807.9)(/1000(80 2

232

ρ

3

The buoyancy force acting on the balloon is

Nsmkg

Nmsmmkg

gVF

mmrV

balloonairB

balloon

5.5956/1

1)6.523)(/807.9)(/16.1(

6.5233/)5(43/4

2323

333

=

⋅=

=

===

ρ

ππ

The total mass is

kgmmm

kgmmkgVm

peopleHetotal

HeHe

8.2267028.86

8.86)6.523(/716.1 33

=×+=+=

=

== ρ

The total weight is

Nsmkg

NsmkggmW total 2.2224

/11

)/807.9)(8.226(2

2 =

⋅==

Thus the net force acting on the balloon is F F W Nnet B= − = − =5956 5 2224 2 3732 3. . .

Then the acceleration becomes

2sm516 /.1

/18.2262.3732 2

=

⋅==

Nsmkg

kgN

mF

atotal

net

4

The density of the oil is obtained by multiplying its specific gravity by the density of water which is given to be 1000 kg/m3,

ρ ρ ρ= = =( )( ) ( . )( / ) /s H O kg m kg m2

0 85 1000 8503 3

The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids,

[ ]kPa790.

/10001)5)(/807.9)(/1000()5)(/807.9)(/850(

)()(

22323

=

+=

+=∆+∆=∆

mNkPamsmmkgmsmmkg

ghghPPP wateroilwateroiltotal ρρ

5

Atmospheric pressure is acting on all surfaces of the petcock, which balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock (SFy = 0) yields

kg04080.1

/1000/807.9

)104)(100( 2

2

26

=

⋅×==

=−

kPasmkg

smmkPa

gAP

m

APW

gage

gage

6

7

8

9

10

11

Tablo A4, A5, A6 yardımı ile

T, °C P, kPa v, m3 / kg Phase description 50 12.352 4.16 Doymuş sıvı buhar karışımı

120.21 200 0.88578 Doymuş buhar 250 400 0.59520 Kızgın buhar 110 600 0.001052 Sıkıştırılmış sıvı

Su için aşağıdaki tabloyu tamamlayınız.

T, °C P, kPa h, kJ / kg x Phase description 325 0.4

160 1682 950 0.0

80 500 800 3161.7

Tablo A4, A5, A6 yardımı ile

T, °C P, kPa h, kJ / kg x Phase description 136.27 325 1435.35 0.4 Doymuş sıvı buhar karışımı

160 618.23 1682 0.483 Doymuş sıvı buhar karışımı 177.66 950 752.74 0.0 Doymuş sıvı

80 500 335.02 - - - Sıkıştırılmış sıvı 350 800 3162.2 - - - Kızgın buhar

12

R 134a için aşağıdaki tabloyu tamamlayınız. T, °C P, kPa h, kJ / kg x Faz açıklaması

240 81 4 0.27

-20 500 1400 362

20 1.0 Tablo A11, A12, A13 yardımı ile

T, °C P, kPa h, kJ / kg x Faz açıklaması -5.38 240 81 0.179 Doymuş sıvı buhar karışımı

4 337.90 110.03 0.27 Doymuş sıvı buhar karışımı -20 500 25.49 - - - Sıkıştırılmış sıvı 130 1400 363.02 - - - Kızgın buhar 20 572.07 261.59 1.0 Doymuş buhar

The pressure at the bottom of the 5-cm pan is the saturation pressure corresponding to the boiling temperature of 98°C: P P kPasat C= =@ .98 94 63o

The pressure difference between the bottoms of two pans is

kPasmkg

kPamsmmkghgP 43.3

/10001

)35.0)(/8.9)(/1000(2

23 =

⋅==∆ ρ

Then the pressure at the bottom of the 40-cm deep pan is

P = 94.63 + 3.43 = 98.06 kPa

Then the boiling temperature becomes

T Tboiling sat kPa= =@ . .98 06 99 0 Co

40 cm

5 cm

13

The specific volume of the water is

v = = =Vm

mkg

m kg0 510

0 053

3. . /

At -20°C, vf = 0.0007362 m3/kg and vg = 0.14729 m3/kg. Thus the tank contains saturated liquid-vapor mixture since vf < v < vg , and the pressure must be the saturation pressure at the specified temperature,

kPa132.8220@

==− Csat

PP o

(b) The quality of the refrigerant-134a and its total internal energy are determined from

kJ

vvv

9.903)/39.90)(10(/39.90)39.2584.218(336.039.25

336.00007362.014729.0

0007362.005.0

===

=−×+=+=

=−

−=

−=

kgkJkgmuUkgkJxuuu

x

fgf

fg

f

(c) The mass of the liquid phase and its volume are determined from

3m0.0048v

kg6.6

9)/0007362.0)(64.6(

410)336.01()1(3 ===

=×−=−=

kgmkgmV

mxm

fff

tf

(a) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the temperature in the tank must be the saturation temperature at the specified pressure,

T Tsat kPa= =@ 800 170.43 Co (Table A-5)

(b) The total mass in this case can easily be determined by adding the mass of each phase,

R-134a 10 kg -20°C

14

m V mm kg

kg

mV m

m kgkg

m m m

ff

f

gg

g

t f g

= = =

= = =

= + = + =

3v

v

010 001115

89 69

0 90 2404

374

89 69 374

3

3

3

.. /

.

.. /

.

. . 93.43 kg

(c) At the final state water is superheated vapor, and its specific volume is

kgmCTkPaP

/3544.0v350800 3

22

2 =

==

o (Table A-6)

Then, V m kg m kgt2 2

39343 0 3544= = =v ( . )( . / ) 33.1m3

Nitrogen at specified conditions can be treated as an ideal gas. Then the boundary work for this polytropic process can be determined from Eq. 3-12,

kJ89.0−=−

−⋅=

−−

=−−

== ∫

4.11)300360)(/2968.0)(2(

1)(

1122

1

1122

KKkgkJkgnTTmR

nVPVP

dVPWb

P

v

21

P

V

1

2

PVn =

15

(a) The term 10 2/ v must have pressure units since it is added to P. Thus the quantity 10 must have the unit kPa·m6/kmol2. (b) The boundary work for this process can be determined from

P R Tv v

R TV N V N

NR TV

NV

u u u= − = − = −10 10 10

2 2

2

2/ ( / )

and

kJ863.8=

⋅

−⋅+⋅=

−+=

−== ∫∫

333226

3

312

2

1

22

1 2

22

1

11

21

41)5.0)(/10(

24ln)300)(/314.8)(5.0(

1110ln10

mkPakJ

mmkmolkmolmkPa

mmKKkmolkJkmol

VVN

VVTNRdV

VN

VTNRdVPW uu

b

P

V

T = 300 K

2 4

16

We take the water in the piston-cylinder device as our system. This is a closed system since no mass enters or leaves. The conservation of energy equation for this case reduces to

)()()(

12

12

000

hhmWtVIhhmWW

PEKEUWWQ

b

be

bother

−=−∆−−−=−−

∆+∆+∆=−+

since P = constant and ∆U + Wb = ∆H for constant pressure quasi-equilibrium processes.

( )

kg75.4kg/m0010528.0

m005.0vV

m

kg/kJ36.15805.22265.011.467hxhh5.0xkPa150P

kg/m0010528.0vvkg/kJ11.467hh

liquid.satkPa150P

3

3

1

1

fg2f22

1

3kPa150@f1

kPa150@f11

===

=×+=+=

==

====

=

Substituting, v

P

1 2

150

17

V230.9s/kJ1

VA1000)s6045)(A8(

kJ4988V

kJ4988tVIkg/kJ)11.46736.1580)(kg75.4()kJ300(tVI

=

×

=

=∆−=−−∆

We take the water in the tank as our system. This is a closed system since no mass enters or leaves.

( )[ ]

( )

kgkJuuvaporsat

kgmvv

kgkJuxuu

kgmvxvv

kgkJuukgmvv

xkPaP

kgmg

fgf

fgf

fgf

gf

/7.2556.

/42428.0

/5.9397.208825.036.417

/42428.0001043.06940.125.0001043.0

/7.2088,36.417/6940.1,001043.0

25.0100

/42428.0@2

312

11

311

3

1

1

3 ====

=×+=+=

=−×+=+=

====

==

The conservation of energy equation for this case reduces to

)()()(

12

12

0000

uumtVIuumW

PEKEUWWQ

e

be

−=∆−−−=−

∆+∆+∆=−−

v

T

1

2

18

Substituting,

min153.2≅=∆

−=∆

stskJVA

kgkJkgtAV

9189/1

1000/)5.9397.2556)(5()8)(110(

(a) We take the steam in the piston-cylinder device as our system. This is a closed system since no mass enters or leaves.

kgkJuT

kgmv

kPaP

kgmkgm

mV

v

kgkgm

mvV

m

kgkJukgmv

CT

kPaP

/9.4321/296.1

500

/296.1463.06.0

463.0/0803.1

5.0

/4.2654/0803.1

200

200

2

23

2

2

33

22

3

3

1

1

1

31

1

1

==

=

=

===

===

==

=

=

C1131 o

o

P

v

2

1

19

(b) The pressure of the gas changes linearly with the gas, and thus the process curve on a P-V diagram will be a straight line. The boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus,

( ) kJ35=

⋅−

+=−

+== 3

312

21

11

)5.06.0(2

)500200(2 mkPa

kJm

kPaVV

PPAreaWb

(c) The conservation of energy equation for this case reduces to

b

b

WuumQPEKEUWQ

+−=∆+∆+∆=−

)( 12

00

Substituting, Q = (0.463 kg)(4321.9 - 2654.4)kJ/kg + 35 kJ = 807 kJ

At specified conditions air can be treated as an ideal gas. We take the air in the room as our system. Assuming the room is well-sealed and no air leaks out, we have a constant volume closed system. The conservation of energy equation for this case reduces to

Q W U KE PE

W m u u mC T Te

e v ave

0 0 0

2 1 2 1

− = + +− = − = −

∆ ∆ ∆( ) ( ),

or, − = −& ( ),W t mC T Te v ave∆ 2 1 The mass of air is

V m

m PVRT

kPa mkPa m kg K K

kg

= × × =

= =⋅ ⋅

=

4 5 6 120

100 1200 287 280

149 3

3

1

1

3

3( )( )

( . / )( ).

Using Cv value at room temperature from Table A-2a , − × = ⋅ − = −& ( ) ( . )( . / )( )W s kg kJ kg C Ce 15 60 149 3 0 718 23 7o o 1.91 kW The negative sign indicates electrical work is done on the system.

4×5×6 m3 7°C

AIR We

20

At specified conditions air can be treated as an ideal gas. We take the air in the room as our system. Assuming the room is well-sealed and no air leaks out, we have a constant volume closed system. The conservation of energy equation for this case reduces to

Q W W U KE PE

W m u u mC T Te b

e v ave

0 0 0 0

2 1 2 1

− − = + +− = − = −

∆ ∆ ∆( ) ( ),

The mass of air is

V m

m PVRT

kPa mkPa m kg K K

kg

= × × =

= =⋅ ⋅

=

4 6 6 144

100 1440 287 288

174 2

3

1

1

3

3( )( )

( . / )( ).

The electrical work done by the fan is W W t kJ s s kJe e= = × =& ( . / )( )∆ 015 10 3600 5400 Substituting and using Cv value at room temperature from Table A-2a, - (-5400 kJ) = (174.2 kg)(0.718 kJ/kg°C)(T2 - 15)°C T2 = 58.2°C

ROOM

4m × 6m × 6m

Fan

21

We take the entire tank as our system. This is a constant volume closed system since no mass enters or leaves the system. The conservation of energy equation for this ideal gas reduces to

Q W U KE PE

m u u mC T Tv ave

0 0 0 0

2 1 2 10− = + +

= − = −∆ ∆ ∆

( ) ( ),

Thus, T2 = T1 = 50°C and

PVT

PVT

P VVP kPa1 1

1

2 1

22

2

11

12

800= → = = =( ) 400 kPa

At specified conditions air can be treated as an ideal gas. We take the air in the piston-cylinder device as our system. This is a closed system since no mass enters or leaves. The conservation of energy equation for this case reduces to

Q W W U KE PEQ W m h he b

e

− − = + +− = −

∆ ∆ ∆0 0

2 1( )

or W m h h Qe = − − +( )2 1 since ∆U + Wb = ∆H for constant pressure quasi-equilibrium expansion or compression

IDEAL GAS

800 kPa

50°C

Evacuated

22

processes. From Table A-18,

h h kJ kg

h h kJ kgK

K

1 298

2 350

29818

350 49

= =

= =@

@

. /

. /

Substituting, We = -(15 kg)(350.49 - 298.18)kJ/kg + (-60 kJ) = -844.7 kJ or,

kWh0.235−=

−=

kJ3600kWh1

)kJ7.844(We

(a) The pressure of the gas changes linearly with volume during this process, and thus the process curve on a P-V diagram will be a straight line. Then the boundary work during this process is simply the area under the process curve, which is a trapezoidal. Thus,

( )

( ) ( )

kJ150=

⋅−

+=

−+

==

33

1221

b

mkPa1kJ1

m2.05.02

kPa800200

VV2

PPAreaW

(b) If there were no spring, we would have a constant pressure process at P = 200 kPa. The work done during this process is

( ) ( )( ) kJ60mkPa1

kJ1kg/m2.05.0kPa200VVPdVPW 3

312

2

1springno,b =

⋅−=−== ∫

W W Wspring b b no spring= − = − =, 150 60 90 kJ

V (m3

P (kPa)

200 1

2800

0.2 0.5

23

Helium at specified conditions can be treated as an ideal gas. The mass of helium and the exponent n are determined from

( )( )( )( )

536.1n264.0

5.0150400

VV

PP

VPVP

m264.0m5.0kPa400kPa150

K293K413V

PTPT

VRT

VPRT

VP

kg123.0K293Kkg/mkPa0769.2

m5.0kPa150RT

VPm

nn

2

1

1

2n11

n22

331

21

122

2

22

1

11

3

3

1

11

=→

=→

=

→=

=××==→=

=⋅⋅

==

Then the boundary work for this polytropic process can be determined from

( )

( )( )kJ57.2−=

−−⋅

=

−−

=−−

== ∫

536.11K)293413(Kkg/kJ0769.2kg123.0

n1TTmR

n1VPVP

dVPW 1211222

1b

We take the helium in the piston-cylinder device as our system. This is a closed system since no mass enters or leaves. The conservation of energy equation for this closed system reduces to

b

b

WuumQPEKEUWQ

+−=∆+∆+∆=−

)( 12

00

or , Q = mCv(T2 - T1) + Wb Using Cv value from Table A-2a, Q = (0.123 kg)(3.1156 kJ/kg·K)(413 - 293)K + (-57.2 kJ) = -11.2 kJ