Friction losses in Expansion, Contraction & Pipe Fittings

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Presentation transcript:

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 Sudden contraction losses Losses in fittings and valves

We know Bernoulli’s eqn… Total friction losses to be used in Bernoulli’s equation……

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)

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

Contraction loss @ tank exit Friction in 4” pipe

Friction in 4” elbow 4. Contraction loss from 4” to 2” pipe

Friction in 2” pipe

6. Friction in the two 2” elbow

p1 =p2 v1<<v2

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

Velocity = (5x10-3 ) / CSA = 0.6083 m/s Frictional losses are…….. 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 2. Friction in straight pipe MOODY chart.ppt

3. Friction in the two elbows Total losses = 6.837 J/kg 4. Expansion loss @ tank entrance

By Bernoulli’s eqn…

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 cross-sectional area divided by the wetted perimeter. rH = (cross sectional area of channel) / (wetted perimeter of channel)

Few Examples…… For circular tube…. Annulus between two concentric pipes… For square duct…. For rectangular duct…..

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. 983 and viscosity 8 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 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/-

Write short notes on following: 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

Write short notes on following: 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