ENGR 2345
THERMODYNAMICS
FINAL EXAM
DUE: 9 DEC 2014
Work two (2) problems from each chapter using the following list:
2.111, 2.118, and 2-125
3-114, 3-117, and 3-127
4-125, 4-139, and 4-148
5-150, 5-170, and 5-177
2.11) A hollow spherical iron container is filled with iced water at 0°C. The rate of heat loss from the sphere
and the rate at which ice melts in the container are to be determined.
Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant
at the specified values. 2 Heat transfer through the shell is one-dimensional. 3 Thermal properties of the
iron shell are constant. 4 The inner surface of the shell is at the same temperature as the iced water, 0°C.
Properties The thermal conductivity of iron is k = 80.2 W/mC (Table 2-3). The heat of fusion of water is at 1
atm is 333.7 kJ/kg.
5C
Q
T
 ice 

Find the surface area: A = D2 then Q cond  kA
and find m
L
hif
Iced
0.4 cm
water
2-118 The heating and cooling costs of a poorly insulated house can be reduced by up to 30 percent by
adding adequate insulation. The time it will take for the added insulation to pay for itself from the energy it
saves is to be determined.
Assumptions It is given that the annual energy usage of a house is $1200 a year,
and 46% of it is used for heating and cooling. The cost of added insulation is given to
be $200.
Heat loss
Analysis The amount of money that would be saved per year is determined
directly from Money saved and then find the simple Payback Period.
House
2-125E The power required to pump a specified rate of water to a specified elevation is to be determined.
Properties The density of water is taken to be 62.4 lbm/ft3 (Table A-3E).
Analysis The required power is determined from
 g ( z2  z1 )  Vg ( z2  z1 )
W  m
ENGR 2345
THERMODYNAMICS
FINAL EXAM
DUE: 9 DEC 2014
3-114 Carbon dioxide flows through a pipe at a given state. The volume and mass flow rates and the density
of CO2 at the given state and the volume flow rate at the exit of the pipe are to be determined.
Analysis
3 MPa
CO2
450 K
500 K
(a) The volume and mass flow rates may be determined from ideal gas relation as
V1 
N Ru T1
P
1 
then find m
P1V1
m
and then the density 1  1
V1
RT1
(b) The volume flow rate at the exit is V2 
N RuT2
P
3-117 A rigid container that is filled with R-13a is heated. The initial pressure and the final temperature are
to be determined.
Analysis The initial specific volume is 0.090 m3/kg. Using this
with the initial temperature reveals that the initial state is a
mixture. The initial pressure is then the saturation pressure,
R-134a
-40°C
T1  40C

 P  Psat @ -40C
v 1  0.090 m /kg  1
1 kg
3
P
This is a constant volume cooling process (v = V /m = constant).
The final state is superheated vapor and the final temperature is
then
2
P2  280 kPa
1

T 
3
v 2  v 1  0.090 m /kg  2
v
3-127 The pressure in an automobile tire increases during a trip while its volume remains constant. The
percent increase in the absolute temperature of the air in the tire is to be determined.
Assumptions 1 The volume of the tire remains constant. 2 Air is an ideal gas.
Properties The local atmospheric pressure is 90 kPa.
Analysis The absolute pressures in the tire before and after
PV
PV
P1 = Pgage,1 + Patm, P2 = Pgage,2 + Patm, 1 1  2 2
T1
T2
TIRE
200 kPa
ENGR 2345
THERMODYNAMICS
FINAL EXAM
DUE: 9 DEC 2014
4-125 A cylinder equipped with an external spring is initially filled with air at a specified state. Heat is
transferred to the air, and both the temperature and pressure rise. The total boundary work done by the air,
and the amount of work done against the spring are to be determined, and the process is to be shown on a
P-v diagram.
Assumptions 1 The process is quasi-equilibrium. 2 The spring is a linear spring.
Analysis
(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,
Wb,out  Area 
P1  P2
V 2 V1 
2
(b) If there were no spring, we would have a constant pressure
process at P = 100 kPa. The work done during this process is
Wb,out,no spring   PdV  PV2 V1  , Wspring  Wb  Wb,no spring
2
Q
Air
100 kPa
1
4-139 A man drinks one liter of cold water at 3°C in an effort to cool down. The drop in the average body
temperature of the person under the influence of this cold water is to be determined.
Assumptions 1 Thermal properties of the body and water are constant. 2 The effect of metabolic heat
generation and the heat loss from the body during that time period are negligible.
Properties The density of water is very nearly 1 kg/L, and the specific heat of water at room temperature is
c = 4.18 kJ/kg·°C (Table A-3). The average specific heat of human body is given to be 3.6 kJ/kg.°C.
Analysis. The mass of the water is mw  V
We take the man and the water as our system, and disregard any heat and mass transfer and chemical
reactions. Of course these assumptions may be acceptable only for very short time periods, such as the time
it takes to drink the water. Then the energy balance can be written as
E  Eout
in



Net energy transfer
by heat, work, and mass
Esystem



Change in internal,kinetic,
potential,etc. energies
0  U  U body  U water
or
mcT2  T1 body  mcT2  T1 water  0
4-148 A piston-cylinder device contains an ideal gas. An external shaft connected to the piston exerts a
force. For an isothermal process of the ideal gas, the amount of heat transfer, the final pressure, and the
distance that the piston is displaced are to be determined.
Assumptions 1 The kinetic and potential energy changes are
negligible, ke  pe  0 . 2 The friction between the piston
W
and the cylinder is negligible.
Analysis (a) We take the ideal gas in the cylinder to be the
system. This is a closed system since no mass crosses the
system boundary. The energy balance for this stationary
closed system can be expressed as
GAS
Q
1 bar
ENGR 2345
THERMODYNAMICS
E E
inout


Net energy transfer
by heat, work, and mass
FINAL EXAM
DUE: 9 DEC 2014
E system



Change in internal,kinetic,
potential,etc. energies
W b,in  Qout  U ideal gas  mc v (T2  T1 ) ideal gas )  0 (since T2  T1 and KE  PE  0)
W b,in  Qout
(b) The relation for the isothermal work of an ideal gas may be used to determine the final volume in the
cylinder. But we first calculate initial volume
V1 
V 
L1 , Wb,in   P1V1 ln  2  , P1V1  P2V2
4
 V1 
D 2
(c) The final position of the piston and the distance that the piston is displaced are
V2 
D 2
4
L2 , find L
5-150 Steam expands in a turbine whose power production is 9000 kW. The rate of heat lost from the
turbine 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.
Properties From the steam tables (Tables A-6 and A-4)
P1  1.6 MPa 
 h1 ,
T1  350C 
T2  30C 
 h2 
x2  1

Analysis We take the turbine as the system, which is a control volume since
mass crosses the boundary. Noting that there is one inlet and one exiti the
energy balance for this steady-flow system can be expressed in the rate
form as
E  E out
in


Rate of net energy transfer
by heat, work, and mass
E system0 (steady)



1.6 MPa
Heat
350°C
Turbine
30°C
sat. vap.
0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
m 1 h1  m 2 h2  W out  Q out
Find Q out
5-170E Steam is mixed with water steadily in an adiabatic device. The temperature of the water leaving this
device 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 and heat interactions. 4 There is no heat transfer
between the mixing device and the surroundings.
Analysis We take the mixing device as the system, which is a control volume. The energy balance for this
steady-flow system can be expressed in the rate form as
ENGR 2345
THERMODYNAMICS
E  E
inout

E system0 (steady)



Rate of net energy transfer
by heat, work, and mass
FINAL EXAM
DUE: 9 DEC 2014
0
Steam
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
m 1 h1  m 2 h2  m 3 h3
60 psia
From a mass balance
350F
1
Water
2
3
m 3  m 1  m 2
The enthalpies of steam and water are (Table A-6E and A-4E)
P1  60 psia 
P2  60 psia 
 h1 
 h2  h f @ 40 F
T1  350F 
T2  40F

60
60 psia
40F
Substituting into the energy balance equation solving for the exit enthalpy gives h3
At the exit state P3=60 psia and h3=65.19 kJ/kg. An investigation of Table A-5E reveals that this is
compressed liquid state. Approximating this state as saturated liquid at the specified temperature, the
temperature may be determined from Table A-4E to be
P3  60 psia

 T3  T f @ h65.19 Btu/lbm
h3  65.19 Btu/lbm 
5-177 A fan is powered by a 0.5 hp motor, and delivers air at a rate of 85 m3/min. The highest possible air
velocity at the fan exit is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus
mCV  0 and ECV  0 . 2 The inlet velocity and the change in potential energy are negligible,
V1  0 and pe  0 . 3 There are no heat and work interactions other than the electrical power consumed by
the fan motor. 4 The efficiencies of the motor and the fan are 100% since best possible operation is
assumed. 5 Air is an ideal gas with constant specific heats at room temperature.
Properties The density of air is given to be  = 1.18 kg/m3. The constant pressure specific heat of air at room
temperature is cp = 1.005 kJ/kg.C (Table A-2).
Analysis We take the fan-motor assembly as the system. This is a control volume since mass crosses the
system boundary during the process. We note that there is only one inlet and one exit, and thus m 1  m 2  m
The velocity of air leaving the fan will be highest when all of the entire electrical energy drawn by the motor
is converted to kinetic energy, and the friction between the air layers is zero. In this best possible case, no
energy will be converted to thermal energy, and thus the temperature change of air will be zero, T2  T1 .
Then the energy balance for this steady-flow system can be expressed in the rate form as
E  E out
in


E system0 (steady)


Rate of net energy transfer
by heat, work, and mass
0
Rate of change in internal,kinetic,
potential,etc. energies
E in  E out
We,in  m h1  m (h2 + V22 /2) (since V1  0 and pe  0)
Noting that the temperature and thus enthalpy remains constant, the
relation above simplifies further to
W e,in  m V22 /2 where
m  V solve for V2
0.5 hp
85 m3/min