gr.3
DESCRIPTION
sdfgsgTRANSCRIPT
Overall Mechanical Energy Balance
• A more useful type of mechanical balance for fluids, especially liquids, is a modification of the total energy balance to deal with mechanical energy.
• Mechanical energy involves work, kinetic energy, potential energy, and the flow work part of the enthalpy term.
• Mechanical energy is a form of energy that is either work or a form that can be directly converted to work.
Kinetic Energy change
Potential Energy change
Pressure change
Sum of Friction losses
Shaft work
Note:α= 1 for turbulent flow, 0.5 for laminar flow
Reynold’s Number
D
N Re
sPas
m
mDs
mm
kg
Units
.
:
2
3
If:Re < 2000, Laminar (straight flow)Re > 4000, Turbulent (eddying)
Friction Loses in Expansion, Contraction, and Pipe Fittings
• Skin friction losses in flow through straight pipe are calculated by using the Fanning friction factor. However, if the velocity of the fluid is changed in direction or magnitude, additional friction loses occur. This results from additional turbulence which develops because of vortices and other factors.
Methods to estimate the losses
1. Sudden enlargement losses. If the cross section of a pipe enlarges very gradually, very little or no extra losses are incurred. If the change is sudden, it results in additional losses due to eddies formed by the jet expanding in the enlarged section. The friction loss can be calculated by:
Where:
exh Is the friction loss in J/kg
tcoefficien lossexpansion the
,)1( 2
2
1
A
AKex
2. Sudden contraction losses. When the cross section of the pipe is suddenly reduced, the stream cannot follow around the sharp corner, and additional friction losses due to eddies occur. For turbulent flow, this is given by:
tcoefficien lossn contractio theis
section downstreamor smaller in the velocity average theis
laminarfor 0.5 flow,ent for turbul 1
lossfriction theis
2
c
c
K
h
3. Losses in fittings and valves. Pipe fittings and valves also disturb the normal flow lines in a pipe and cause additional friction losses. In a short pipe with many fittings, the friction loss from these fittings could be greater than in the straight pipe. The friction loss from these fittings could be greater than in the straight pipe. The friction loss for the fittings and valves is given by the following equation:
fitting the toleading pipe in the velocity average theis
or valve fitting for thefactor loss theis
:
1fK
where
Note: Values for Kf are given in Table 2.10-1 for turbulent flow and 2.10-2 for laminar flow.
4. Frictional losses in mechanical energy balance equation. The frictional losses from the friction in the straight pipe (Fanning friction), enlargement losses, contraction losses, and losses in fittings and valves are all incorporated in the ΣF term from the mechanical energy balance equation, so that
• If all velocities, v, v1, and v2 are the same,
Pump Horsepower in Flow System
pump the todelivered shaft workor energy
efficiency fractional
pump by the fluid the todeliveredenergy mechanical
:
p
s
W
n
Ws
where
nWpW
Friction Losses and Pump Horsepower
Hot water in an open storage tank at 82.2⁰C is being pumped at a rate of 0.379 from this storage tank. The line from the storage tank to the pump suction is 6.1m of 2-in. schedule 40 steel pipe and it contains three elbows. The discharge line after the pump is 61m of 2-in pipe and contains two elbows. The water discharges to the atmosphere at a height of 6.1m above the water level in the storage tank.
a. Calculate the frictional losses ΣFb. Make a mechanical energy balance and
calculate Ws of the pump in .c. What is the kW power of the pump if its
efficiency is 75%?
Solution
s
m
mx
smxVelocity
mx
sPaxC
xs
xm
rate
918.21065.21
/10317.6
1065.21 Area
diameter) (inside m 0.05250or
mm 52.50 40 schedule in) D(2 A.5,Appendix on Based
.103474.02.82@ A.2,Appendix on Based
m
kg970.4C82.2 ion,interpolat Using
10317.660
min1
min379.0
24
233
24
3
3
33
1
FlowTurbulent ,10279.4103474.0
)4.970)(918.2)(0525.0(
Number, sReynold' Using
5
3
3
Re
x
smkg
x
mkg
sm
mDvN
Mechanical Energy Balance
02
becomesequation
)0,,(
0)()(2
1
2
22
1211
1212
21
22
s
s
WFgz
ppzv
WFpp
zzg
Calculation for ΣF
1. Contraction loss- tank to pipe
kg
Jh
smxh
K
A
A
A
AK
vKh
c
c
c
c
cc
33.2
2
)/918.2(55.0
55.0)01(55.0
0 );1(55.0
)2
(
2
1
2
1
2
22
2. Friction loss in 5 elbows
kg
Jh
xsm
xh
vKh
f
f
ff
965.15
52
)/918.2(75.0
elbows) of no.(2
75.0K 1,-2.10 Tableon Based
2
22
f
3. Friction loss in straight pipe
kg
JF
smx
m
mF
v
D
LfF
f
x
D
D
xN
mmmL
f
f
f
47.104
2
)/918.2()
0525.0
1.67)(0048.0(4
24
0048.0
000876.00525.0
106.4
0525.0
10279.4
1.67611.6
2
22
5
5Re
• For ΣF
kg
JF
kg
JF
FhhF ffc
77.122
)47.104965.1523.2(
• For shaft work:
kg
JW
W
Wkg
Jmsm
sm
WFgzv
s
s
s
s
9.186
09.186
077.122)1.6)(/81.9(2
)/918.2(
02
22
2
22
• kW power of pump if efficiency is 75%
kWWkW
s
kg
kg
JkW
kg
JWp
Wp
nWpWs
s
kgm
m
kg
s
mxvm
v
m
53.117.1527
)130.6(13.249
13.249
))(75.0(85.186
130.6
)4.970)(10317.6(3
33