flow through pipes - sri venkateswara college of … of... · • shear stress and pressure...
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FLOW THROUGH PIPES
UNIT – 3
MOF
Contents• Viscous Flow
• Shear stress and pressure gradient relationship in Laminar Flow
– Parallel Plates
– Circular Pipes (Hagen Poiseulle’s equation)
• Losses in Pipes
– Major and Minor losses
• Darcy Weisbach’s equation• Darcy Weisbach’s equation
– Pipe roughness
– Friction factor
– Moody’s diagram
• Connection of pipes
– Pipes in series
– Pipes in parallel
ObjectivesDifferentiate between laminar and turbulent flows in pipelines.
Describe the velocity profile for laminar and turbulent flows.
Compute Reynolds number for flow in pipes.
Define the friction factor, and compute the friction losses in Define the friction factor, and compute the friction losses in pipelines.
Recognize the source of minor losses, and compute minor losses in pipelines.
Analyze simple pipelines, pipelines in series, parallel, and simple pipe networks.
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Types of Flow• Laminar Flow or Viscous Flow
• Transition Flow
• Turbulent Flow
Based on Reynolds No
Re =(ρVD/µ) - No unit
(<2000)
(2000 to 4000)
(>4000)
Re =(ρVD/µ) - No unit
Reynolds Experiment
Reynolds Experiment
Types of Flow Based on Re
Flow of Viscous Fluid in a circular pipeHagen – Poiseulle Equation
• Step -1: To determine the
– Shear Stress Distribution
– Velocity Distribution– Velocity Distribution
• Maximum Velocity
• Average Velocity
– Pressure Difference
• Step 2- Assumptions
– Fluid Follows Newton’s law of Viscosity
– There is no slip between the particles at the boundary
(The fluid particles adjacent to the pipe will have zero velocity)
Hagen- Poiseulle derivation
• Step -3 :Diagram
To determine Shear stress Distribution
• Step – 4 : Forces acting on the Fluid
To determine Shear stress Distribution
• Step -5 Equate the Forces
To determine Shear stress Distribution
• Step – 6 Boundary Condition
To determine Velocity Distribution • Step-1: Shear stress is indirectly have the velocity
component.
• Step -2 Apply Boundary condition To determine Maximum Velocity
• Step-3 : To determine discharge
To determine Average Velocity
• Step 4- Determine Avg velocity
To determine Average Velocity
To determine the Pressure Difference
∂p
Forces acting on fluid particle
Force = pressure x area= p x area
Sum of all the Forces acting on fluid particle, ∑F=0
Boundary conditions – to find constants C1 and C2.
Average Velocity = discharge / Total C/s area.
Discharge = Actual Velocity xarea of each strip
Average Velocity = discharge / Total C/s area.
Problems-1
• An oil of viscosity 9 poise and specific gravity 0.9 is flowing through a horizontal pipe of 60mm diameter. If the pressure drop in a 100m length of pipe is 1800 KN/m2 . Determine
– Rate of Flow
Centre line velocity– Centre line velocity
– Frictional Drag over the length of pipe
– Power required to maintain the flow
– Type of flow
Given Data: Viscosity = 9 poise = 0.9 Ns/m2
Specific Gravity S = 0.9 …. Density = 900 Kg/m3
Diameter of pipe = 0.06mLength of Pipe L = 100 mPressure Difference p1-p2 = 1800 kN/m2
Formula Used:
I. Find average velocity u. II. Then Discharge Q = Area x Avg VelocityIII. Centre Line velocity Umax = 2 x Avg VelocityIV. Frictional Drag Force F = shear stress x Area = IV. Frictional Drag Force F = shear stress x Area =
V. Power required to maintain flow = Work Done/time=Force x Distance / time
Power = Force x avg. velocityOr Power = Q x Pr Difference(vi) Type of flow – Reynolds No
Answers: Avg Velocity = 2.25 m/sQ = 6.636 lt/sUmax = 4.5 m/sτo = 270 N/m2
F = 5.089 NP = 11.45 kWRe = 135 <2000 , Laminar FLow
Pressure Forces
Friction Forces
Equate the Forces
m
Frictional Factor f• Laminar Flow
– f depends only on Reynolds No
• Transition Flow
– f depends on both Reynolds No and Roughness of pipe(Re and R/k)
• Turbulent Flow
– Smooth Pipe(Re)– Smooth Pipe(Re)
– Rough Pipe(R/k)
• Given Data:
Head Losses in a pipe
Problem on losses.
Section A-A
Section B-B
Applying Bernoulli Equation between sections A-A and B-B
hzVp
zVp
LB
BB
A
AA
gggg
22
22
Here, pA =pB = 0 Since it is open to atmosphereVA = 0 since fluid is static at section 1 and VB=V2 velocity in pipe 2considering datum as the centre line of pipe.ZB = 0 and ZA= 8m
PiPES in SERIES
• Total Head Loss ∆H = head loss in pipe 1 + head loss in pipe 2 + head loss in pipe 3
• Discharge will be same in all pipes.
Pipes in parallel- Penstock• Total Discharge Q = Q1 + Q2 +..+ Qn
• Head Loss will be same in all pipes.
Pipe Network
• A water distribution system consists of complex interconnected pipes, service reservoirs and/or pumps, which deliver water from the treatment plant to the consumer.
• Water demand is highly variable, whereas supply is normally constant. Thus, the distribution system must include storage elements, and must be capable of flexible operation.
• Pipe network analysis involves the determination of the pipe flow rates and pressure heads at the outflows points of the network. The flow rate and pressure heads must satisfy the continuity and energy equations.pressure heads must satisfy the continuity and energy equations.
• The earliest systematic method of network analysis (Hardy-Cross Method) is known as the head balance or closed loop method. This method is applicable to system in which pipes form closed loops. The outflows from the system are generally assumed to occur at the nodes junction.
• For a given pipe system with known outflows, the Hardy-Cross method is an iterative procedure based on initially iterated flows in the pipes. At each junction these flows must satisfy the continuity criterion, i.e. the algebraic sum of the flow rates in the pipe meeting at a junction, together with any external flows is zero.
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Hydraulic Transients:
• Rapid pressure changes inside a closed conduit in unsteady flow conditions.
Control of Hydraulic Transients:
• Changing profile of penstock
• Increasing diameter of conduit• Increasing diameter of conduit
• Provision of surge tank and pressure relief valves
Water hammer:
Change in pressure above or below normal pressure caused by sudden changes in the rate of flow of water.
Experienced in penstocks and closed hydraulic conduits.
Caused by Sudden closure Of valves or gates
Caused byOf valves or gates
Conversion of
Dynamic headKinetic head
Sudden closure Of valves or gates
Increase in pressurehead
Speed of closure
Velocity of flow Penstock length
Elastic propertyOf pipe material
To reduce water hammer :
• Penstocks should be of short length.
• Valves of turbine should be closed slowly.
• To install pressure release valve.
SURGE TANK :
• Artificial reservoir induced along the pressure conduit system.
• Introduced U/S or D/S.
• Handles excessive pressure changes in the pipe system
FUNCTIONS :
• To absorb water hammer pressure from elastic shock waves • To absorb water hammer pressure from elastic shock waves arising from sudden closure of gates or valves in the penstock.
• To provide free reservoir surface.
• To temporarily store water during load rejection.
• To provide water to turbine to pick up new load safely.
Types of Surge tanks:
1.Simple surge Tank :
References
1. Bansal, R.K., “Fluid Mechanics and Hydraulics Machines”,5th edition, Laxmi Publications Pvt. Ltd, New Delhi, 2008
2. Modi P.N and Seth "Hydraulics and Fluid Mechanicsincluding Hydraulic Machines", Standard Book House Newincluding Hydraulic Machines", Standard Book House NewDelhi. 2015.
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