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Lec 26: Frictionless flow with work, pipe flow
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• For next time:– Read: § 8-14 to 8-18– HW 13 due Monday, December 1, 2003
• Outline:– Bernoulli equation in pipe flow– Accounting for pumps and turbines– Accounting for friction in pipes
• Important points:– Don’t forget the conservation of mass equation– Be careful how you account for pump and
turbine losses– Understand the Moody Diagram
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Fluid Mechanics
• A venturi tube or meter--converging-diverging nozzle frequently used to measure the volumetric flowrate of a fluid. It must be inserted into a pipe or duct as a part of the pipe or duct.
1 2
4
TEAMPLAYTEAMPLAY
• For a venturi such as that shown before, the following data apply: dia1 = 6.0 in,
dia2 = 4.0 in. The pressure difference
P1 – P2 = 3 psi. Water with a density of
62.4 lbm/ft3 is flowing. Find the rate of flow in ft3/min.
• Hint: use flowrate Qv = A1V1 = A2V2
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Pipe flow
• We previously had the Bernoulli equation in this form and we observed that the terms are energy per unit mass.
• In a pipe (or duct), this equation represents the mechanical energy per unit mass at a flow cross section.
constant
2
2P
gz
v
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Pipe flow
• For frictionless flow, the mechanical energy will be the same at every cross section of the pipe and
2
22
1
21 gz
2
Pgz
2
P
21 vv
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Pipe flow
Cross section 1
Cross section 2
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Pipe flow
• In real pipe or duct flows, energy must be used to overcome friction, and so at subsequent cross sections the energy is less.
• Pipe flow--examples are water systems and petroleum pipeline systems.
• Duct flow--examples are air conditioning ducts in buildings.
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Pipe and duct flow
• We had the following equation for incompressible steady flow:
• or, replacing v by and rearranging
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21
22
1212 22)()( zzguuPPvwq
vv
)(22 1212
21
2212 uuzzg
PPwq
vv
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Pipe and duct flow
)()( 1212 TTcuu v •The term represents the increase in internal energy that occurs due to friction as the fluid flows in a pipe or duct. The textbook calls this a mechanical loss term, emech,
loss.
•The text also splits the work term into two:
•wpump represents pumping power input or the power required by a pump or fan to move the flow.
•Wturbine represents the work done by a water turbine, for example.
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Pipe and duct flow
• Thus the book arrives at the following equation for adiabatic flow:
• or
)(22 1212
21
2212 uuzzg
PPw
vv
lossmechturbine
shaftpump
ewgzP
wgzP
,2
222
,1
211
2
2
v
v
12
Pipe and duct flow
• The previous equation
• is equation 11-31 in the text book, where the equation (incorrectly) appears thusly
lossmechturbineupump ewgz
Pwgz
P,2
222
,1
211
22
vv
lossmechturbineshaftpump ewgz
Pwgz
P,2
222
,1
211
22
vv
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Pipe and duct flow
• The shaft work includes the work to increase the mechanical energy of the fluid and to overcome losses. Eq 11-31 and 11-32 should read:
• where if we forget the wturbine for the moment, wpump,u raises the mechanical energy of the flow.
turbineupump wgz
Pwgz
P 2
222
,1
211
22
vv
turbineupump WgzP
mWgzP
m
2
222
,1
211
22
vv
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Pipe and duct flow
• Divide the previous equation by g to get an equation in terms of head:
lossmech
turbine
shaftpump
g
e
gz
gg
P
gz
gg
P
,
2
222
,1
211
w
2
w
2
v
v
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Pipe and duct flow
• The term (e/g)mech,loss is identified as the head loss term, and we calculate it later.
L,
Hlossmechg
e
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Pipe and duct flow
• Not all of the shaft power going into the pump is converted into useful mechanical energy supplied to the fluid, such as raising water up into a water tower.
• Some is lost in frictional heating that manifests itself as a slight temperature rise of the fluid.
• Thus wpump,shaft= wpump,u+emech,losses
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Pipe and duct flow
• This loss gives rise to the following pump efficiency
in shaft,
upump,ump w
w
input work Mechanical
fluid the tosuppliedenergyMechanical
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TEAMPLAYTEAMPLAY
• Consider a water pump in a horizontal, constant area pipe. The mass flow rate is 50 kg/s, and the pump receives 17.0 kW of power from its driver. The delta P (pressure change) across the pump is 250 kPa.
• Determine the mechanical efficiency of the pump and the temperature rise of the fluid.
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Pipe and duct flow
L2
22
1
21 Hz
g2g
Pz
g2g
P
21 vv
•Using the equation in terms of head loss and omitting the work terms (meaning we do not have a pump, turbine or compressor in the system),
•Where the previous equation in terms of flow energy per unit mass has been divided by g (not gc) to yield units of head (length).
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Pipe and duct flow
• It can be shown that the the head loss term can be determined in terms of the shear stress at the wall, o.
• Where is the length of the pipe and D is the diameter of the pipe.
oL gD
4H
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Pipe and duct flow
• The shear stress at the wall is a function of five independent variables.
o = o(,,V,D,e) is the dynamic or absolute viscosity• V is the mean velocity of the flow• e (or ) is the surface roughness of the
pipe, and can be described as a length.
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Pipe and duct flow
• With five independent variables, the solution is very complex.
• However, the complexity is reduced to two variables by use of only two non-dimensional variables.
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Pipe and duct flow
• The non-dimensional shear stress can be expressed as
D
e,f
2
2o VD
V
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Pipe and duct flow
VD
= Re, the Reynolds number (non- dimensional)
e/D = relative roughness
= ratio of inertia forces/viscous forces
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Pipe and duct flow
•The non-dimensional shear stress is tabulated as a function of the Reynolds number and the surface roughness.
•The non-dimensional shear stress is called a friction factor.
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Pipe and duct flow
• Fanning friction factor, , used in heat transfer:
• Moody friction factor, f, used in fluid mechanics:
2
2o
V
4f
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Pipe and duct flow
• Moody friction factor, f, used in fluid mechanics is typically given as
• and appears on page 974 of the textbook.
2o8
4fV
2
4 2o
V
28
29
Pipe and duct flow
• So,
g2D
f8
f
gD
4
gD
4H
22
oL
VV
•And
D
eRe,offunctionf
30
T hus
L2
22
1
21 Hz
g2g
Pz
g2g
P
21 vv
gDf
2
2V
P i p e a n d d u c t fl o w
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TEAMPLAYTEAMPLAY
• Determine the pressure drop over 1000 ft of pipe carrying water if the flowrate is 70 cfs and the pipe diameter is 30 inches. Assume that the pipe has a roughness, e, of 0.1 inches and the temperature is 25 C. Water has a density of 62.4 lbm/cubic foot.
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TEAMPLAYTEAMPLAY
A 2 ft diameter cylindrical aire conditioning duct hands horizontally from a ceiling and carries 10,000 CFM of 55 F air. Relative roughness factor e/D = 0.0002. Find the pressure drop der 100 ft of duct length in the units of inches of water if 27.7 in of water is 1 psia.
1. Write and simplify the basic equation.2. Find the density, velocity of flow, Reynold’s
number and friction factor.3. Solve the problem.