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Chapter 5 Mass, Bernoulli, and Energy Equations
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted
only to teachers and educators for course preparation. If you are a student using this Manual, you are using it
without permission.
5-39
Energy Equation
5-63C It is impossible for the fluid temperature to decrease during steady, incompressible, adiabatic flow
since this would require the entropy of an adiabatic system to decr

Transcript

Chapter 5
Mass, Bernoulli, and Energy Equations
PROPRIETARY MATERIAL
. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
5-39
Energy Equation
5-63C
It is
impossible
for the fluid temperature to decrease during steady, incompressible, adiabatic flow since this would require the entropy of an adiabatic system to decrease, which would be a violation of the 2
nd
law of thermodynamics.
5-64C
Yes, the
frictional effects
are negligible if the fluid temperature remains constant during steady, incompressible flow since any irreversibility such as friction would cause the entropy and thus temperature of the fluid to increase during adiabatic flow.
5-65C
Head loss
is the loss of mechanical energy due to friction in piping expressed as an equivalent column height of fluid, i.e., head. It is related to the mechanical energy loss in piping by
g m E g eh
L
&&
pipingloss,mech pipingloss,mech
==
.
5-66C
Useful pump head is the useful power input to the pump expressed as an equivalent column height of fluid. It is related to the useful pumping power input by
g mW g wh
&&
u pump,u pump,
pump
==
5-67C
The
kinetic energy correction factor
is a correction factor to account for the fact that kinetic energy using average velocity is not the same as the actual kinetic energy using the actual velocity profile. Its effect is usually negligible (the square of a sum is not equal to the sum of the squares of its components).
5-68C
By Bernoulli Equation, the maximum theoretical height to which the water stream could rise is the tank water level, which is 20 meters above the ground. Since the water rises above the tank level, the tank cover must be airtight, containing pressurized air above the water surface. Otherwise, a pump would have to pressurize the water somewhere in the hose.
Chapter 5
Mass, Bernoulli, and Energy Equations
PROPRIETARY MATERIAL5-69
Underground water is pumped to a pool at a given elevation. The maximum flow rate and the pressures at the inlet and outlet of the pump are to be determined.
√
Assumptions
1
The flow is steady and incompressible.
2
The elevation difference between the inlet and the outlet of the pump is negligible.
3
We assume the frictional effects in piping to be negligible since the
maximum
flow rate is to be determined,
4
The effect of the kinetic energy correction factors is negligible,
α
= 1. .0
pippingloss,mech
=
E
&
Properties
We take the density of water to be 1 kg/L = 1000 kg/m
3
.
Analysis
(
a
) The pump-motor draws 3-kW of power, and is 70% efficient. Then the useful mechanical (shaft) power it delivers to the fluid is kW1.2kW)3)(70.0(
electricmotor - pumpu pump,
===
W W
&&
η
We take point 1 at the free surface of underground water, which is also taken as the reference level (
z
1
= 0), and point 2 at the free surface of the pool. Also, both 1 and 2 are open to the atmosphere (
P
1
=
P
2
=
P
atm
), the velocities are negligible at both points (
V
1
≅
V
2
≅
0), and frictional losses in piping are disregarded. Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to
lossmech,turbine2
2222 pump12111
22
E W gz V P mW gz V P m
&&&&&
++ ++=+
++
α ρ α ρ
1 2
Pump Pool30 m
In the absence of a turbine, and
pipingloss,mech pumploss,mechlossmech,
E E E
&&&
+=
pumploss,mech pumpu pump,
E W W
&&&
−=
. Thus,
W
2u pump,
gz m
&&
=
Then the mass and volume flow rates of water become kg/s14.7
kJ1/sm1000m))(30m/s(9.81
kJ/s1.2
2222 u pump,
= ==
gz W m
&&
/sm107.14
33
−
×===
3
kg/m1000 kg/s14.7
ρ
m
&&
V
(
b
) We take points 3 and 4 at the inlet and the exit of the pump, respectively, where the flow velocities are m/s86.1
4/m)(0.07
/sm1014.7
4/
2332333
=×===
−
π π
D AV
V V
&&
, m/s64.3
4/m)(0.05
/sm1014.7
4/
2332444
=×===
−
π π
D A
V V
&&
V
We take the pump as the control volume. Noting that
z
3
=
z
4
, the energy equation for this control volume reduces to
pumploss,mechturbine4
2444 pump32333
22
E W gz V P mW gz V P m
&&&&&
++ ++=+
++
α ρ α ρ
→
V
&&
u pump,242334
2)(
W V V P P
+−=−
ρα
Substituting,
kPa289.2
=+−=
⋅×+ ⋅−=−
233-2
22334
kN/m)1.2949.4(
kJ1mkN1/sm107.14
kJ/s1.2
m/skg1000 kN12]m/s)(3.64)m/s86.1()(1.0)[kg/m(1000
P P
Discussion
In an actual system, the flow rate of water will be less because of friction in pipes. Also, the effect of flow velocities on the pressure change across the pump is negligible in this case (under 2%) and can be ignored.
. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
5-40
Chapter 5
Mass, Bernoulli, and Energy Equations
PROPRIETARY MATERIAL5-70
Underground water is pumped to a pool at a given elevation. For a given head loss, the flow rate and the pressures at the inlet and outlet of the pump are to be determined.
√
Assumptions
1
The flow is steady and incompressible.
2
The elevation difference between the inlet and the outlet of the pump is negligible.
3
The effect of the kinetic energy correction factors is negligible,
α
= 1.
Properties
We take the density of water to be 1 kg/L = 1000 kg/m
3
.
Analysis
(
a
) The pump-motor draws 3-kW of power, and is 70% efficient. Then the useful mechanical (shaft) power it delivers to the fluid is
kW1.2kW)3)(70.0(
electricmotor - pumpu pump,
===
W W
&&
η
We take point 1 at the free surface of underground water, which is also taken as the reference level (
z
1
= 0), and point 2 at the free surface of the pool. Also, both 1 and 2 are open to the atmosphere (
P
1
=
P
2
=
P
atm
), and the velocities are negligible at both points (
V
1
≅
V
2
≅
0). Then the energy equation for steady incompressible flow through a control volume between these two points that includes the pump and the pipes reduces to
lossmech,turbine2
2222 pump12111
22
E W gz V P mW gz V P m
&&&&&
++ ++=+
++
α ρ α ρ
In the absence of a turbine, and
W
and thus
pipingloss,mech pumploss,mechlossmech,
E E E
&&&
+=
pumploss,mech pumpu pump,
E W
&&&
−=
pipingloss,mech2u pump,
E gz mW
&&&
+=
5-41
Noting that , the mass and volume flow rates of water become
L
ghm E
&&
=
lossmech,
kg/s116.6
kJ1 /sm1000m)5)(30m/s(9.81
kJ/s1.2
)(
2222u pump,2u pump,
= +=+=+=
L L
h z g W gh gz W m
&&&
/sm106.12
33
−
×≅===
smm
/116.6
kg/m1000kg/s116.6
33
ρ
&&
V
(
b
) We take points 3 and 4 at the inlet and the exit of the pump, respectively, where the flow velocities are 21
Pump Pool30 m
m/s589.1
4/m)(0.07
/sm10116.6
4/
2332333
=×===
−
π π
D AV
V V
&&
,
m/s115.3
4/m)(0.05
/sm10116.6
4/
2332444
=×===
−
π π
D A
V V
&&
V
We take the pump as the control volume. Noting that
z
3
=
z
4
, the energy equation for this control volume reduces to
pumploss,mechturbine4
2444 pump32333
22
E W gz V P mW gz V P m
&&&&&
++ ++=+
++
α ρ α ρ
→
V2)(
u pump,242334
&&
W V V P P
+−=−
ρα
Substituting,
kPa340
≅=+−=
⋅×+ ⋅−=−
kPa P P
8.339kN/m)4.3436.3(
kJ1mkN1/sm106.116
kJ/s1.2
m/skg1000kN12]m/s)(3.115)m/s589.1()(1.0)[kg/m(1000
233-2
22334
Discussion
Note that frictional losses in pipes causes the flow rate of water to decrease. Also, the effect of flow velocities on the pressure change across the pump is negligible in this case (about 1%) and can be ignored. . © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Chapter 5
Mass, Bernoulli, and Energy Equations
PROPRIETARY MATERIAL5-71E
In a hydroelectric power plant, the elevation difference, the power generation, and the overall turbine-generator efficiency are given. The minimum flow rate required is to be determined.
√
Assumptions
1
The flow is steady and incompressible.
2
The water levels at the reservoir and the discharge site remain constant.
3
We assume the flow to be
frictionless
since the
minimum
flow rate is to be determined,
.0
lossmech,
=
E
&
Properties
We take the density of water to be
ρ
= 62.4 lbm/ft
3
.
Analysis
We take point 1 at the free surface of the reservoir and point 2 at the free surface of the discharge water stream, which is also taken as the reference level (
z
2
= 0). Also, both 1 and 2 are open to the atmosphere (
P
1
=
P
2
=
P
atm
), the velocities are negligible at both points (
V
1
=
V
2
= 0), and frictional losses are disregarded. Then the energy equation in terms of heads for steady incompressible flow through a control volume between these two points that includes the turbine and the pipes reduces to
1eturbine,eturbine,2
2222 u pump,12111
22
z hhh z
g V g P h z g V g P
L
=→++++=+++
α ρ α ρ
Substituting and noting that
W
, the extracted turbine head and the mass and volume flow rates of water are determined to be
eturbine,gen-turbineelectturbine,
ghm
&&
η
=
ft240
1eturbine,
==
z h
lbm/s 370
= ==
kW1Btu/s9478.0
Btu/lbm1/sft037,25
ft))(240ft/s0.83(32.2
kW100
222turbinegen-turbine
electturbine,
ghW m
η
&&
/sft5.93
3
===
3
lbm/ft62.4lbm/s370
ρ
m
&&
V
2 1Generator
Turbine 240 ftWater
Therefore, the flow rate of water must be at least 5.93 ft
3
/s to generate the desired electric power while overcoming friction losses in pipes.
Discussion
In an actual system, the flow rate of water will be more because of frictional losses in pipes. . © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
5-42

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