Chapter 4 The First Law of Thermodynamics

4-32 One part of an insulated rigid tank contains an ideal gas while the other side is evacuated. The finaltemperature and pressure in the tank are to be determined when the partition is removed.Assumptions 1 The kinetic and potential energy changes are negligible. Me = Ape = O .2 The tank is

insulated and thus heat transfer is negligible.

Analysis We take the entire tank as the system. This is a closed system since no mass crosses the boundaries

of the system. The energy balance for this system can be expressed as

E -E = t!JE tin out sys em'---v---' '---v---'

Net energy transfer Change in internal, kinetic,by heat, work, and mass potential, etc energies

O=~U=m(~-ul)

~=uI

Therefore,

i

ifr

i.t

}ti

~

4-22

Chapter 4 The First Law of Thermodynamics

4-36 A cylinder is initially filled with air at a specified state. Air is heated electrically at constant pressure,and some heat is lost in the process. The amount of electrical energy supplied is to be determined. "'

Assumptions 1 The cylinder is stationary and thus the kinetic and potential energy changes are zero. 2 Air isan ideal gas with variable specific heats. 3 The thermal energy stored in the cylinder itself and theresistance wires is negligible. 4 The compression or expansion process is quasi-equilibrium.

Properties The initial and final enthalpies of air are (Table A-17)

h) = h@298K = 298.18 kJ /kg

hz = h@ 350 K = 350.49 kJ / kg

Analysis We take the contents of the cylinder as the system. This is a closed system since no mass enters or

leaves. The energy balance for this closed system can be expressed as

since I1U + Wb = I1H during a constant pressure quasi-equilibrium process.

Substituting,We.in = (15 kg)(350.49 -298.18)kJ/kg + (60 kJ) = 845 kJ

Of,

1 kWh= 0.235 kWhWe in = (845kJ) (.3600 kJ

Alternative solution The specific heat of air at the average temperature of T aye = (25+ 77)/251°C = 324 K is, from Table A-2b, Cp.aye = 1.0065 kJ/kg.oC. Substituting,

We,in= mC p(T2 -Ti )+QOUI

= (15 kg)(1.0065 kJ/kg.oC)(77 -25)OC + 60 kJ = 845 kJ

4-26

Discussion Note that for small temperature differences, both approaches give the same result.

Chapter 4 The First Law of Thermodynamics

I2

I:=::aAIR

1

4-87 Air is compressed by a compressor. The mass flow rate of air through the compressor is 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 Air is an ideal gas with variable specific heats.

Properties The inlet and exit enthalpies of air are (Table A-17)

Ti = 25°C = 298 K ~ h1 = h@ 298 K = 298.2 kJ/kg

T2 = 347°C = 620 K ~ h2 = h@ 620K = 628.07 kJ/kg

Analysis We take the compressor as the system, which is a control volume since

mass crosses the boundary. The energy balance for this steady-flow system can

\Je expressed in the rate form as

...710 (steady)E;n -Eou, = L\Esystem = 0' ' ' v'

Rate of net energy transfer Rate of change in internal, kinetic,by heat, work, and mass potential. etc. energies

Ein = Eou,

Win + ni(h1 + V( 12) = Qout + ni(h2 + V i 12) (since tJ.pe = 0)

... ( vi -V(Win -QOUI =m h2 -hl + 2

Substituting, the mass flow rate is determined to be

250 kJ/s -(1500/60 kJ/s) = i 62807- 298.2 +~0m/s)2 -0 ( IkJ/kg

m = 0.674 kg/s

2

4-62

~

1 ,500 kJlmin

Chapter 4 The First Law of Thermodynamics

4-246 An insulated cylinder is divided into two parts. One side of the cylinder contains N2 gas and the otherside contains He gas at different states. The final equilibrium temperature in the cylinder when thermalequilibrium is established is to be determined for the cases of the piston being fixed and moving freely.

Assumptions 1 Both N2 and He are ideal gases with constant specific heats. 2 The energy stored in thecontainer itself is negligible. 3 The cylinder is well-insulated and thus heat transfer is negligible.

Properties The gas constants and the constant volume specific heats are R = 0.2968 kPa.m3/kg.K is Cv =0.743 kJ/kg. °C for N2, and R = 2.0769 kPa.m3/kg.K is Cv = 3.1156 kJ/kg. °C for He (Tables A-I and

A-2)

= 4.77 kgmN2 = (

R~ ?53 K )

Analysi" The mass of each gas in the cylinder is

/~V11 -(500 kPaXl m3-) N2 = (0.2968kPa .m3/k; .K,:

, -(500 kPa)(I m3

He = (2.0769 kPa .m3/kg .K~ 298 K)

Taking the entire contents of the cylinder as our system, the 1 st law relation can be written as

Ein -Eou, Msystemc , '--v--'Net energy transfer Change in internal, kinetic,

by heat, work, and mass potential, etc energies

O=tJ.U=(tJ.U)Nl +(tJ.U)He

O=[mCv(T2 -T1)]Nl +[mCv(T2 -T])]He

~

R~

= 0.808 kgmHe =

Substituting,

(4. 77kgXo. 743kJ/kg.o C Xr f -80 1 c + (0.808kg){1.1156kJ/kgoO C Xr f -251 c = 0

It gives Tr= 57.2°C

where T f is the final equilibrium temperature in the cylinder .

The answer would be the same if the piston were not free to move since it would effect only

pressure, and not the specific heats.

Discussion Using the relation PV = NRu T , it can be shown that the total number of moles in the cylinder is0.170 + 0.202 = 0.372 krnol, and the final pressure is 510.6 kPa.

4-187

Chapter 6 Entropy

6.122 An egg is dropped into boiling water. The amount of heat transfer to the egg by the time it is cookedand the amount of entropy generation associated with this heat transfer process are to be determined.

Assumptions 1 The egg is spherical in shape with a radius of ro = 2.75 cm. 2 The thermal properties of theegg are constant. 3 Energy absorption or release associated with any chemical and/or phase changes withinthe egg is negligible. 4 There are no changes in kinetic and potential energies.

Properties The density and specific heat of the egg are given to be p = 1020 kg/m3 and Cp = 3.32 kJ/kg.oC.

Analysis We take the egg as the system. This is a closes system since no mass enters or leaves the egg. Theenergy balance for this closed system can be expressed as

E. -E = M Boiling

~ -Oll~ .syslern, Watery Net energy transfer Change in internal, kinetic,

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

Q .= /j.U = m(u 2 -u ] ) = mC(T2 -7j )In ~gg

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

m = pV = p~= (1020kg/m3) ,,(0.055 m)3 = 0.0889 kg

6 6

Qin = mC p (T 2 -TI) = (0.0889 kg)(3.32 kJ/kg.oC)(70 -8)OC = 18.3 kJ

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°C at all times:

S,n -SOUl + Sgen = ~system'--v--"-,.-'

Net entropy transfer Entropyby heat and mass generation

' '

Changein entropy

Q;n+ s = !:J.ssystem

-gen

Tb

s Q;n A "-.gen = --+ => system

Th

where

Substituting,

Chapter 6 Entropy

6-127 A 1000- W iron is left on the iron board with its base exposed to the air at 20°C. The rate of entropy

generation is to be determined in steady operation.

Assumptions Steady operating conditions exist.

Analysis We take the iron to be the system, which is a closed system. Considering that the iron experiencesno change in its properties in steady operation, including its entropy, the rate form of the entropy balance

for the iron simplifies to Ir on

l000W<POSin -SOUl + S gen: = !!JSsystem = 0 ~ -..

'--v---' ' "---v---'Rate of net entropy transfer Rate of entropy Rate of change -..

by heat and mass generation of entropy -..

-..

=0Qout +Sgen.iron

Tb,out

Therefore,

~

Tb.ou,

l000W

673K= 1.486 W/K=s gen,iron =

The rate of total entropy generation during this process is determined by applying the entropy balance on anextended system that includes the iron and its irnrnediate surroundings so that the boundary temperature of

the extended system is at 20°C at all times. It givesQ" Q 100 OW

/K-!!!!!- = -= --= 3.413 Ws gen.totaJ =

TSU" 293KTb.ou,

Discussion Note that only about one-third of the entropy generationoccurs within the iron. The rest occurs in the air surrounding the iron as

the temperature drops from 400°C to 20°C without serving any useful

purpose.

6-67

Chapter 6 Entropy

6-130 Steam expands in a turbine from a specified state to another specified state. The rate of entropygeneration during this process is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible.

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

PI = 8MPa 1 hl = 3272.0kJ/kg

Ti = 450°C f SI = 6.5551kJ/kg .K

P2 = 50kPa } h2 = 2645.9kJ/kg

sat.vapor S2 = 7.5939kJ/kg .K

Analysis There is only one inlet and one exit, and thus 11tt = ~ = 111 .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 PI = 8 MPa

E. -E M 710 (steady) = O Ti = 450°CIn out system

'---v ' '---~-Rate of net energy transfer Rate of change in internal, kinetic,by heat. work. and mass potential. etc. energies

Ein = Eout

mhl = Qout + Wout +m~

Qout = m(hl -~)- Wout

~

~

4MW

Substituting, P2 = 50 kPa

sat. vaporQOUI = (25,000/3600 kg/s)(3272 -2645.9) kJ /kg -4000 kJ /s = 347.9 kJ /s

+ Sgen' '-.-'

Rate of net entropy transfer Rate of entropyby heat and mass generation

The rate of total entropy generation during this process is determined by applying the entropy

balance on an extended system that includes the turbine and its irnrnediate surroundings so that the boundary

temperature of the extended system is 25°C at all times. It gives

...<10Sin -Saul =Msystem =0~- ' "

Rate of changeof entropy

..QOUIs "

0ms l -ms 2 --+ =

gen

T b,surr

Substituting, the rate of entropy generation during this process is determined to be

.. ( ) QOUI ( X "\1, 347.9kW

Sgen = m $2- $1 + ~= 25,OOO/3600kg/s 7.5939- 6.5551}l"J/kg .K + = 8.38kW/K

T b,surr 298K

,'1

\

-""'

6-70