# Fluid Mechanics Notes pdf

Views:

Category: Education

## Presentation Description

Module 2 for MG University

## Presentation Transcript

### slide 1:

MODULE -3 Equations of Motion : According to Newton’s Second Law of Motion the net force acting on a fluid element in the direction x is equal to mass m of the fluid element multiplied by the acceleration a x in the x-direction. Mathematically F x m. a x In the fluid flow the following forces are present: i F g gravity force ii F p the pressure force iii F v force due to viscosity iv F t force due to turbulence v F c force due to compressibility The net force F x F g x + F p x + F v x + F t x + F c x i For flow where F c is negligible the resulting equations of motion are called Reynold’s Equations of Motion. i.e F x F g x + F p x + F v x + F t x ii For flow where F t is negligible the resulting equations of motion are called Navier-Stokes Equation. i.e F x F g x + F p x + F v x + F c x iii For flow where F v is negligible the resulting equations of motion are called Euler’s Equation of Motion. i.e F x F g x + F p x + F v x + F c x Euler’s Equation for Motion Consider a stream-line flow taking place in the direction as shown in the figure below. Consider a small cylindrical element LM of Cross-Section dA and length ds. The forces acting on the cylindrical element are: P – the pressure of the fluid element at section L P+ dp – pressure at M dA - area of crossection of the fluid element

### slide 2:

2 1. Net pressure force in the direction of flow p. dA - p+dpdA - dp.dA 2. Componenet of the weight of the fluid element in the direction of flow - g. dA. ds cos dz - g. dA. ds ds r q r - g. dA. dz mass of the fluid element .dA.ds dV Acceleration of the fluid element . dt According to Netons law F m. -dp. dA - . dA. dz .dA.ds . Deviding by . dA dv ds dv v ds dt ds a dv g v ds r r r r r · \ · dp - v. dv dp i.e. v. dv + 0 is Eulers equation gdz gdz r r - + Let Bernoulli’s Equation Bernoullis Theorem states that in a steady ideal flow of an incompressible fluid the total energy at a point of the fluid is constant. The total energy consists of pressure energy kinetic energy and potential energy or datum energy. 2 2 p V constant w 2 p V Where - pressure energy head Kinetic energy head z - Datum head w 2 Mathematically z g g + + - Bernoullis’ Equation is obtained by integrating Euler’s Equation Of Motion. 2 2 dp i.e. v. dv + Constant p V i.e Constant 2 p V Constant 2g gdz z g z g r r r + + + + + ∫ ∫ ∫ Assumptions The following assumptions are made in the derivation of Bernoulli’s Equation: i The Fluid is Ideal ie. Viscosity is zero. ii The flow is Steady iii The flow is Incompressible iv The flow is Irrotational v The flow is Streamline.

### slide 3:

3 vi The only forces acting on the fluid are gravity forces and pressure forces. Orifice An Orifice is an opening in the wall or base of a vessel through which the fluid flows. Classification of Orifices Orifices are classified as follows: 1. According to size i Small Orifice ii Large Orifice 2. According to shape i Circular Orifice ii Rectangular Orifice iii Square Orifice iv Triangular Orifice 3. Shape of the upstream edge i Sharpe-edge Orifice ii Bell-Mouthed Orifice. 4. According to discharge conditions i Free Discharge Orifice ii Drowned or Submerged Orifice a Fully Submerged Orifice b Partially Submerged Orifice Flow Through an Orifice Consider a small circular Orifice with sharp edge in the sidewall of a tank discharging free in to the atmosphere. Let the center of the Orifice be at a depth of H below the free surface. As the fluid flows through the Orifice it contracts and attains a parallel form at distance d/2 from the plane of the Orifice. The point at which the streamlines first become parallel is called Vena Contracta. The cross- sectional area at the Vena Contracta is less than that at the Orifice. Considering points at 1 and 2 and applying Bernoulli’s Theorem p 1 + v 1 2 + z 1 p 2 + v 2 2 + z 2 w 2g w 2g But p 1 p 2 p a Atmospheric Pressure z 1 z 2 + H

### slide 4:

4 and v 1 0 Liquid at Section 1 is still Thus v 2 2 H 2g v 2 2gH Flow Through mouth piece Mouth pieces classified according to 1. Position : Internal and external 2. Shape : cylindrical convergent and convergent divergent 3. Nature of discharge : mouth piece running full and running free. Hydraulic Co-efficients 1. Co-efficient of contraction C c 2. Co-efficient of velocity C v 3. Co-efficient of Discharge C d 4. Co-efficient of Resistance C r 1. Co-efficient of contraction C c It is the area of the jet at Vena Contracta to the area of the Orifice. The value of C c varies from 0.61 to 0.69 depending up on the size shape of the orifice and head under which the flow takes place. Let a c Area of the Jet at Vena Contracta a Area of the Orifice Then C c a c A 2. Co-efficient of velocity C v It is the ratio of Actual Velocity V of the jet at Vena Contracta to the Theoretical Velocity V th . The value of C v varies from 0.95 to 0.99 depending up on the size shape of the orifice and head under which the flow takes place. C v Actual Velocity of the jet at Vena Contracta V Theoretical VelocityV th 3. Co-efficient of Discharge C d It is the ratio of Actual Discharge Q through an Orifice to the Theoretical DischargeV th . The value of C d varies from 0.61 to 0.65 depending up on the size shape of the orifice and head under which the flow takes place. C d Actual DischargeD Theoretical DischargeD th

### slide 5:

5 Actual Area · Actual Velocity Theoretical Area · Theoretical Velocity \ C d C c · C v 4. Co-efficient of Resistance C r The loss of head or loss of Kinetic Energy in the orifice to the head of water actual Kinetic Energy available at the exit of the Orifice is known as Co-efficient of Resistance. C r Loss Of Head in the Orifice Head Of Water Problems : 1.An orifice 50mm in diameter is discharging water under a head of 10m. If C d 0.6 and C v 0.97 find: 1. Actual Discharge 2.Actual Velocity of Jet at Vena Contracta Sol n : C d 0.6 C v 0.97 D 1 50cm.05m Area A 1 π.05 2 . 001963 m 2 4 C v Actual DischargeD 0.6 Theoretical DischargeD th Theoretical Discharge Area of Orifice · Theoretical Velocity a · 2gH .5 0.02749 m 3 /s Actual Discharge 0.6 · 0.02749 0.01649 m 3 /s C v Actual Velocity Theoretical Velocity Actual Velocity C v · Theoretical Velocity 0.97 · 2gH .5 13.58m/s Notch It is an opening provided in the side of a tank or a vessel such that the liquid surface in the tank is always below the top edge of the opening. Types of Notches 1. Rectangular Notch 2. Triangular Notch 3. Trapezoidal Notch 4. Stepped Notch Weir A Weir is a regular obstruction in an open stream over which the flow takes place. Discharge Over a Rectangular Notch or Weir

### slide 6:

6 d Let H Head of water above the apex of the notch length of the Notch c Co-efficient of Discharge Area of strip L dh Theoretical Velocity of the water flowing through the strip 2gh L d d d discharge through the strip dQ c x Area of the strip x Theoretical velocity c L. dh 2gh Q c · · H 0 3/2 d L. 2gh dh 2 c L. 2g H 3 · · · · ∫ Problems : 1.A rectangular weir 2.0m wide has a constant head of 500m. Find the discharge over the notch if C d 0.62. Sol n : Q 2 C d 2g 1/2 L H 3/2 3 1.294 m 3 /s Discharge Over a Triangular Notch or Weir

### slide 7:

7 d Let H Head of water above the apex of the notch angle of the Notch c Co-efficient of Discharge Consider a horizontal strip of water of thickness dh at a depth h from free surface From f q d ig. LN H-h tan /2 Area of the strip 2 H-h tan /2 . dh Theoretical Velocity of the water flowing through the strip 2gh discharge through the strip dQ c x Area of the strip x Theore q q d H d 0 tical velocity c 2 H-h tan /2 . dh 2gh Q c 2 H-h tan /2 . 2gh dh q q · · · ∫ 5/2 d 8 c 2g tan /2 H 15 q Problems : 1.Find the discharge over a triangular notch of angle 60 when the head over the triangular notch is 0.2m. Assume C d 0.6. Sol n : Q 8 C d tan θ/2 2g 1/2 H 5/2 15 0.01462 m 3 /s Reynolds Number R e : It is defined as the ratio of Inertia Force of a fluid and the viscous force of a fluid. Inertia Force F i Mass · Acceleration of the flowing fluid. r · Volume · Velocity Time r · Volume · Velocity Volume per Sec A · V Time r · A · V · V r AV 2 Viscous Force F v Shear Stress · Area t · A m V · A L \ R e Inertia Force r AV 2 Viscous Force m V · A L r VL m

### slide 8:

8 In case of pipe flow the linear Dimension L is taken as diameter d. Hence Reynold’s Number for pipe flow r Vd m Kinematic Viscosity ν m r R e Vd ν When R e 2000 ….then the flow is laminar When R e 4000 ….then the flow is turbulent When R e between 2000 and 4000 ….then the flow is unpredictable Problems : 1.Water is flowing through a pipe having diameters 30cm and 20cm at the bottom and upper end respectively. The intensity of pressure at the bottom end is 24.5· 10 4 N/m 2 and the upper end is 9.8· 10 4 N/m 2 . Determine the difference in datum head if the rate of flow through the pipe is 40litre/sec. Sol n : 1000 litre 1 m 3 At section 1 D 1 30cm.30m Area A 1 π.3 2 . 07065 m 2 4 V 1 p 1 24.5· 10 4 N/m 2 At section 2 D 2 20cm.20m Area A 2 π.2 2 .0314 m 2 4 V 2

### slide 9:

9 p 2 9.8· 10 4 N/m 2 A 1 V 1 A 2 V 2 40litre/sec 40 · 10 -3 m 3 /sec \ V 1 40 · 10 -3 0.5661 m/sec .07065 V 2 40 · 10 -3 1.274 m/sec .0314 Applying Bernoulli’s Equation p 1 + v 1 2 + z 1 p 2 + v 2 2 + z 2 w 2g w 2g 24.5· 10 4 + 0.5661 2 + z 1 9.8· 10 4 + 1.274 2 + z 2 9800 2 · 9.8 9800 2 · 9.8 z 2 - z 1 13.70 m. 2.Water is flowing through a tapering pipe having diameters 300mm and 150mm at sections 1 and 2 respectively. The discharge through the pipe is 40 litres/sec. The section 1 is 10 m above datum and section 2 is 6m above datum. Find the intensity of pressure at section 2 if that at section 1 is 400kN/m 2 . Sol n : 1000 litre 1 m 3 At section 1 D 1 300mm.3m Area A 1 π.3 2 . 0707 m 2 4 z 1 10m V 1 p 1 400· 10 3 N/m 2 At section 2 D 2 150cm.15m Area A 2 π.15 2 .01767 m 2 4 z 2 6m

### slide 10:

10 V 2 p 2 A 1 V 1 A 2 V 2 40litre/sec 40 · 10 -3 m 3 /sec \ V 1 40 · 10 -3 0.566 m/sec .0707 V 2 40 · 10 -3 2.264 m/sec .01767 Applying Bernoulli’s Equation p 1 + v 1 2 + z 1 p 2 + v 2 2 + z 2 w 2g w 2g 400· 10 3 + 0.566 2 + 10 p 2 + 1.274 2 + 6 9800 2 · 9.8 w 2 · 9.8 p 2 436.8 kN/m 2 Applications of Bernauli’s Principle 1. Venturi meter 2. Orifice meter 3. Pitot tube 4. Rotameter Venturimeter Venturimeter is a devise used for measuring the rate of flow of a fluid through a pipe. It consists of three parts. 1. A short Converging part 2. Throat 3. Diverging part Expression for Rate of Flow Through Venturimeter Consider a Venturimeter fitted in a horizontal pipe through which a fluid is flowing. Let

### slide 11:

11 d 1 Diameter at the inlet or Section 1 p 1 Pressure at Section 1 v 1 Velocity of fluid at section 1 and d 2 p 2 v 2 are corresponding values at Section 2 Applying Bernoulli’s Equation p 1 + v 1 2 + z 1 p 2 + v 2 2 + z 2 w 2g w 2g As the pipe is horizontal z 1 z 2 \ p 1 + v 1 2 p 2 + v 2 2 w 2g w 2g p 1 – p 2 v 2 2 – v 1 2 ----------------------1 w 2g 2g p 1 – p 2 is the difference of pressure heads at Sections 1 and 2 and is equal to h w p 1 – p 2 h w Substituting the value of h in the above equation h v 2 2 – v 1 2 ----------------------2 2g 2g a 1 v 1 a 2 v 2 \ v 1 a 2 v 2 a 1 Substituting the value of v 1 in the above equation h v 2 2 – v 1 2 2g 2g v 2 2 – a 2 v 2 2 2g 2g a 1 2 v 2 2 1 - a 2 2 2g a 1 2 v 2 2 a 1 2 - a 2 2 2g a 1 2 \ v 2 2 2gh · a 1 2 a 1 2 -a 2 2 v 2 2gh · a 1 2

### slide 12:

12 a 1 2 -a 2 2 v 2 2gh · a 1 a 1 2 -a 2 2 Discharge Q a 2 v 2 2gh · a 1 a 2 ------------------3 a 1 2 -a 2 2 Equation 3 gives the discharge under ideal conditions and is called Theoretical Discharge. Actual Discharge will be less that Theoretical Discharge. Q act C d 2gh · a 1 a 2 a 1 2 -a 2 2 Value of h given by differential U-Tube Manometer Case 1 : If the differential manometer contains a liquid which is heavier than the liquid flowing through the pipe. Let S h Specific Gravity of the heavier liquid in U-Tube S o Specific Gravity of the liquid flowing through the pipe x Difference of the heavier liquid column in U-Tube then h x S h – 1 S o Proof For Reference Only

### slide 13:

13 Let B and A be the throat and inlet of the Venturimeter which are at same levels from a horizontal reference. Let P 2 be the pressure at throat and P 1 be the pressure at Inlet. Consider the datum line at X- X. Let x Difference of mercury level in the U-Tube y Distance of center of B from the mercury level in the right limb S l Specific Gravity of Light liquid at A and B S h Specific Gravity of Heavy Liquid w l Specific Gravity of liquid flowing through the venturimeter. w w Specific Gravity of water. Pressure Head above X-X in the left limb x S l w w + y S l w w + P 2 Pressure Head above X-X in the right limb x S h w w + y S l w w + P 1 Equating the two pressure head x S h w w + y S l w w + P 1 x S l w w + y S l + P 2 P 1 - P 2 x S h w w - x S l w w x w w S h - S l ie. Difference of pressure head at A and B P 1 - P 2 x w w S h - S l S l w l w l w l w w P 1 - P 2 x w w S h - S l w l S l w w w l w l w l P 1 - P 2 x w w S h - S l w l S l w w S l w w P 1 - P 2 h xS h - 1 Meters of Flowing Liquid w l S l Note Expression h A - h B hS g - S 1 is in Meters Of Water. But above expression is used for finding the differential head whose value is in Meters of Flowing Liquid. To find the difference in pressure using both the cases 1. h A - h B hS g - S 1 in Meters Of Water so Pressure difference hS g - S 1 · w w 2. P 1 - P 2 h xS h - 1 Meters of Flowing Liquid w l S l so Pressure difference xS h - 1 · w liquid S l Case 2 :

### slide 14:

14 Let the differential manometer contains a liquid which is heavier than the liquid flowing through the pipe. Let Sl h Specific Gravity of the light liquid in U-Tube S o Specific Gravity of the liquid flowing through the pipe x Difference of the heavier liquid column in U-Tube then h x1- S l S o Types of Venturimeters Venturimeters are classified as 1. Horizontal Venturimeter 2. Vertical Venturimeter 3. Inclined Venturimeter Vertical and Inclined Venturimeter The same formula for discharge as used for Horizontal Venturimeter holds good in these cases as well. But h p 1 – p 2 + z 2 - z 1 w Problems : 1.A horizontal venturimeter with inlet diameter 200mm and throat diameter 100mm is used to measure the flow of water. The pressure at inlet is 0.18N/mm 2 and the vacuum pressure head at the throat is 280mm of mercury. Find the rate of flow. The value of C d may be taken as 0.98. Sol n : C d 0.98 D 1 200mm.20m Area A 1 π.2 2 . 0314 m 2 4 D 2 100mm.10m Area A 2 π.1 2 . 00785 m 2 4 Pressure at Inlet p 1 180· 10 3 N/m 2 p 1 180· 10 3 18.3m w 9800 w water · h water w mercury · h mercury Dividing by w water Specific Gravity of Water · h water Specific Gravity of Mercury · h mercury h water Specific Gravity of Mercury · h mercury Specific Gravity of Water

### slide 15:

15 p 2 -280 mm of Mercury w -28 m of Mercury -0.28 · 13.6 -3.8 m of water h p 1 – p 2 18.3 – -3.8 22.1 m w Q act C d 2gh · a 1 a 2 a 1 2 -a 2 2 Q 0.165m 3 /s 2.A horizontal venturimeter with inlet diameter 200 and throat diameter 100mm is used to measure the flow of water. The reading of the differential manometer connected to the inlet is 180mm of Mercury. Find the rate of flow. The value of C d may be taken as 0.98. Sol n : C d 0.98 D 1 200mm.20m Area A 1 π.2 2 . 0314 m 2 4 D 2 100mm.10m Area A 2 π.1 2 . 00785 m 2 4 h p 1 – p 2 h x S h – 1 0.18 13.6 – 1 2.268m w S o 1 Q act C d 2gh · a 1 a 2 a 1 2 -a 2 2 Q 0.0528m 3 /s 3.A 300· 150mm venturimeter is provided in a vertical pipeline carrying oil of specific gravity 0.9 flow being upward. The difference in elevation of the throat section and entrance section of the venturimeter is 300mm. The differential U-Tube mercury manometer shows a gauge deflection of 250mm. Calculate 1 The Discharge of oil and 2The pressure difference between entrance section and throat section. Sol n : C d 0.98 D 1 300mm.3m Area A 1 π.3 2 . 07 m 2 4 D 2 150mm.15m Area A 2 π.15 2 . 01767 m 2 4 Discharge h p 1 – p 2 h x S h – 1 0.25 13.6 – 1 3.53m of oil

### slide 16:

16 w S o .9 Q act C d 2gh · a 1 a 2 a 1 2 -a 2 2 Q 0.1489m 3 /s Pressure Difference p 1 – p 2 + z 2 - z 1 3.53 w z 2 - z 1 300 p 1 – p 2 - 0.3 3.53 w p 1 – p 2 33.8 kN/ m 2 Orifice Meter Orifice Meter is a device used for measuring discharge of fluid through a pipe. It consists of a flat circular plate having a circular hole which is concentric with the pipe. The orifice diameter may vary from 0.4 to 0.8 times the diameter of the pipe. Let d 1 Diameter at the inlet or Section 1 p 1 Pressure at Section 1 v 1 Velocity of fluid at section 1 and d 2 p 2 v 2 are corresponding values at Section 2 Applying Bernoulli’s Equation p 1 + v 1 2 + z 1 p 2 + v 2 2 + z 2 w 2g w 2g p 1 + z 1 – p 2 + z 2 v 2 2 – v 1 2 ----------------------1 w w 2g 2g

### slide 17:

17 p 1 + z 1 – p 2 + z 2 h Differential Head w w Substituting the value of h in the above equation h v 2 2 – v 1 2 ----------------------2 2g 2g C c a 2 a 0 C c Coefficient of contraction \ a 2 C c a 0 Also a 1 v 1 a 2 v 2 v 1 a 2 v 2 a 1 \ v 1 C c a 0 v 2 a 1 Substituting the value of v 1 in the above equation h v 2 2 – C c a 0 v 2 2 2g 2g a 1 2 h v 2 2 1 - a 0 2 C c 2 2g a 1 2 v 2 2gh 1 - a 0 2 C c 2 a 1 2 The Discharge Q a 2 v 2 C c a 0 v 2 ----------------------4 C d C c 1 - a 0 2 a 1 2 1 - a 0 2 C c 2 a 1 2 C c C d 1 - a 0 2 C c 2 a 1 2 1 - a 0 2 a 1 2 Substituting the value of C c and v 2 in equation 4 Q a 0 C d 1 - a 0 2 C c 2 · 2gh a 1 2 1 - a 0 2 1 - a 0 2 C c 2 a 1 2 a 1 2

### slide 18:

18 Q a 0 C d 2gh a 0 a 1 C d 2gh 1 - a 0 2 C c 2 a 1 2 - a 0 2 a 1 2 Problems : 1.Water is flowing through a pipeline of 50cm at 30 C. An orifice is placed in the pipe line to measure the flow rate. Orifice diameter is 20cm. If the manometer reads 30cm of Mercury calculate the water flow rate and velocity of the fluid through the pipe. Sol n : C d 0.6 D 1 50cm.5m Area A 1 π.5 2 . 1963 m 2 4 D 0 20mm.20m Area A 2 π.2 2 . 0314 m 2 4 Q act C d 2gh · a 1 a 0 a 1 2 -a 0 2 Q 0.1655m 3 /s Velocity of water through the pipe Q 0.1655 0.843 m/s a 1 0.1938 Difference Between Venturimeter and Orifice Meter Venturimeter Orifice Meter Used for Measuring the Flow Rate of all Incompressible Fluids. Used for Measuring the Flow Rate of Liquids. C d Value is More C d Value is Less More Accurate Less Accurate Expensive Less Expensive Pitot Tube Pitot tube is a device used to measure the velocity of flow at any point in a pipe or channel. It works on the principle that if the velocity of flow at any point becomes zero the pressure there is increased due to the conversion of the kinetic energy into pressure energy.

### slide 19:

19 Pitot Tube consists of a glass tube whose lower end is bent at right angle. The lower end is directed in the upstream direction. The liquid rises in the tube due to conversion of kinetic energy into pressure energy. The velocity is determined by measuring the rise of liquid in the tube. Consider two points 1 and 2 at the same level in such a way that point 2 is just at inlet of the Pitot Tube and point 1 is far away from the tube. Let p 1 Intensity of Pressure at point 1 v 1 Velocity of Flow at point 1 p 2 Intensity of Pressure at point 2 v 2 Velocity of Flow at point 2 H Depth of tube in the liquid h Rise of liquid in the tube above the free surface. Applying Bernoulli’s Equation p 1 + v 1 2 + z 1 p 2 + v 2 2 + z 2 ----------------------A w 2g w 2g z 1 z 2 0 as 1 and 2 are on the same line p 1 H Pressure Head at 1 w p 2 h + H Pressure Head at 2 w Substituting these values in Equation A H + v 1 2 h + H 2g \ h v 1 2 2g v 1 2gh

### slide 20:

20 This is the theoretical velocity and the actual velocity is given by v 1 act C v 2gh C v is the Co-efficient of Pitot Tube. Simple Pitot Tube a with static tube only b with Static and Stagnation Tube Problems : 1.A sub-marine fitted with a pitot tube moves horizontally in sea. Its axis is 12m below the surface of water. The pitot tube fixed infront of the sub-marine and along its axis is connected to the two limbs of a U-Tube containing mercury the reading of which is found to be 200mm. Find the speed of the sub-marine. Take specific-gravity of sea water 1.025 times fresh water. Sol n : Reading of the manometer y 200 mm 0.2m of mercury. Specific Gravity of mercury S h 13.6 Specific Gravity of sea water S o 1.025 h x S h – 1 0.2 13.6 - 1 2.45 S o 1.025 Velocity of sub-marine v 1 2gh 6.93 m/s 2.Petroleum oil of specific gravity 0.93 and viscosity flows through a horizontal 5 cm pipe. A pitot tube is inserted at the center of the pipe and its leads are filled with the same oil and attached to a U- Tube containing water. The reading on the manometer is 10cm. Calculate the volumetric flow of oil. The co-efficient of pitot tube is 0.98. Sol n : Reading of the manometer y 10 cm 0.1m of mercury. Specific Gravity of mercury S h 13.6 Specific Gravity of oil S o .09 h x S h – 1 0.1 13.6 - 1 1.411 S o 0.9 Velocity of sub-marine

### slide 21:

21 v 1 2gh 5.156 m/s Q A · V π.05 2 · 5.156 0.01m 3 /s 4 Rotameter : Rotameter consists of a transparent tube kept in vertical position. Inside the tube there is a float which moves up and down in the tube depending upon the intensity of the flow. A calibrated scale on the side of the tube indicates the rates of flow. Slots on the head of the float keep the float in vertical central position. Measurement of discharge using Rotameter is not accurate and so it is used for small variations of discharge.