Homework 11 Solutions

#1 2-75 Water is pumped from a lake to a storage tank at a specified rate. The overallefficiency of the pump-motor unit and the pressure difference between the inlet and theexit of the pump are to be determined.Assumptions 1 The elevations of the tank and the lake remain constant. 2 Frictionallosses in the pipes are negligible. 3 The changes in kinetic energy are negligible. 4 Theelevation difference across the pump is negligible.Properties We take the density of water to be r = 1000 kg/m3.Analysis (a) We take the free surface of the lake to bepoint 1 and the free surfaces of the storage tank to bepoint 2. We also take the lake surface as the referencelevel (z1 = 0), and thus the potential energy at points 1and 2 are pe1 = 0 and pe2 = gz2. The flow energy at bothpoints is zero since both 1 and 2 are open to theatmosphere (P1 = P2 = Patm). Further, the kinetic energyat both points is zero (ke1 = ke2 = 0) since the water atboth locations is essentially stationary. The mass flowrate of water and its potential energy at point 2 are

kg/s70/s)m 070.0)( kg/m1000( 33 === V&& rm

kJ/kg196.0/sm 1000

kJ/kg1m) 20)(m/s (9.81

222

22 =˜¯

ˆÁË

Ê== gzpe

Then the rate of increase of the mechanical energy of water becomes

kW 13.7kJ/kg) 6kg/s)(0.19 70()0()( 22inmech,outmech,fluidmech, ===-=-=D pempemeemE &&&&

The overall efficiency of the combined pump-motor unit is determined from its definition,

67.2%or 0.672 kW20.4

kW7.13

inelect,

fluidmech,motor-pump ==

D=

W

E&

&h

(b) Now we consider the pump. The change in the mechanical energy of water as it flows through the pumpconsists of the change in the flow energy only since the elevation difference across the pump and thechange in the kinetic energy are negligible. Also, this change must be equal to the useful mechanical energysupplied by the pump, which is 13.7 kW:

PPP

meemE D=-

=-=D V&&&&r

12inmech,outmech,fluidmech, )(

Solving for DP and substituting,

kPa 196=˜̃¯

ˆÁÁË

Ê ⋅=

D=D

kJ1

m kPa1

/sm 0.070

kJ/s13.7 3

3fluidmech,

V&&E

P

Therefore, the pump must boost the pressure of water by 196 kPa in order to raise its elevation by 20 m.

Discussion Note that only two-thirds of the electric energy consumed by the pump-motoris converted to the mechanical energy of water; the remaining one-third is wastedbecause of the inefficiencies of the pump and the motor.

20m

Pump

Storagetank

2

1

#2 3-36 A piston-cylinder device that is filled with R-134a is heated. The final volume isto be determined.Analysis This is a constant pressure process. The initial specific volume is

/kgm 1595.0kg 10

m 595.1 33

1 ===m

Vv

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

12)-A (Table kPa 100C26.4- @sat 1 == °PP

The final state is superheated vapor and the specific volumeis

13)-A (Table/kgm 30138.0 C100

kPa 100 32

2

2 =˛˝¸

°=

=v

T

P

The final volume is then3m 3.0138=== /kg)m 30138.0)(kg 10( 3

22 vV m

#3 4-74 Argon is compressed in a polytropic process. The work done and the heattransfer are to be determined.Assumptions 1 Argon is an ideal gas since it is at a high temperature and low pressurerelative to its critical point values of 151 K and 4.86 MPa. 2 The kinetic and potentialenergy changes are negligible, 0peke @D@D .Properties The properties of argon are R = 0.2081kJ/kg⋅K and cv = 0.3122 kJ/kg⋅K(Table A-2a).Analysis We take argon as the system. This is a closedsystem since no mass crosses the boundaries of the system.The energy balance for this system can be expressed as

)( 12out,in

energies etc. potential, kinetic, internal,in Change

system

mass and work,heat,by nsferenergy traNet

outin

TTmcUWQ

EEE

b -=D=-

D=-

v

4342143421

Using the boundary work relation for the polytropicprocess of an ideal gas gives

kJ/kg 147.51120

1200

1.2-1

K) 303)(KkJ/kg 2081.0(1

1

2.1/2.0/)1(

1

21out, -=

˙˙˚

˘

ÍÍÎ

È-˜

¯

ˆÁË

Ê⋅=

˙˙

˚

˘

ÍÍ

Î

È-˜̃

¯

ˆÁÁË

Ê

-=

- nn

b P

P

n

RTw

Thus,kJ/kg 147.5=in,bw

The temperature at the final state is

K 444.7kPa 120

kPa 1200K) 303(

2.1/2.01(

1

212 =˜

¯

ˆÁË

Ê=˜̃

¯

ˆÁÁË

Ê=

)/nn-

P

PTT

R-134a-26.4°C10 kg

1.595 m3

P

v

21

QArgon

120 kPa30°CPv n =

constant

From the energy balance equation,kJ/kg 103.3K)303K)(444.7kJ/kg 3122.0(kJ/kg 5.147)( 12out,in -=-⋅+-=-+= TTcwq b v

Thus,kJ/kg 103.3=outq

#4 4-60 The enthalpy changes for neon and argon are to be determined for a giventemperature change.Assumptions At specified conditions, neon and argon behave as an ideal gas.Properties The constant-pressure specific heats of argon and neon are 0.5203 kJ/kg⋅K and1.0299 kJ/kg⋅K, respectively (Table A-2a).Analysis The enthalpy changes are

kJ/kg 156.1=-⋅=D=D K)100K)(400kJ/kg (0.5203argon Tch p

kJ/kg 309.0=-⋅=D=D K)100K)(400kJ/kg (1.0299neon Tch p

#5 5-119 R-134a is condensed in a condenser. The heat transfer per unit mass is to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kineticand potential energy changes are negligible. 3 There are no work interactions.Analysis We take the pipe in which R-134a is condensed as the system, which is acontrol volume. The energy balance for this steady-flow system can be expressed in therate form as

21out

21out

out21

outin

energies etc. potential, kinetic, internal,in change of Rate

(steady) 0system

mass and work,heat,by nsferenergy tranet of Rate

outin

)(

0

hhq

hhmQ

Qhmhm

EE

EEE

-=

-=

+=

=

=D=-

&&

&&&

&&

444 344 21&

43421&& ä

The enthalpies of R-134a at the inlet and exit of the condenser are (Table A-12, A-13).

kJ/kg 77.1170

kPa 1200

kJ/kg 39.311C80

kPa 1200

kPa 1200@22

11

1

=@˛˝¸

=

=

=˛˝¸

°=

=

fhhx

P

hT

P

Substituting,kJ/kg 193.6=-= 77.11739.311outq

#6 5-123 An insulated rigid tank is evacuated. A valve is opened, and air is allowed to fillthe tank until mechanical equilibrium is established. The final temperature in the tank isto be determined.

1200kPa

sat. liq.

R134a

1200 kPa80°C

qout

Assumptions 1 This is an unsteady process since the conditions within the device arechanging during the process, but it can be analyzed as a uniform-flow process since thestate of fluid at the inlet remains constant. 2 Air is an ideal gas with constant specificheats. 3 Kinetic and potential energies are negligible. 4 There are no work interactionsinvolved. 5 The device is adiabatic and thus heat transfer is negligible.Properties The specific heat ratio for air at room temperature is k = 1.4 (Table A-2).Analysis We take the tank as the system, which is a control volumesince mass crosses the boundary. Noting that the microscopic energiesof flowing and nonflowing fluids are represented by enthalpy h andinternal energy u, respectively, the mass and energy balances for thisuniform-flow system can be expressed as

Mass balance:

)0 (since initialout2systemoutin ===ÆD=- mmmmmmm i

Energy balance:

)0 (since initialout22

energies etc. potential, kinetic, internal,in Change

system

mass and work,heat,by nsferenergy traNet

outin

@@==@@=

D=-

pekeEEWQumhm

EEE

ii

4342143421

Combining the two balances:iipipi kTTccTTcTchu ==Æ=Æ= )/(222 vv

Substituting, T2 = ¥ = =1.4 290 K 406 K 133 Co

initiallyevacuated

Air