Thermo_5th_Chap02P001
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Thermo_5th_Chap02P001


DisciplinaTermodinâmica Fundamental86 materiais456 seguidores
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2inmech,outmech,fluidmech,
VV
PP
V
Pv
V
PvmeemE \u3c1V&&&& 
since , and there is no change in the potential 
energy of the fluid. Also, 
vVV /&&& == \u3c1m
1
2
Motor 
35 kW
Pump 
inlet 
PUMP
m/s 9.19
4/m) (0.08
/sm 1.0
4/ 2
3
2
11
1 ==== \u3c0\u3c0DAV
VV && 
m/s 84.8
4/m) (0.12
/sm 1.0
4/ 2
3
2
22
2 ==== \u3c0\u3c0DAV
VV && 
 
Substituting, the useful pumping power is determined to be 
 
 kW3.26 
m/s kN1
 kW1
m/s kg1000
 kN1
2
m/s) (19.9)m/s 84.8( ) kg/m860( kN/m400/s)m (0.1 2
22
323
fluidmech,upump,
=
\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb
\u22c5\uf8f7
\uf8f7
\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb
\uf8f7\uf8f7\uf8f8
\uf8f6
\uf8ec\uf8ec\uf8ed
\uf8eb
\u22c5
\u2212+=
\u2206= EW &&
 
Then the shaft power and the mechanical efficiency of the pump become 
kW 5.31kW) 35)(90.0(electricmotorshaftpump, === WW && \u3b7 
83.6%==== 836.0
 kW31.5
 kW3.26
shaft pump,
 upump,
pump W
W
&
&
\u3b7 
Discussion The overall efficiency of this pump/motor unit is the product of the mechanical and motor 
efficiencies, which is 0.9×0.836 = 0.75. 
 
 
2-36 
2-78E Water is pumped from a lake to a nearby pool by a pump with specified power and efficiency. The 
mechanical power used to overcome frictional effects is to be determined. 
Assumptions 1 The flow is steady and incompressible. 2 The elevation difference between the lake and the 
free surface of the pool is constant. 3 The average flow velocity is constant since pipe diameter is constant. 
Properties We take the density of water to be \u3c1 = 62.4 lbm/ft3. 
Analysis The useful mechanical pumping power delivered to water is 
 hp76.8 hp)12)(73.0(pumppumpupump, === WW && \u3b7 
The elevation of water and thus its potential energy 
changes during pumping, but it experiences no 
changes in its velocity and pressure. Therefore, the 
change in the total mechanical energy of water is 
equal to the change in its potential energy, which is 
gz per unit mass, and for a given mass flow 
rate. That is, 
gzm&
2
1
35 ft 
Pump
Lake
Pool 
 zgzgmpememE \u2206=\u2206=\u2206=\u2206=\u2206 V&&&&& \u3c1mechmech
Substituting, the rate of change of mechanical energy of water becomes 
hp 76.4
ft/slbf 550
hp 1
ft/slbm 32.2
lbf 1)ft 35)(ft/s /s)(32.2ft )(1.2lbm/ft (62.4 2
233
mech =\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb
\u22c5\uf8f7\uf8f8
\uf8f6\uf8ec\uf8ed
\uf8eb
\u22c5=\u2206E
& 
Then the mechanical power lost in piping because of frictional effects becomes 
hp 4.0 =\u2212=\u2206\u2212= hp 76.476.8mechu pump,frict EWW &&&
 Discussion Note that the pump must supply to the water an additional useful mechanical power of 4.0 hp to 
overcome the frictional losses in pipes. 
 
 
	Forms of Energy
	Energy Transfer by Heat and Work
	Mechanical Forms of Work
	The First Law of Thermodynamics
	Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room.
	Assumptions 1 The room is well sealed, and heat loss from the room is negligible. 2 All the appliances are kept on.
	Assumptions 1 The house is well-sealed, so no air enters or heaves the house. 2 All the lights and appliances are kept on. 3 The house temperature remains constant.
	Energy Conversion Efficiencies