Modern Engineering Thermodynamics

5.5 Incompressible Liquids 153
The unknown is T 2 . The system is closed, and the material is the 1.00 qt
of liquid water.
The system states and processes are
1.00 quart of water
State 1
p 1 = 14:7 psia
T 1 = 60:0°F
Process: Constant pressure
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mechanical mixing
State 2
p 2 = p 1 = 14:7 psia
System boundary
Note that, in this problem, we do not know the values of two independent
properties in the second state, nor does the process path give
us any information about an additional second-state property. This
example illustrates how the energy balance itself can be used to find
the value of a state property.
0.250 hp electric motor
The basic energy balance (EB) equation for this system is
1Q 2 − 1 W 2 = mðu 2 − u 1 Þ + KE 2 − KE 1 + PE 2 − PE 1
Since we are given no information about the kinetic or potential energies
of the system, we assume that they do not change during the process
under analysis; that is,
FIGURE 5.4
Example 5.4.
KE 2 − KE 1 = PE 2 − PE 1 = 0
The auxiliary equations needed here are for the heat and work energy transport modes. They are 1 Q 2 = 0 (insulated system),
and in this case, the work mode is shaft work, but it can be calculated from the definition of power as 1 W 2 = _WðΔtÞ, where Δt
is the time interval of the process.
Since 14.7 psia is much greater than the saturation pressure at 60.0°F (which is 0.2563 psia), state 1 is seen to be a compressed
liquid. We could find u 1 by interpolating the pressure between the saturation and compressed liquid tables at 60.0°F,
then use the energy balance to find u 2 .Withu 2 and p 2 , we could presumably find T 2 by again interpolating in the tables. Or
else, we could treat the water as a simple incompressible material and use the auxiliary equation for specific heat (Eq. (3.33)),
u 2 − u 1 = cðT 2 − T 1 Þ. The latter approach is the simplest in this case, soweuseitandtakethespecificheatofwatertobe
1.00 Btu/ðlbm⋅RÞ
The mass of 1 qt of water at 60.0°F is given by m = pV = V/v, where
V = ð1:00 qtÞ 1 gal
0:133 68 ft 3 /gal = 0:0334 ft 3
4qt
and v = v f ð60:0°FÞ = 0:01603 ft 3 /lbm (from Table C.1a of Thermodynamic Tables to accompany Modern Engineering Thermodynamics).
Therefore,
Then, the energy balance gives
m =
0:0334 ft 3
0:01603 ft 3 = 2:08 lbm
/lbm
and
u 2 − u 1 = cðT 2 − T 1 Þ = 1 Q 2
m − 1 W 2
m
T 2 = T 1 + 1 Q 2
mc − 1 W 2
mc
ð− 0:250 hpÞð10:0 minÞð1 h/60 minÞ½2545 Btu=ðhp⋅hÞŠ
= 60:0°F þ 0 −
ð2:08 lbmÞ½1:00 Btu=ðlbm⋅RÞŠ
= 111°F
Exercises
1. Is the result in Example 5.4 of T 2 = 111°F an unreasonably high fluid temperature, since a blender of this type is never
actually insulated (adiabatic)? In other words, what is the direction of the heat transfer in an uninsulated blender and
what effect does this heat transfer have on the temperature T 2 ? Answer: An uninsulated blender has heat transfer from
the mixing fluid to the surroundings, producing a lower value of T 2 than that calculated in Example 5.4.
2. What properties other than those used in the solution to Example 5.4 affect the designer’s choice of the motor power for
the blender? Answer: The fluid viscosity and the blender speed.
3. If the fluid temperature in Example 5.4 reaches only 85.0°F instead of 111°F after 10.0 min of operation, determine the
power delivered by the motor. Answer: _W motor = 0:123 hp.