FLOW THROUGH PIPES - Sri Venkateswara College of … of... · • Shear stress and pressure...

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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