# 2 Internal Fluid Flow

Size: px
Start display at page:

Transcription

1 Internal Fluid Flow.1 Definitions Fluid Dynamics The study of fluids in motion. Static Pressure The pressure at a given point exerted by the static head of the fluid present directly above that point. Static pressure is related to motion on a molecular scale. Dynamic or Velocity Pressure Dynamic pressure is related to fluid motion on a large scale i.e. fluid velocity. 1

2 Stagnation Pressure Total Pressure The sum of the static pressure plus the dynamic pressure of a fluid at a point. Streamline An imaginary line in a moving fluid across which, at any instant, no fluid is flowing. ie it indicates the instantaneous direction of the flow. Figure.1 the stream tube Stream tube A bundle of neighbouring streamlines may be imagined to form a stream tube (not necessarily circular) through which the fluid flows.

3 Control volume A fixed volume in space through which a fluid is continuously flowing. The boundary of a control volume is termed the control surface. The size and shape is entirely arbitrary and normally chosen such that it encloses part of the flow of particular interest. Classification of Fluid behaviour a) Steady or unsteady A flow is termed steady if its properties do not vary with time. A flow is termed unsteady if properties at a given point vary with time. Quasi-steady flow is essentially unsteady but its properties change sufficiently slowly with respect to time, at a given point, that a series of steady state solutions will approximately represent the flow. b) Uniform or Non-uniform A uniform flow is one in which properties do not vary from point to point over a given cross-section. Non-uniform flow has its properties changing with respect to space in a given cross-section. c) One-dimensional or Multi-dimensional One-dimensional flow, is one in which the direction and magnitude of the velocity at all points are identical. Variation of velocity in other directions is so small that they can be neglected. eg. flow of water in small bore pipe at low flow rates. Two-dimensional flow is one in which the velocity has two main components. Three-dimensional flow is one in which the flow velocity has significant components in all three directions. d) Viscid or Inviscid This some time distinguished as Viscid and inviscid flow in relation to the viscous forces whether they are neglected or taken into account e) Compressible or Incompressible If the changes in density are relatively small, the fluid is said to be incompressible. If the changes in density are appreciable, in case of the fluid being subjected to relatively high pressures, the fluid has to be treated as Compressible. f) Ideal or Real An ideal fluid is both inviscid and incompressible. This definition is useful in forming analytical solution to fluid flow problems. Fluids in reality are viscous and compressible. Thus, the effect of compressibility and viscosity must be considered for accurate analysis. It must be stressed that in most common engineering applications at standard pressure and temperature, water can be assumed incompressible and inviscid. The assumption of ideal fluid can help to formulate a solution, an approximate solution, still better than no solution. 3

4 . Conservation of Mass The continuity equation applies the principle of conservation of mass to fluid flow. Consider a fluid flowing through a fixed volume tank having one inlet and one outlet as shown below. Figure. conservation of mass If the flow is steady i.e no accumulation of fluid within the tank, then the rate of fluid flow at entry must be equal to the rate of fluid flow at exit for mass conservation. If, at entry (or exit) having a cross-sectional area A (m ), a fluid parcel travels a distance dl in time dt, then the volume flow rate (V, m 3 /s) is given by: V = (A. dl)/ t but since dl/ t is the fluid velocity (v, m/s) we can write: Q = V x A The mass flow rate (m, kg/s) is given by the product of density and volume flow rate i.e. m = ρ.q = ρ.v.a Between two points in flowing fluid for mass conservation we can write: m 1 = m or ρ 1 V 1 A 1 = ρ V A (.1) If the fluid is incompressible i.e. ρ 1 = ρ then: V 1 A 1 = V A (.) Hence an incompressible flow in a constant cross-section will have a constant velocity. For branched systems the continuity equation implies that the sum of the incoming fluid mass (or volume) flow rates must equal the sum of the outgoing mass (or volume) flow rates. Worked Example.1 Air enters a compressor with a density of 1. kg/m 3 at a mean velocity of 4 m/s in the 6 cm x 6 cm square inlet duct. Air is discharged from the compressor with a mean velocity of 3 m/s in a 5 cm diameter circular pipe. Determine the mass flow rate and the density at outlet. 4

5 Solution: Given: ρ 1 = 1. kg/m 3, V 1 = 4 m/s, V = 3 m/s A 1 = 0.06 x 0.06 = m A = πd = 4 = m The mass flow rate is: m = ρ 1 A 1 V 1 = 1. x x 4 = 17.8 x 10-3 kg/s Conservation of mass between sections 1 and implies that: ρ 1 A 1 V 1 = ρ A V 5

6 Hence the density at section is calculated:.3 Conservation of Energy There are three forms of non-thermal energy for a fluid at any given point:- The kinetic energy due to the motion of the fluid. The potential energy due to the positional elevation above a datum. The pressure energy, due to the absolute pressure of the fluid at that point. Conservation of energy necessitates that the total energy of the fluid remains constant, however, there can be transformation from one form to another. If all energy terms are written in the form of the head (potential energy), ie in metres of the fluid, then: p ρg represents the pressure head (sometimes known as flow work ) represents the velocity head (also known as kinetic energy) The energy conservation, thus, implies that between any two points in a fluid This equation is known as the Bernoulli equation and is valid if the two points of interest 1 & are very close to each other and there is no loss of energy. (.3a) In a real situation, the flow will suffer a loss of energy due to friction and obstruction between stations 1 &, hence where h L is the loss of energy between the two stations. (.3b) When the flow between stations 1 & is caused by a pump situated between the two stations, the energy equation becomes: (.3c) Where h p is the head gain due to the pump. 6

7 Worked Example. A jet of water of 0 mm in diameter exits a nozzle directed vertically upwards at a velocity of 10 m/s. Assuming the jet retains a circular cross - section, determine the diameter (m) of the jet at a point 4.5 m above the nozzle exit. Take ρ water = 1000 kg/m 3. Solution: Bernoulli equation: Given: v, z 1 = 0 (Datum) z = 4.5 m. p 1 = p (both atmospheric). The energy equation reduces to: From continuity equation: mywbut.com 7

8 Hence.4 Flow Measurement There are a large number of devices for measuring fluid flow rates to suit different applications. Three of the most commonly encountered restriction methods will be presented here. Restriction methods of fluid flow are based on the acceleration or deceleration of the fluid through some kind of nozzle, throat or vena contracta. The theoretical analysis applies the continuity and Bernoulli equations to an ideal fluid flow between points 1 and thus:- Start with Bernoulli equation: Rearranging p1 v1 + ρg g + z 1 p v = + ρg g + z 8

9 Then use the continuity equation V 1. A 1 = V. A Therefore (V 1 /V ) = (A /A 1 ) Substituting into the rearranged Bernoulli equation and solving for V we have:- (.4) The theoretical volume flow rate is Q = A V And the theoretical mass flow rate is m = ρ A V The above values are theoretical because ideal fluid flow conditions were assumed. Actual flow rate values are obtained by multiplying the theoretical values by a meter discharge coefficient C d to account for frictional and obstruction losses encountered by the fluid in its passage through the meter. The energy losses manifest themselves as a greater pressure drop (P 1 - P ) then that predicted by the theory. It can be shown that (.5) (a) The Venturi meter The Venturi meter has a converging section from the initial pipe diameter down to a throat, followed by a diverging section back to the original pipe diameter. See figure. Figure. the Venturi tube Differential pressure measurements are taken between the inlet (1) and throat () positions. The geometry of the meter is designed to minimize energy losses (C d > 0.95). 9

10 (b) The Orifice meter An orifice meter is a flat plate, with a hole which may be square edged or bevelled, inserted between two flanges in a pipe line. In this instance positions (1) and () are as shown below. Orifice plates have a simple construction and are therefore inexpensive but they suffer from high energy losses (C d = 0.6). Figure.3 the Orifice meter (c) The Pitot-static tube A slender concentric tube arrangement, aligned with the flow, used to measure flow velocity by means of a pressure difference. See figure below. The outer tube is closed in the flow direction but has sidewall holes to enable the measurement of static pressure. The inner tube is open in the direction of the fluid flow and is thus experiencing the total (static + dynamic) pressure of the fluid flow. It is assumed that the fluid velocity is rapidly brought to zero upon entry to the inner tube with negligible friction (C d ~1). The pressure difference between the tubes is applied to a U tube manometer which will therefore indicate the velocity pressure. Start with Bernoulli equation: p 1 ρg v1 + g + z 1 = p ρg v + g + z Since the Pitot-static tube is mounted horizontally, the z-terms will cancel out, and the static end is motionless, ie V = 0. It can be shown that the duct velocity V 1 is given by:- (.6) 10

11 Figure.4 Pitot-static tube Worked Example.3 A Venturi meter fitted in a 15 cm pipeline has a throat diameter of 7.5 cm. The pipe carries water, and a U-tube manometer mounted across the Venturi has a reading of 95. mm of mercury. Determine: 1. the pressure drop in Pascal s, indicated by the manometer. the ideal throat velocity (m/s) 3. the actual flow rate (l/s) if the meter C D is

12 Solution: (i) p 1 - p = ρ m x g x h = x 9.81 x = 1701 Pa (ii) v ( p1 p ) ( 1701) = = = m / s ρ [ 1 ( A / A ) ] 1000[ 1 ( )] 1 (iii) Q = C d V A = x 5.06 x = 0.04 m 3 /s =.4 l/s.5 Flow Regimes Consider the variation in velocity across the cross-section of a pipe containing a fluid in motion. There is no motion of fluid in direct contact with the pipe wall, and the velocity of the fluid stream increases in a direction away from the walls of the pipe. In 1839, Hagen (USA) observed that the fluid moves in layers with a velocity gradient. He observed that the velocity gradient in a circular pipe follows a parabolic law, at low flow rates. This type of flow is termed LAMINAR. When the flow rate of the fluid stream is high, the velocity distribution had a much flatter shape and this type of flow is known as TURBULENT. The average velocity producing turbulent flow is greater than that for a laminar flow of a given fluid in a given duct. For both flows, the build-up of velocity is along the radius of the pipe, and the maximum velocity occurs at the centre line. 1

13 Osborne Reynolds demonstrated experimentally in 1883 (Manchester) that under laminar flow, the fluid streamlines remain parallel. This was shown with the aid of a dye filament injected in the flow which remained intact at low flow velocities in the tube. As the flow velocity was increased (via a control valve), a point was reached at which the dye filament at first began to oscillate then broke up so that the colour was diffused over the whole cross-section indicating that particles of fluid no longer moved in an orderly manner but occupied different relative positions in successive cross-sections downstream. Reynolds also found that it was not only the average pipe velocity V which determined whether the flow was laminar or turbulent, but that the density (ρ) and viscosity (µ) of the fluid and the pipe diameter (D), also determined the flow regime. He proposed that the criterion which determined the type of regime was the dimensionless group (ρvd/µ). This group has been named the Reynolds number (Re) as a tribute to his contribution to Fluid Mechanics. Re = ρ V D/µ (.7) Based on Reynolds number the flow can be distinguished into three regimes for pipe flow: Laminar if Re < 000 Transitional if 000 < Re < 4000 Turbulent if Re > 4000 Re = 000, 4000 are the lower and upper critical values..6 Darcy Formula Consider a duct of length L, cross-sectional area A c, surface area A s, in which a fluid of density, is flowing at mean velocity V. The forces acting on a segment of the duct are that due to pressure difference and that due to friction at the walls in contact with the fluid. If the acceleration of the fluid is zero, the net forces acting on the element must be zero, hence According to Newton s Second Law of Motion for a constant velocity flow: F = 0 The force due to pressure on either side of the section is equal to the friction force resisting the flow: (P 1 -P ). A c - (f ρ V /). A s = 0 13

14 Where the pressures act normal to the flow direction on the area of cross-section Ac, and the frictional force acts on the circumferential wall area As, separating the fluid and the pipe s surface. Let h f denote the head lost (m) due to friction over a duct length L, ie p 1 - p = ρ g h f Substituting we get h f = f. (A s /A c ). V /g For a pipe A s /A c = π D L /π D /4 = 4L/D h f = (4 fl/d).v /g (.8) This is known as Darcy formula..7 The Friction factor and Moody diagram The value of the friction factor (f) depends mainly on two parameters namely the value of the Reynolds number and the surface roughness. 14

15 For laminar flow (ie Re < 000), the value of the friction factor is given by the following equation irrespective of the nature of the surface: (.9) While for a smooth pipe with turbulent (i.e. Re > 4000) flow, the friction factor is given by: (Blasius equation) (.10) For Re > 000 and Re < 4000, this region is known as the critical zone and the value of the friction factor is uncertain and not quoted on the Moody diagram (Figure.5). In the turbulent zone, if the surface of the pipe is not perfectly smooth, then the value of the friction factor has to be determined from the Moody diagram. The relative roughness (k/d) is the ratio of the average height of the surface projections on the inside of the pipe (k) to the pipe diameter (D). In common with Reynolds number and friction factor this parameter is dimensionless. Values of k are tabled on the Moody chart for a sample of materials. 15

16 Figure.5 the Moody Chart/diagram 16

17 Worked Example.4 Water flows in a 40mm diameter commercial steel pipe (k = x 10-3 m) at a rate of 1 litre/s. Determine the friction factor and head loss per metre length of pipe using: 1. The Moody diagram. Smooth pipe formulae. Compare the results. Take: ρ = 1000 kg/m 3, µ = 1 x 10-3 kg/ms Solution: V = Q/A = /( 1.56 x 10-3 ) = m Re = ρ V D/µ = 1000 x x 0.04/ 1 x 10-3 = i.e. turbulent 1. Moody diagram k/d = x 10-3 /0.04 = From intersection of k/d and Re values on Moody diagram read off f =

18 Therefore. Using Blasius equation for smooth pipe: i.e. 9% less than Moody. Note that if the pipe is assumed smooth, the friction factor from the Moody diagram would be f = which is closer to the Blasius value..8 Flow Obstruction Losses When a pipe changes direction, changes diameter or has a valve or other fittings there will be a loss of energy due to the disturbance in flow. This loss of energy (h o ) is usually expressed by: (.11) Where V is the mean velocity at entry to the fitting and K is an empirically determined factor. Typical values of K for different fittings are given in the table below: 18

19 Obstruction tank exit tank entry smooth bend Mitre bend Mitre bend with guide vanes 90 degree elbow 45 degree elbow Standard T Return bend Strainer Globe valve, wide open Angle valve, wide open Gate valve, wide open 3 / 4 open 1 / open 1 / 4 open Sudden enlargement Conical enlargement: 6 o (total included angle) 10 o 15 o 5 o Sudden contractions: area ratio 0. (A /A 1 ) K Table.1: Obstruction Losses in Flow Systems.9 Fluid Power The fluid power available at a given point for a fluid is defined as the product of mass, acceleration due to gravity and the fluid head, and since the mass flow rate is defined as the volume flow rate multiplied by the fluid density, the Fluid power therefore can be expressed as: P = ρ. g. Q.h tot (.1a) For a pump, h tot represents the head required to overcome pipe friction (h f ), obstruction losses (h o ) and to raise the fluid to any elevation required (h z ). 19

20 ie h tot = h z + h f + h o (.13a) Note: If the delivery tank operates at pressure in excess of the supply tank an additional term (h p ) must be added to the required head equation as this pressure rise must also be supplied by the pump. If the pump efficiency η p is introduced, the actual pump head requirement is: P = ρ. g. Q.h tot / η p (.1b) For a turbine with efficiency η t, the power output is given by: P = ρ. g. Q.h tot xη t (.1c) Where h tot = h z - (h f + h o ) (.13b) Worked Example.5 Determine the input power to an electric motor (η m = 90%) supplying a pump (η p = 80%) delivering 50 l/s of water (ρ = 1000 kg/m 3, µ = kg/ms) from tank1 to tank as shown below if the pipeline length is 00m long, of 150 mm diameter galvanised steel ( assumed surface roughness k=0.15mm). 0

21 Solution: Three bends (each K=0.9), tank entry (k=0.5), exit loss (k=1) and one valve (k=5) Input power.10 Fluid Momentum Knowledge of the forces exerted by moving fluids is important in the design of hydraulic machines and other constructions. The Continuity and Bernoulli relationships are not sufficient to solve forces acting on bodies in this case and the momentum principle derived from Newton s Laws of motion is also required. 1

22 Momentum is defined as the product of mass and velocity, and represents the energy of motion stored in the system. Momentum is a vector quantity and can only be defined by specifying its direction as well as magnitude. Newton s Second Law of Motion The rate of change of momentum is proportional to the net force acting, and takes place in the direction of that force. i.e. F = dm dt (.14) Since M = m. V Newton s Third Law of Motion To every action there is a reaction equal in magnitude and opposite in direction. Application of Momentum Equation Consider horizontal jet impinging a surface tangentially at a steady state. Resolving horizontally we have for the x-component Resolving vertically we have for the y-component

23 The resultant force acting on the solid surface due to the jet is given by If smooth, then k = 1 and F x = ρa V 1 (1 - cos θ), F y = ρa V 1 sin θ Special cases: The angle of the striking jet has a very important effect on the force, 3 different angles are illustrated below: 3

24 Worked Example.6 A jet of water having a diameter of 7.5 cm and a velocity of 30 m/s strikes a stationary a flat plate at angle θ = 30 o as shown below. Determine the magnitude and direction of the resultant force on the plate assuming there is no friction between the jet and the plate. Take ρ water = 1000 kg/m 3. Solution: A = m Smooth i.e. k = 1, V 1 = V = 30 m/s F x = ρav 1 (1 - cos θ) = 1000 x x 30 (1 - cos 30) = 533 N 4

25 F y = ρa V 1 sin θ = 1000 x x 30 x sin 30 = 1989 N Flow forces on a Reducer Bend The change of momentum of a fluid flowing through a pipe bend induces a force on the pipe. The pressures are to be considered in this case since the reducer bend is part of flowing system which is not subjected to atmospheric conditions. x - Momentum y - Momentum -F x + p 1 A 1 - p A cos θ = ρ V (V cos θ -V 1 ) -F y + p A sin θ = ρ Q (-V sin θ - 0) The total force From continuity equation: V = V1x A A 1 For A 1 = A and θ = 90 o 5

26 p F = ( ρ Q 1 (ρ V 1 + p 1 + p ) + p 1 A 1 [1 - p Fy The force acting at an angle θ R = tan -1 with the x - axis F x Worked Example.7 A bend in a horizontal pipeline reduces from 600 mm to 300 mm whilst being deflected through 60 o. If the pressure at the larger section is 50 kpa, for a water flow rate of 800 l/s determine the magnitude and direction of the resulting force on the pipe. Take ρ water = 1000 kg/m 3 ]) 1/ 6

27 Solution: From Continuity Q= V 1 A 1 V 1 = 0.8/0.8 =.83 m/s Q = V A V = 0.8/0.07 = 11.4 m/s From Bernoulli equation: From Momentum equation: F x = p 1 A 1 - p A cos θ - ρq (V cos θ - V 1 ) = [50 x 10 3 x 0.8] - [188.8 x 10 3 x 0.07 x 0.5] - [1000 x 0.8((11.4 x 0.5) -.83)] = N F y = p A sin θ - ρq (- V sin θ - 0) = [ x 10 3 x 0.07 x 0.866] - [1000 x 0.8(-9.889)] = N Worked Example.8 A siphon has a uniform circular bore of 75 mm diameter and consists of a bent pipe with its crest 1.4 m above water level and a discharge to the atmosphere at a level m below water level. Find the velocity of flow, the discharge and the absolute pressure at crest level if the atmospheric pressure is 98.1 kn/m. Neglect losses due to friction. 7

28 Solution: Bernoulli equation between 1-3: z g v g p z g v g p + + = + + ρ ρ z 1 =, z 3 = 0 (level 3 is assumed datum). p 1 = p 3 (both atmospheric). And V 1 = 0 The energy equation reduces to: The flow rate is calculated: Applying Bernoulli equation between 1 and : mywbut.com 8

29 .11 Tutorial Problems.1 A 0 mm dam pipe forks, one branch being 10 mm in diameter and the other 15 mm in diameter. If the velocity in the 10 mm pipe is 0.3 m/s and that in the 15 mm pipe is 0.6 m/s, calculate the rate of flow in cm 3 /s and velocity in m/s in the 0 mm diameter pipe. (19.6 cm 3 /s, m/s). Water at 36 m above sea level has a velocity of 18 m/s and a pressure of 350 kn/m. Determine the potential, kinetic and pressure energy of the water in metres of head. Also determine the total head. Ans (35.68 m, 16.5 m, 36 m, 88. m).3 The air supply to an engine on a test bed passes down a 180 mm diameter pipe fitted with an orifice plate 90 mm diameter. The pressure drop across the orifice is 80 mm of paraffin. The coefficient of discharge of the orifice is 0.6 and the densities of air and paraffin are 1. kg/m 3 and 830 kg/m 3 respectively. Calculate the mass flow rate of air to the engine. Ans (0.16 kg/s) 9

30 .4 Determine the pressure loss in a 100 m long, 10 mm diameter smooth pipe if the flow velocity is 1 m/s for: a) air whose density 1.0 kg/m 3 and dynamic viscosity 1 x 10-5 Ns/m. b) water whose density kg/m 3 and dynamic viscosity 1 x 10-3 Ns/m. Ans: (30 N/m, 158 kn/m )..5 Determine the input power to an electric motor (η m = 90%) supplying a pump (η p = 90%) delivering 50 l/s of water (ρ = 1000 kg/m 3, µ = kg/ms) between two tanks with a difference in elevation of 50m if the pipeline length is 100m long in total of 150 mm diameter, assume a friction factor of and neglect minor losses. Ans: (33.6 kw)..6 A jet of water strikes a stationary flat plate perpendicularly, if the jet diameter is 7.5 cm and its velocity upon impact is 30 m/s, determine the magnitude and direction of the resultant force on the plate, neglect frictional effect and take water density as 1000 kg.m 3. Ans (3970 N).7 A horizontally laid pipe carrying water has a sudden contraction in diameter from 0.4m to 0.m respectively. The pressure across the reducer reads 300 kpa and 00 kpa respectively when the flow rate is 0.5 m 3 /s. Determine the force exerted on the section due to the flow, assuming that friction losses are negligible. Ans: (5.5 kn)..8 A siphon has a uniform circular bore of 75 mm diameter and consists of a bent pipe with its crest 1.8 m above water level and a discharge to the atmosphere at a level 3.6 m below water level. Find the velocity of flow, the discharge and the absolute pressure at crest level if the atmospheric pressure is 98.1 kn/m. Neglect losses due to friction. Ans ( m 3 /s, 45.1 kn/m ) 30

### EXPERIMENT No.1 FLOW MEASUREMENT BY ORIFICEMETER

EXPERIMENT No.1 FLOW MEASUREMENT BY ORIFICEMETER 1.1 AIM: To determine the co-efficient of discharge of the orifice meter 1.2 EQUIPMENTS REQUIRED: Orifice meter test rig, Stopwatch 1.3 PREPARATION 1.3.1

### Lesson 6 Review of fundamentals: Fluid flow

Lesson 6 Review of fundamentals: Fluid flow The specific objective of this lesson is to conduct a brief review of the fundamentals of fluid flow and present: A general equation for conservation of mass

### 5 ENERGY EQUATION OF FLUID MOTION

5 ENERGY EQUATION OF FLUID MOTION 5.1 Introduction In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics. The pertinent laws

### Chapter Four fluid flow mass, energy, Bernoulli and momentum

4-1Conservation of Mass Principle Consider a control volume of arbitrary shape, as shown in Fig (4-1). Figure (4-1): the differential control volume and differential control volume (Total mass entering

### Mass of fluid leaving per unit time

5 ENERGY EQUATION OF FLUID MOTION 5.1 Eulerian Approach & Control Volume In order to develop the equations that describe a flow, it is assumed that fluids are subject to certain fundamental laws of physics.

### V/ t = 0 p/ t = 0 ρ/ t = 0. V/ s = 0 p/ s = 0 ρ/ s = 0

UNIT III FLOW THROUGH PIPES 1. List the types of fluid flow. Steady and unsteady flow Uniform and non-uniform flow Laminar and Turbulent flow Compressible and incompressible flow Rotational and ir-rotational

### Chapter 4 DYNAMICS OF FLUID FLOW

Faculty Of Engineering at Shobra nd Year Civil - 016 Chapter 4 DYNAMICS OF FLUID FLOW 4-1 Types of Energy 4- Euler s Equation 4-3 Bernoulli s Equation 4-4 Total Energy Line (TEL) and Hydraulic Grade Line

### FE Fluids Review March 23, 2012 Steve Burian (Civil & Environmental Engineering)

Topic: Fluid Properties 1. If 6 m 3 of oil weighs 47 kn, calculate its specific weight, density, and specific gravity. 2. 10.0 L of an incompressible liquid exert a force of 20 N at the earth s surface.

### UNIT I FLUID PROPERTIES AND STATICS

SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code : Fluid Mechanics (16CE106) Year & Sem: II-B.Tech & I-Sem Course & Branch:

### FLUID MECHANICS D203 SAE SOLUTIONS TUTORIAL 2 APPLICATIONS OF BERNOULLI SELF ASSESSMENT EXERCISE 1

FLUID MECHANICS D203 SAE SOLUTIONS TUTORIAL 2 APPLICATIONS OF BERNOULLI SELF ASSESSMENT EXERCISE 1 1. A pipe 100 mm bore diameter carries oil of density 900 kg/m3 at a rate of 4 kg/s. The pipe reduces

### CHAPTER 3 BASIC EQUATIONS IN FLUID MECHANICS NOOR ALIZA AHMAD

CHAPTER 3 BASIC EQUATIONS IN FLUID MECHANICS 1 INTRODUCTION Flow often referred as an ideal fluid. We presume that such a fluid has no viscosity. However, this is an idealized situation that does not exist.

### Experiment- To determine the coefficient of impact for vanes. Experiment To determine the coefficient of discharge of an orifice meter.

SUBJECT: FLUID MECHANICS VIVA QUESTIONS (M.E 4 th SEM) Experiment- To determine the coefficient of impact for vanes. Q1. Explain impulse momentum principal. Ans1. Momentum equation is based on Newton s

### Experiment (4): Flow measurement

Experiment (4): Flow measurement Introduction: The flow measuring apparatus is used to familiarize the students with typical methods of flow measurement of an incompressible fluid and, at the same time

### FACULTY OF CHEMICAL & ENERGY ENGINEERING FLUID MECHANICS LABORATORY TITLE OF EXPERIMENT: MINOR LOSSES IN PIPE (E4)

FACULTY OF CHEMICAL & ENERGY ENGINEERING FLUID MECHANICS LABORATORY TITLE OF EXPERIMENT: MINOR LOSSES IN PIPE (E4) 1 1.0 Objectives The objective of this experiment is to calculate loss coefficient (K

### Fluid Mechanics. du dy

FLUID MECHANICS Technical English - I 1 th week Fluid Mechanics FLUID STATICS FLUID DYNAMICS Fluid Statics or Hydrostatics is the study of fluids at rest. The main equation required for this is Newton's

### Reynolds, an engineering professor in early 1880 demonstrated two different types of flow through an experiment:

7 STEADY FLOW IN PIPES 7.1 Reynolds Number Reynolds, an engineering professor in early 1880 demonstrated two different types of flow through an experiment: Laminar flow Turbulent flow Reynolds apparatus

### FE Exam Fluids Review October 23, Important Concepts

FE Exam Fluids Review October 3, 013 mportant Concepts Density, specific volume, specific weight, specific gravity (Water 1000 kg/m^3, Air 1. kg/m^3) Meaning & Symbols? Stress, Pressure, Viscosity; Meaning

### Consider a control volume in the form of a straight section of a streamtube ABCD.

6 MOMENTUM EQUATION 6.1 Momentum and Fluid Flow In mechanics, the momentum of a particle or object is defined as the product of its mass m and its velocity v: Momentum = mv The particles of a fluid stream

### Lecture 3 The energy equation

Lecture 3 The energy equation Dr Tim Gough: t.gough@bradford.ac.uk General information Lab groups now assigned Timetable up to week 6 published Is there anyone not yet on the list? Week 3 Week 4 Week 5

### Hydraulics. B.E. (Civil), Year/Part: II/II. Tutorial solutions: Pipe flow. Tutorial 1

Hydraulics B.E. (Civil), Year/Part: II/II Tutorial solutions: Pipe flow Tutorial 1 -by Dr. K.N. Dulal Laminar flow 1. A pipe 200mm in diameter and 20km long conveys oil of density 900 kg/m 3 and viscosity

CE 6303 MECHANICS OF FLUIDS L T P C QUESTION BANK 3 0 0 3 UNIT I FLUID PROPERTIES AND FLUID STATICS PART - A 1. Define fluid and fluid mechanics. 2. Define real and ideal fluids. 3. Define mass density

### Part A: 1 pts each, 10 pts total, no partial credit.

Part A: 1 pts each, 10 pts total, no partial credit. 1) (Correct: 1 pt/ Wrong: -3 pts). The sum of static, dynamic, and hydrostatic pressures is constant when flow is steady, irrotational, incompressible,

### Chapter 3 Bernoulli Equation

1 Bernoulli Equation 3.1 Flow Patterns: Streamlines, Pathlines, Streaklines 1) A streamline, is a line that is everywhere tangent to the velocity vector at a given instant. Examples of streamlines around

### Q1 Give answers to all of the following questions (5 marks each):

FLUID MECHANICS First Year Exam Solutions 03 Q Give answers to all of the following questions (5 marks each): (a) A cylinder of m in diameter is made with material of relative density 0.5. It is moored

### Lesson 37 Transmission Of Air In Air Conditioning Ducts

Lesson 37 Transmission Of Air In Air Conditioning Ducts Version 1 ME, IIT Kharagpur 1 The specific objectives of this chapter are to: 1. Describe an Air Handling Unit (AHU) and its functions (Section 37.1).

### Objectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation

Objectives Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation Conservation of Mass Conservation of Mass Mass, like energy, is a conserved

### COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour. Basic Equations in fluid Dynamics

COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour Basic Equations in fluid Dynamics Course teacher Dr. M. Mahbubur Razzaque Professor Department of Mechanical Engineering BUET 1 Description of Fluid

### S.E. (Mech.) (First Sem.) EXAMINATION, (Common to Mech/Sandwich) FLUID MECHANICS (2008 PATTERN) Time : Three Hours Maximum Marks : 100

Total No. of Questions 12] [Total No. of Printed Pages 8 Seat No. [4262]-113 S.E. (Mech.) (First Sem.) EXAMINATION, 2012 (Common to Mech/Sandwich) FLUID MECHANICS (2008 PATTERN) Time : Three Hours Maximum

### Approximate physical properties of selected fluids All properties are given at pressure kn/m 2 and temperature 15 C.

Appendix FLUID MECHANICS Approximate physical properties of selected fluids All properties are given at pressure 101. kn/m and temperature 15 C. Liquids Density (kg/m ) Dynamic viscosity (N s/m ) Surface

### Chapter 8: Flow in Pipes

Objectives 1. Have a deeper understanding of laminar and turbulent flow in pipes and the analysis of fully developed flow 2. Calculate the major and minor losses associated with pipe flow in piping networks

### 1-Reynold s Experiment

Lect.No.8 2 nd Semester Flow Dynamics in Closed Conduit (Pipe Flow) 1 of 21 The flow in closed conduit ( flow in pipe ) is differ from this occur in open channel where the flow in pipe is at a pressure

### Applied Fluid Mechanics

Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and

### LECTURE 6- ENERGY LOSSES IN HYDRAULIC SYSTEMS SELF EVALUATION QUESTIONS AND ANSWERS

LECTURE 6- ENERGY LOSSES IN HYDRAULIC SYSTEMS SELF EVALUATION QUESTIONS AND ANSWERS 1. What is the head loss ( in units of bars) across a 30mm wide open gate valve when oil ( SG=0.9) flow through at a

### Rate of Flow Quantity of fluid passing through any section (area) per unit time

Kinematics of Fluid Flow Kinematics is the science which deals with study of motion of liquids without considering the forces causing the motion. Rate of Flow Quantity of fluid passing through any section

### Fluid Mechanics c) Orificemeter a) Viscous force, Turbulence force, Compressible force a) Turbulence force c) Integration d) The flow is rotational

Fluid Mechanics 1. Which is the cheapest device for measuring flow / discharge rate. a) Venturimeter b) Pitot tube c) Orificemeter d) None of the mentioned 2. Which forces are neglected to obtain Euler

### 2.The lines that are tangent to the velocity vectors throughout the flow field are called steady flow lines. True or False A. True B.

CHAPTER 03 1. Write Newton's second law of motion. YOUR ANSWER: F = ma 2.The lines that are tangent to the velocity vectors throughout the flow field are called steady flow lines. True or False 3.Streamwise

### Piping Systems and Flow Analysis (Chapter 3)

Piping Systems and Flow Analysis (Chapter 3) 2 Learning Outcomes (Chapter 3) Losses in Piping Systems Major losses Minor losses Pipe Networks Pipes in series Pipes in parallel Manifolds and Distribution

### equation 4.1 INTRODUCTION

4 The momentum equation 4.1 INTRODUCTION It is often important to determine the force produced on a solid body by fluid flowing steadily over or through it. For example, there is the force exerted on a

### Chapter (6) Energy Equation and Its Applications

Chapter (6) Energy Equation and Its Applications Bernoulli Equation Bernoulli equation is one of the most useful equations in fluid mechanics and hydraulics. And it s a statement of the principle of conservation

### Therefore, the control volume in this case can be treated as a solid body, with a net force or thrust of. bm # V

When the mass m of the control volume remains nearly constant, the first term of the Eq. 6 8 simply becomes mass times acceleration since 39 CHAPTER 6 d(mv ) CV m dv CV CV (ma ) CV Therefore, the control

### If a stream of uniform velocity flows into a blunt body, the stream lines take a pattern similar to this: Streamlines around a blunt body

Venturimeter & Orificemeter ELEMENTARY HYDRAULICS National Certificate in Technology (Civil Engineering) Chapter 5 Applications of the Bernoulli Equation The Bernoulli equation can be applied to a great

### CEE 3310 Control Volume Analysis, Oct. 7, D Steady State Head Form of the Energy Equation P. P 2g + z h f + h p h s.

CEE 3310 Control Volume Analysis, Oct. 7, 2015 81 3.21 Review 1-D Steady State Head Form of the Energy Equation ( ) ( ) 2g + z = 2g + z h f + h p h s out where h f is the friction head loss (which combines

### vector H. If O is the point about which moments are desired, the angular moment about O is given:

The angular momentum A control volume analysis can be applied to the angular momentum, by letting B equal to angularmomentum vector H. If O is the point about which moments are desired, the angular moment

### The Bernoulli Equation

The Bernoulli Equation The most used and the most abused equation in fluid mechanics. Newton s Second Law: F = ma In general, most real flows are 3-D, unsteady (x, y, z, t; r,θ, z, t; etc) Let consider

### R09. d water surface. Prove that the depth of pressure is equal to p +.

Code No:A109210105 R09 SET-1 B.Tech II Year - I Semester Examinations, December 2011 FLUID MECHANICS (CIVIL ENGINEERING) Time: 3 hours Max. Marks: 75 Answer any five questions All questions carry equal

### UNIT II Real fluids. FMM / KRG / MECH / NPRCET Page 78. Laminar and turbulent flow

UNIT II Real fluids The flow of real fluids exhibits viscous effect that is they tend to "stick" to solid surfaces and have stresses within their body. You might remember from earlier in the course Newtons

### William В. Brower, Jr. A PRIMER IN FLUID MECHANICS. Dynamics of Flows in One Space Dimension. CRC Press Boca Raton London New York Washington, D.C.

William В. Brower, Jr. A PRIMER IN FLUID MECHANICS Dynamics of Flows in One Space Dimension CRC Press Boca Raton London New York Washington, D.C. Table of Contents Chapter 1 Fluid Properties Kinetic Theory

### Figure 3: Problem 7. (a) 0.9 m (b) 1.8 m (c) 2.7 m (d) 3.6 m

1. For the manometer shown in figure 1, if the absolute pressure at point A is 1.013 10 5 Pa, the absolute pressure at point B is (ρ water =10 3 kg/m 3, ρ Hg =13.56 10 3 kg/m 3, ρ oil = 800kg/m 3 ): (a)

### INSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad AERONAUTICAL ENGINEERING QUESTION BANK : AERONAUTICAL ENGINEERING.

Course Name Course Code Class Branch INSTITUTE OF AERONAUTICAL ENGINEERING Dundigal, Hyderabad - 00 0 AERONAUTICAL ENGINEERING : Mechanics of Fluids : A00 : II-I- B. Tech Year : 0 0 Course Coordinator

### MASS, MOMENTUM, AND ENERGY EQUATIONS

MASS, MOMENTUM, AND ENERGY EQUATIONS This chapter deals with four equations commonly used in fluid mechanics: the mass, Bernoulli, Momentum and energy equations. The mass equation is an expression of the

### Chapter 6. Losses due to Fluid Friction

Chapter 6 Losses due to Fluid Friction 1 Objectives ä To measure the pressure drop in the straight section of smooth, rough, and packed pipes as a function of flow rate. ä To correlate this in terms of

### Mechanical Engineering Programme of Study

Mechanical Engineering Programme of Study Fluid Mechanics Instructor: Marios M. Fyrillas Email: eng.fm@fit.ac.cy SOLVED EXAMPLES ON VISCOUS FLOW 1. Consider steady, laminar flow between two fixed parallel

### Measurements using Bernoulli s equation

An Internet Book on Fluid Dynamics Measurements using Bernoulli s equation Many fluid measurement devices and techniques are based on Bernoulli s equation and we list them here with analysis and discussion.

### FLUID MECHANICS. Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation

FLUID MECHANICS Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation CHAP 3. ELEMENTARY FLUID DYNAMICS - THE BERNOULLI EQUATION CONTENTS 3. Newton s Second Law 3. F = ma along a Streamline 3.3

### ENGINEERING FLUID MECHANICS. CHAPTER 1 Properties of Fluids

CHAPTER 1 Properties of Fluids ENGINEERING FLUID MECHANICS 1.1 Introduction 1.2 Development of Fluid Mechanics 1.3 Units of Measurement (SI units) 1.4 Mass, Density, Specific Weight, Specific Volume, Specific

### ME3560 Tentative Schedule Fall 2018

ME3560 Tentative Schedule Fall 2018 Week Number 1 Wednesday 8/29/2018 1 Date Lecture Topics Covered Introduction to course, syllabus and class policies. Math Review. Differentiation. Prior to Lecture Read

### Chapter 7 The Energy Equation

Chapter 7 The Energy Equation 7.1 Energy, Work, and Power When matter has energy, the matter can be used to do work. A fluid can have several forms of energy. For example a fluid jet has kinetic energy,

### For example an empty bucket weighs 2.0kg. After 7 seconds of collecting water the bucket weighs 8.0kg, then:

Hydraulic Coefficient & Flow Measurements ELEMENTARY HYDRAULICS National Certificate in Technology (Civil Engineering) Chapter 3 1. Mass flow rate If we want to measure the rate at which water is flowing

### NPTEL Quiz Hydraulics

Introduction NPTEL Quiz Hydraulics 1. An ideal fluid is a. One which obeys Newton s law of viscosity b. Frictionless and incompressible c. Very viscous d. Frictionless and compressible 2. The unit of kinematic

### Pipe Flow. Lecture 17

Pipe Flow Lecture 7 Pipe Flow and the Energy Equation For pipe flow, the Bernoulli equation alone is not sufficient. Friction loss along the pipe, and momentum loss through diameter changes and corners

### ME3560 Tentative Schedule Spring 2019

ME3560 Tentative Schedule Spring 2019 Week Number Date Lecture Topics Covered Prior to Lecture Read Section Assignment Prep Problems for Prep Probs. Must be Solved by 1 Monday 1/7/2019 1 Introduction to

### Hydraulics for Urban Storm Drainage

Urban Hydraulics Hydraulics for Urban Storm Drainage Learning objectives: understanding of basic concepts of fluid flow and how to analyze conduit flows, free surface flows. to analyze, hydrostatic pressure

### BERNOULLI EQUATION. The motion of a fluid is usually extremely complex.

BERNOULLI EQUATION The motion of a fluid is usually extremely complex. The study of a fluid at rest, or in relative equilibrium, was simplified by the absence of shear stress, but when a fluid flows over

### Atmospheric pressure. 9 ft. 6 ft

Name CEE 4 Final Exam, Aut 00; Answer all questions; 145 points total. Some information that might be helpful is provided below. A Moody diagram is printed on the last page. For water at 0 o C (68 o F):

### Viscous Flow in Ducts

Dr. M. Siavashi Iran University of Science and Technology Spring 2014 Objectives 1. Have a deeper understanding of laminar and turbulent flow in pipes and the analysis of fully developed flow 2. Calculate

### PROPERTIES OF FLUIDS

Unit - I Chapter - PROPERTIES OF FLUIDS Solutions of Examples for Practice Example.9 : Given data : u = y y, = 8 Poise = 0.8 Pa-s To find : Shear stress. Step - : Calculate the shear stress at various

### CEE 3310 Control Volume Analysis, Oct. 10, = dt. sys

CEE 3310 Control Volume Analysis, Oct. 10, 2018 77 3.16 Review First Law of Thermodynamics ( ) de = dt Q Ẇ sys Sign convention: Work done by the surroundings on the system < 0, example, a pump! Work done

### Lecture 2 Flow classifications and continuity

Lecture 2 Flow classifications and continuity Dr Tim Gough: t.gough@bradford.ac.uk General information 1 No tutorial week 3 3 rd October 2013 this Thursday. Attempt tutorial based on examples from today

### Detailed Outline, M E 320 Fluid Flow, Spring Semester 2015

Detailed Outline, M E 320 Fluid Flow, Spring Semester 2015 I. Introduction (Chapters 1 and 2) A. What is Fluid Mechanics? 1. What is a fluid? 2. What is mechanics? B. Classification of Fluid Flows 1. Viscous

### MECHANICAL PROPERTIES OF FLUIDS:

Important Definitions: MECHANICAL PROPERTIES OF FLUIDS: Fluid: A substance that can flow is called Fluid Both liquids and gases are fluids Pressure: The normal force acting per unit area of a surface is

### 11.1 Mass Density. Fluids are materials that can flow, and they include both gases and liquids. The mass density of a liquid or gas is an

Chapter 11 Fluids 11.1 Mass Density Fluids are materials that can flow, and they include both gases and liquids. The mass density of a liquid or gas is an important factor that determines its behavior

### Fluid Mechanics Answer Key of Objective & Conventional Questions

019 MPROVEMENT Mechanical Engineering Fluid Mechanics Answer Key of Objective & Conventional Questions 1 Fluid Properties 1. (c). (b) 3. (c) 4. (576) 5. (3.61)(3.50 to 3.75) 6. (0.058)(0.05 to 0.06) 7.

### CH.1 Overview of Fluid Mechanics/22 MARKS. 1.1 Fluid Fundamentals.

Content : 1.1 Fluid Fundamentals. 08 Marks Classification of Fluid, Properties of fluids like Specific Weight, Specific gravity, Surface tension, Capillarity, Viscosity. Specification of hydraulic oil

### Major and Minor Losses

Abstract Major and Minor Losses Caitlyn Collazo, Team 2 (1:00 pm) A Technovate fluid circuit system was used to determine the pressure drop across a pipe section and across an orifice. These pressure drops

### Water Circuit Lab. The pressure drop along a straight pipe segment can be calculated using the following set of equations:

Water Circuit Lab When a fluid flows in a conduit, there is friction between the flowing fluid and the pipe walls. The result of this friction is a net loss of energy in the flowing fluid. The fluid pressure

### LOSSES DUE TO PIPE FITTINGS

LOSSES DUE TO PIPE FITTINGS Aim: To determine the losses across the fittings in a pipe network Theory: The resistance to flow in a pipe network causes loss in the pressure head along the flow. The overall

### Signature: (Note that unsigned exams will be given a score of zero.)

Neatly print your name: Signature: (Note that unsigned exams will be given a score of zero.) Circle your lecture section (-1 point if not circled, or circled incorrectly): Prof. Dabiri Prof. Wassgren Prof.

### Applied Fluid Mechanics

Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and

### Applied Fluid Mechanics

Applied Fluid Mechanics 1. The Nature of Fluid and the Study of Fluid Mechanics 2. Viscosity of Fluid 3. Pressure Measurement 4. Forces Due to Static Fluid 5. Buoyancy and Stability 6. Flow of Fluid and

### 6. Basic basic equations I ( )

6. Basic basic equations I (4.2-4.4) Steady and uniform flows, streamline, streamtube One-, two-, and three-dimensional flow Laminar and turbulent flow Reynolds number System and control volume Continuity

### Hydraulics and hydrology

Hydraulics and hydrology - project exercises - Class 4 and 5 Pipe flow Discharge (Q) (called also as the volume flow rate) is the volume of fluid that passes through an area per unit time. The discharge

### AEROSPACE ENGINEERING DEPARTMENT. Second Year - Second Term ( ) Fluid Mechanics & Gas Dynamics

AEROSPACE ENGINEERING DEPARTMENT Second Year - Second Term (2008-2009) Fluid Mechanics & Gas Dynamics Similitude,Dimensional Analysis &Modeling (1) [7.2R*] Some common variables in fluid mechanics include:

### FLOW MEASUREMENT IN PIPES EXPERIMENT

University of Leicester Engineering Department FLOW MEASUREMENT IN PIPES EXPERIMENT Page 1 FORMAL LABORATORY REPORT Name of the experiment: FLOW MEASUREMENT IN PIPES Author: Apollin nana chaazou Partner

### Chapter 8: Flow in Pipes

8-1 Introduction 8-2 Laminar and Turbulent Flows 8-3 The Entrance Region 8-4 Laminar Flow in Pipes 8-5 Turbulent Flow in Pipes 8-6 Fully Developed Pipe Flow 8-7 Minor Losses 8-8 Piping Networks and Pump

### Lecture23. Flowmeter Design.

Lecture23 Flowmeter Design. Contents of lecture Design of flowmeter Principles of flow measurement; i) Venturi and ii) Orifice meter and nozzle Relationship between flow rate and pressure drop Relation

### mywbut.com Hydraulic Turbines

Hydraulic Turbines Hydro-electric power accounts for up to 0% of the world s electrical generation. Hydraulic turbines come in a variety of shapes determined by the available head and a number of sizes

### Chapter 5 Control Volume Approach and Continuity Equation

Chapter 5 Control Volume Approach and Continuity Equation Lagrangian and Eulerian Approach To evaluate the pressure and velocities at arbitrary locations in a flow field. The flow into a sudden contraction,

### VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur

VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF CIVIL ENGINEERING QUESTION BANK III SEMESTER CE 8302 FLUID MECHANICS Regulation 2017 Academic Year 2018 19 Prepared by Mrs.

### 10.52 Mechanics of Fluids Spring 2006 Problem Set 3

10.52 Mechanics of Fluids Spring 2006 Problem Set 3 Problem 1 Mass transfer studies involving the transport of a solute from a gas to a liquid often involve the use of a laminar jet of liquid. The situation

### Lecture 13 Flow Measurement in Pipes. I. Introduction

Lecture 13 Flow Measurement in Pipes I. Introduction There are a wide variety of methods for measuring discharge and velocity in pipes, or closed conduits Many of these methods can provide very accurate

### Aerodynamics. Basic Aerodynamics. Continuity equation (mass conserved) Some thermodynamics. Energy equation (energy conserved)

Flow with no friction (inviscid) Aerodynamics Basic Aerodynamics Continuity equation (mass conserved) Flow with friction (viscous) Momentum equation (F = ma) 1. Euler s equation 2. Bernoulli s equation

### 4 Mechanics of Fluids (I)

1. The x and y components of velocity for a two-dimensional flow are u = 3.0 ft/s and v = 9.0x ft/s where x is in feet. Determine the equation for the streamlines and graph representative streamlines in

### 2 Navier-Stokes Equations

1 Integral analysis 1. Water enters a pipe bend horizontally with a uniform velocity, u 1 = 5 m/s. The pipe is bended at 90 so that the water leaves it vertically downwards. The input diameter d 1 = 0.1

### EXPERIMENT NO. 4 CALIBRATION OF AN ORIFICE PLATE FLOWMETER MECHANICAL ENGINEERING DEPARTMENT KING SAUD UNIVERSITY RIYADH

EXPERIMENT NO. 4 CALIBRATION OF AN ORIFICE PLATE FLOWMETER MECHANICAL ENGINEERING DEPARTMENT KING SAUD UNIVERSITY RIYADH Submitted By: ABDULLAH IBN ABDULRAHMAN ID: 13456789 GROUP A EXPERIMENT PERFORMED

### Signature: (Note that unsigned exams will be given a score of zero.)

Neatly print your name: Signature: (Note that unsigned exams will be given a score of zero.) Circle your lecture section (-1 point if not circled, or circled incorrectly): Prof. Dabiri Prof. Wassgren Prof.

### Principles of Convection

Principles of Convection Point Conduction & convection are similar both require the presence of a material medium. But convection requires the presence of fluid motion. Heat transfer through the: Solid

### Fluid Mechanics Lab (ME-216-F) List of Experiments

Fluid Mechanics Lab (ME-216-F) List of Experiments 1. To determine the coefficient of discharge C d, velocity C v, and contraction C c of various types of orifices 2. Determine of discharge coefficients

### When water (fluid) flows in a pipe, for example from point A to point B, pressure drop will occur due to the energy losses (major and minor losses).

PRESSURE DROP AND OSSES IN PIPE When water (luid) lows in a pipe, or example rom point A to point B, pressure drop will occur due to the energy losses (major and minor losses). A B Bernoulli equation: