Friction losses in Expansion,
Contraction & Pipe Fittings
• Friction losses in flow thro straight pipe are
calculated by using Fanning friction factor ‘f ’
• If the velocity of the fluid is changed in direction
or magnitude, additional friction losses occur.
• This results from additional turbulence which
develops bec’z of vortices and other factors.
– Sudden enlargement losses
– Sudden contraction losses
– Losses in fittings and valves
–Sudden enlargement losses
2
a
v
J
hex  K ex          in
2
kg
–Sudden contraction
losses
2
vb
J
hc  K c         
2
kg
–Losses in fittings and valves
2
a
v
J
h ff  K f
       in
2
kg
• We know Bernoulli’s eqn…
va2
pb
vb2
 gZa   WP 
 gZb   ( Frictionlosses)T

2

2
pa
• Total friction losses to be used in Bernoulli’s
equation……
2
2
2
va
vb
vb
L v2
4f
 K ex
 Kc
 Kf
D 2
2
2
2
Prob 2
• An elevated storage tank contains water at
82.2°C as shown in Fig. It is desired to
have a discharge rate at point 2 of
0.223ft3/s. What must be the height H in ft
of the surface of the water in the tank
relative to the discharge point? The pipe
used is commercial steel pipe, schedule
40, and the lengths of the straight portions
of pipe are shown.
• Density = 0.97 g/cc ; viscosity = 0.347 cP
For Schedule 40 pipe,
4” = 4.026”
2” = 2.067” and ε = 4.6x10-5m
Kc = 0.55 (for tank – 4”pipe)
Kf = 0.75 (4”elbow & 2”elbow)
Kc = 0.405 (4” – 2” pipe)
1.
2.
3.
4.
5.
6.
Contraction loss @ tank exit
Friction in 4” pipe
Friction in 4” elbow
Contraction loss from 4” to 2” pipe
Friction in 2” pipe
Friction in the two 2” elbow
• v3 = (0.223 ft3 /sec) / CSA of 4” pipe
= 0.7688 m/sec
• v4 = v2 = (0.223 ft3 /sec) / CSA of 2” pipe
= 2.9168 m/sec
1. Contraction loss @ tank exit
2
3
2
v
(0.7688)
hc  K c
 0.55
 0.1625J / kg
2
2
2. Friction in 4” pipe
N Re  2.198x105    turbulent
  4.6x10 m(steelpipe)
5
 / D  4.528x10
4
from Moody' s chart
 f  0.0047
4 fLv 2
 F .L 
 0.3312J / kg
2D
3. Friction in 4” elbow
2
3
v
(0.7688) 2
hf  K f
 0.75
 0.2216J / kg
2
2
4. Contraction loss from 4” to 2” pipe
2
4
v
(2.9168) 2
hc  K c
 0.405
 1.722 J / kg
2
2
5. Friction in 2” pipe
N Re  4.28x105    turbulent
  4.6 x105 m( steelpipe)
 / D  8.76x10
4
from Moody' s chart
 f  0.0048
L  125  10  50  185 '  56.388 m
2
4 fLv
F .L 
 87.719J / kg
2D
6. Friction in the two 2” elbow
v42
(2.9168) 2
hf  2 x K f
 2 x (0.75)
 6.3808J / kg
2
2
sum up all frictional losses.........
F .L) T  97.155 J / kg
v12
p2
v22
 gZ1   WP 
 gZ2   ( Frictionlosses) T

2

2
p1
• p1 =p2
• v1<<v2
v22
 gZ1  gZ2 
 ( Frictionlosses) T
2
 H  10.34m
Prob 3
• Water @ 20ºC is being pumped from a
tank at the rate of 5x10-3 m3/s. All of the
piping is 4” schedule 40 pipe. The pump
has an efficiency of 65%. Calculate the kW
power needed for the pump.
Given…for 4” Schedule 40 pipe, D=0.1023m
Kc=0.55
Density = 998.2 kg/m3
Kf=0.75
Viscosity=1.005 cP
Kex=1.0
Frictional losses are……..
1.
2.
3.
4.
Contraction loss @ tank exit
Friction in straight pipe
Friction in the two elbows
Expansion loss @ tank entrance
Velocity = (5x10-3 ) / CSA = 0.6083 m/s
1. Contraction loss @ tank exit
hc  K c
v2
2
 0.1017J / kg
2. Friction in straight pipe
MOODY chart.ppt
N Re  6.181x104    turbulent
  4.6 x105 m( steelpipe)
 / D  0.00045
from
Moody' schart, f  0.0051
L  5  50  15  100  170m
4 fLv 2
F .L 
 6.272J / kg
2D
3. Friction in the two elbows
h f  2K f
v2
2
 0.2774J / kg
4. Expansion loss @ tank entrance
hex  K ex
v2
2
 0.185J / kg
Total losses = 6.837 J/kg
• By Bernoulli’s eqn…
va2
pb
vb2
 gZa   WP 
 gZb   ( Frictionlosses)T

2

2
pa
Wp  236.901J / kg
Wp  1.182kW
EQUIVALENT LENGTH
• In some applications it is convenient
to calculate pressure drops in fittings
from added equivalent lengths of
straight pipe
• ‘Le’ is the equivalent length of st. pipe
in m having the same frictional loss
as the fitting.
• The ‘Le’ values for fittings are simply
added to length of the st. pipe to get
the total length of equivalent st. pipe
to use in (FL)
Water @ 20ºC is pumped from a storage tank thro 100m of
3cm dia pipe. The pipe line has TWO globe valves which
are fully open and THREE 90º elbows. Water is discharged
into another tank thro a spray nozzle. The discharge is @ a
height of 20m above the level of water in the storage tank.
The pressure required @ the nozzle entrance is 4x105
N/m2. Flow rate of water 1kg/sec. Viscosity is 0.975 cP
f= 0.0014 + 0.125/Re0.32
Estimate a). Energy loss due to fricition
b). Pump work required per kg of water
c). Theoretical HP required for the pump
Equivalent length in terms of pipe dia:
Open globe valve = 300D
90ºElbow =30D
•
•
•
•
•
•
•
•
•
V = 1.4147 m/sec
NRe = 43 529.55
f = 0.00549
To calculate FL)T first find total length L’….
L’ = L + Le = 100 + { (2x300x0.03) +
(3x30x0.03) } = 120.7 m
= 2 f L’ v2 / D = 88.413 J/kg
Sub. in Bernoulli’s eqn….assume Pa =1
atm
Wp = 585.617 J/kg
= 0.786HP
Water is to be pumped from a pond to the top of a
tower 1829cm above the water level in the pond.
It is desired to deliver 0.34 cu m/min of water at
a pressure of 2.08atm. The pipe line consists of
122m length of st. pipe of 7.62cm ID with EIGHT
elbows of 90º & FOUR gate valves. Calculate
the HP of the pump having an efficiency of 80%:
f= 0.046 / Re0.2
Equivalent length in terms of pipe dia:
Gate valve = 7D
90ºElbow =32D
•
•
•
•
•
•
•
•
•
V = 1.2425 m/sec
NRe = 90 957.58
f = 0.004688
To calculate FL)T first find total length L’….
L’ = L + Le = 122 + { (8x32x0.0762) +
(4x7x0.0762) } = 143.64 m
FL)T = 2 f L’ v2 / D = 27.239 J/kg
Sub. in Bernoulli’s eqn….assume Pa =1
atm
Wp = 394.29 J/kg
= 2.999HP
Flow in non-circular ducts
• For flow in a duct of non-circular cross-section, the
hydraulic mean diameter may be used in place of the
pipe diameter and the formulae for circular pipes can
then be applied without introducing a large error. This
method of approach is entirely empirical.
• The hydraulic mean diameter DH is defined as four times
the hydraulic mean radius rH.
• Hydraulic mean radius is defined as the flow crosssectional area divided by the wetted perimeter.
rH = (cross sectional area of channel) / (wetted
perimeter of channel)
Few Examples……
• For circular tube….
rH 
D 2
4  D
D
4
• Annulus between two concentric pipes…
 Do 2   Di 2

  
4
4



rH 
Di  Do
• For square duct….
• For rectangular duct…..
rH


  Do  Di
4
a2
a


4a
4
ab
ab
rH 

2a  2b 2(a  b)
prob1
a). Calculate the hydraulic mean diameter
of the annular space between a 40mm and
a 50mm tube.
b). A liquid having a density of 60.58lb/ft3
and a viscosity of 0.347cP is flowing thro a
pipe of dia 0.3355ft. The flow rate of the
liquid is 75cm/sec. Calculate the Reynolds
number.
• A horizontally placed pipe carries fluid (sp.gr.
0.983 and viscosity 8.35cP) at a mass flow rate
of 50kg/s/. Besides the pipe friction there is a
flow restriction in the pipe whose frictional
resistance can be expressed as HR= 4.65V
metres of fluid where V is the ave.linear velocity
of the fluid in m/s. The friction factor in the pipe
is given by
f=0.073/NRe0.226
If the internal dia of pipe is 45cm and its length is
4000m, calculate the total flow resistance
developed during flow, expressed as metres of
the flowing liquid
• 60% sulfuric acid is to be pumped at the
rate of 4000cc/sec thro a pipe 25mm dia
and raised to a height of 25m. The pipe is
30m long and it runs straight. Calculate
the theoretical power required.
The sp.gr of the acid is 1.531. Friction
factor may be assumed to be 0.0047
2.08 HP
• Water at a rate of 200tons/hr has to be pumped
from a river to factory overhead tank placed at a
height of 25m from the river bed; the total length
of pipeline is 1.5km. Cast iron pipe having an ID
of 30cm will be used for the purpose. The
average temperature of water in the river may
be taken as 30ºC, viscosity 0.764cP
• Cal Re
• Head lost due to friction
• Re = 3x105
• f = 0.0036
• h = 2.255m
• Its planned to install a steel pipe with an ID of
20cm to transfer 1000kg/m3, molasses having a
viscosity 500cP and density 1.6g/cc. The line is
to be 1000m long and delivery end is to be 5m
higher than the intake. Calculate
• Pressure drop due to friction
• If overall efficiency of the pump is 60% what will
be the HP required?
• v = 0.33157 m/s
• Re = 212.2
• f = 0.0754
• (FL)T = 82.899 J/kg
• Wp = 219.99 J/kg== 4.92HP
• Determine the cost of pumping 3,00,000L/hr of
an oil, sp.gravity 0.9 and viscosity 30cP thro a
pipeline, 25cm dia and 50km long. It may be
assumed that the efficiency of pump together
with motor is 50% and the power costs
40paise/kW. The pipe line is horizontal and f =
0.046/Re0.20
• v = 1.697m/s
• Re = 12,732
• f = 6.946x10-3
• FL)T = 8002.017 J/kg
• Wp = 16006.91 J/kg
• ==Wp = 1200.518kW
• Therefore cost = Rs.480/-
ASSIGNMENT - II
Write short notes on following:
•
•
•
•
•
•
Boundary layer formation
Boundary layer separation
Drag coefficient
Stream lining
Entry length calculation
Fully developed flow
– Last date of submission: 21th April
ASSIGNMENT - III
•
•
•
•
•
•
Write short notes on following:
Positive displacement pumps
Rotary pumps
Airlift pumps
Jet pumps
Fans, blowers and compressors
Different types of pipe fittings and valves
– Last date of submission: 7th May