6-101
6-136E Saturated steam at a saturation pressure of 0.95 psia and thus at a saturation temperature of Tsat = 100F (Table A–
4E) condenses on the outer surfaces of 144 horizontal tubes by circulating cooling water arranged in a 12 12 square array.
The rate of heat transfer to the cooling water and the average velocity of the cooling water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The tubes are
isothermal. 3 Water is an incompressible substance with constant
properties at room temperature. 4 The changes in kinetic and potential
energies are negligible.
Analysis (a) The rate of heat transfer from the steam to the cooling
water is equal to the heat of vaporization released as the vapor
(b) All of this energy is transferred to the cold water. Therefore, the
Saturated
steam
0.95 psia
Cooling
water
6-103
6-138 Cold water enters a steam generator at 20C, and leaves as saturated vapor at Tsat = 200C. The fraction of heat used
to preheat the liquid water from 20C to saturation temperature of 200C is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Heat losses from the steam generator are negligible. 3 The specific heat
Analysis The heat of vaporization of water represents the
amount of heat needed to vaporize a unit mass of liquid at
a specified temperature. Using the average specific heat,
the amount of heat transfer needed to preheat a unit mass
of water from 20C to 200C is
kJ/kg 752.4=
C20)C)(200kJ/kg 18.4(
preheating
Tcq
and
kJ/kg 2.26921.3648.1939
preheatingboilingtotal
qqq
Therefore, the fraction of heat used to preheat the water is
)(or
2692.2
4.752
preheat oFraction t
total
preheating 28.0% 0.2795 q
q
6-139 Cold water enters a steam generator at 20C and is boiled, and leaves as saturated vapor at boiler pressure. The boiler
pressure at which the amount of heat needed to preheat the water to saturation temperature that is equal to the heat of
vaporization is to be determined.
Analysis The heat of vaporization of water represents the amount of heat needed to vaporize a unit mass of liquid at a
specified temperature, and h represents the amount of heat needed to preheat a unit mass of water from 20C to the
saturation temperature. Therefore,
boilingpreheating
qq
Water
200C
Heater
Steam
Cold water
20C
Steam
6-104
6-140 An ideal gas expands in a turbine. The volume flow rate at the inlet for a power output of 650 kW 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 The device is adiabatic and thus heat transfer is negligible.
Properties The properties of the ideal gas are given as R = 0.30 kPa.m3/kg.K, cp = 1.13 kJ/kg·C, cv = 0.83 kJ/kg·C.
6-106
6-142 Chickens are to be cooled by chilled water in an immersion chiller. The rate of heat removal from the chicken and the
mass flow rate of water are to be determined.
Assumptions 1 Steady operating conditions exist. 2 The thermal properties of chickens and water are constant. 3 Heat gain
of the chiller is negligible.
Properties The specific heat of chicken are given to be 3.54 kJ/kg.°C. The specific heat of water is 4.18 kJ/kg.C (Table A-
Taking the chicken flow stream in the chiller as the system, the energy
balance for steadily flowing chickens can be expressed in the rate
form as
)(
0)peke (since
0
21chickenchickenout
2out1
outin
energies etc. potential,
kinetic, internal,in change of Rate
(steady) 0
sy stem
mass and work,heat,by
nsferenergy tranet of Rate
outin
TTcmQQ
hmQhm
EE
EEE
p
Then the rate of heat removal from the chickens as they are cooled from
15C to 3ºC becomes
chickenwater QQ
Noting that the temperature rise of water is not to exceed 2ºC as it flows through the chiller, the mass flow rate of water
must be at least
kW 13.0
)( water
water
Q
p
If the mass flow rate of water is less than this value, then the temperature rise of water will have to be more than 2C.
Immersion
chilling, 0.5C
Chicken
15C
6-109
6-145 Dough is made with refrigerated water in order to absorb the heat of hydration and thus to control the temperature
rise during kneading. The temperature to which the city water must be cooled before mixing with flour is to be determined
to avoid temperature rise during kneading.
Assumptions 1 Steady operating conditions exist. 2 The dough is at uniform temperatures before and after cooling. 3 The
kneading section is well-insulated. 4 The properties of water and dough are constant.
water. Also, 15 kJ of heat is released for each kg of dough kneaded, and thus 315 = 45 kJ of heat is released from the
dough made using 1 kg of water.
Taking the cooling section of water as the system,
which is a steady-flow control volume, the energy balance for
)(
0)peke (since
21waterwaterout
21
outin
TTcmQQ
hmQhm
p
out
In order for water to absorb all the heat of hydration and end up at a temperature of 15ºC, its temperature before entering
p
That is, the water must be precooled to 4.2ºC before mixing with the flour in order to absorb the entire heat of hydration.
6-146 Long aluminum wires are extruded at a velocity of 8 m/min, and are exposed to atmospheric air. The rate of heat
transfer from the wire is to be determined.
Assumptions 1 Steady operating conditions exist. 2 The thermal properties of the wire are constant.
Analysis The mass flow rate of the extruded wire through the air is
Q
Flour
Dough
6-111
6-148E 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
6-113
6-150 The ventilating fan of the bathroom of an electrically heated building in San Francisco runs continuously. The amount
and cost of the heat “vented out” per month in winter are to be determined.
Assumptions 1 We take the atmospheric pressure to be 1 atm = 101.3 kPa since San Francisco is at sea level. 2 The building
is maintained at 22C at all times. 3 The infiltrating air is heated to 22C before it exfiltrates. 4 Air is an ideal gas with
airair
We can view infiltration as a steady stream of air that is heated as it flows
in an imaginary duct passing through the house. The energy balance for this
6-114
6-151 Outdoors air at -5C and 95 kPa enters the building at a rate of 60 L/s while the indoors is maintained at 25C. The
rate of sensible heat loss from the building due to infiltration is to be determined.
Assumptions 1 The house is maintained at 25C at all times. 2 The latent heat load is negligible. 3 The infiltrating air is
heated to 25C before it exfiltrates. 4 Air is an ideal gas with constant specific heats at room temperature. 5 The changes in
kinetic and potential energies are negligible. 6 Steady flow conditions exist.
Properties The gas constant of air is R = 0.287 kPa.m3/kgK. The specific heat of air at room temperature is cp = 1.005
kJ/kg°C (Table A-2).
3kg/m 235.1
K) 273/kg.K)(-5kPa.m 287.0(
o
oRT
We can view infiltration as a steady stream of air that is heated
as it flows in an imaginary duct passing through the building.
The energy balance for this imaginary steady-flow system can
be expressed in the rate form as
)(
0)peke (since
0
12in
21in
outin
energies etc. potential,
kinetic, internal,in change of Rate
(steady) 0
sy stem
mass and work,heat,by
nsferenergy tranet of Rate
outin
TTcmQ
hmhmQ
EE
EEE
p
Then the sensible infiltration heat load corresponding to an infiltration rate of 60 L/s becomes
kW 2.23=
C5)](C)[25kJ/kg. /s)(1.005m )(0.060kg/m 235.1(
)(
33
oninfiltrati
oipairo TTcQ
V
Therefore, sensible heat will be lost at a rate of 2.23 kJ/s due to infiltration.
Warm
air
25C
Cold air
-5C
95 kPa
60 L/s
Warm air
25C
6-115
6-152 The average air velocity in the circular duct of an air-conditioning system is not to exceed 8 m/s. If the fan converts
80 percent of the electrical energy into kinetic energy, the size of the fan motor needed and the diameter of the main duct
are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus
m E
CV CV
and 0 0
. 2 The inlet velocity is negligible,
.0
1V
3 There are no heat and work interactions other than
the electrical power consumed by the fan motor. 4 Air is an ideal gas with constant specific heats at room temperature.
Properties The density of air is given to be = 1.20 kg/m3. The constant pressure specific heat of air at room temperature is
cp = 1.005 kJ/kg.C (Table A-2).
6-116
6-153 The maximum flow rate of a standard shower head can be reduced from 13.3 to 10.5 L/min by switching to low-flow
shower heads. The ratio of the hot-to-cold water flow rates and the amount of electricity saved by a family of four per year
by replacing the standard shower heads by the low-flow ones are to be determined.
Assumptions 1 This is a steady-flow process since there is no change with time at any point and thus
m E
CV CV
and 0 0
. 2 The kinetic and potential energies are negligible,
ke pe 0
. 3 Heat losses from the system
are negligible and thus
.Q0
4 There are no work interactions involved. 5 .Showers operate at maximum flow conditions
during the entire shower. 6 Each member of the household takes a 6-min shower every day. 7 Water is an incompressible
substance with constant properties. 8 The efficiency of the electric water heater is 100%.
Properties The density and specific heat of water at room temperature are = 1 kg/L and c = 4.18 kJ/kg.C (Table A-3).
6-117
6-154 Problem 6-153 is reconsidered. The effect of the inlet temperature of cold water on the energy saved by using
the low-flow showerhead as the inlet temperature varies from 10°C to 20°C is to be investigated. The electric energy
savings is to be plotted against the water inlet temperature.
Analysis The problem is solved using EES, and the results are tabulated and plotted below.
“Knowns:”
process. We note that there are two inlets and one exit. The mass and energy balances for this steady-flow system can be
expressed in the rate form as follows:”
“Mass balance:”
m_dot_in – m_dot_out = DELTAm_dot_sys
DELTAm_dot_sys=0
h_2 = c_p*T_2
h_3 = c_p*T_3
“(b) The low-flow heads will save water at a rate of “
V_dot_saved = (V_dot_old – V_dot_new)“L/min”*(5“min/person-day”)*(4“persons”)*(365“days/year”) “[L/year]”
m_dot_saved=density*V_dot_saved “[kg/year]”