PRESURE MEASUREMENT 4.2 Manometer – Differential U tube 50
PRESSURE MEASUREMENT 4.2 Manometer – Differential U tube Example 3: A U tube manometer measures the pressure difference between two points A and B in a liquid. The U tube contains mercury. Calculate the difference in pressure if h =1.5 m, h2 = 0.75 m and h1 = 0.5 m. The liquid at A and B is water ( ω = 9.81 × 10 3 N/m2) and the specific gravity of mercury is 13.6. 51
PRESSURE MEASUREMENT 4.3 Manometer – Inverted U tube Pressure at C = Pressure at D 52
PRESSURE MEASUREMENT 4.3 Manometer – Inverted U tube Example 4: The top of an inverted U tube manometer is filled with oil of specific gravity, soil=0.98 and the remainder of the tube with water whose specific weight of water, ω= 9.81×10 3 N/m2 . Find the pressure difference in N/m2 between two points A and B at the same level at the base of the legs where the difference in water level h is 75 mm. 53
PRESSURE MEASUREMENT 4.3 Manometer – Inverted U tube Test your understanding….. An inverted U tube as shown in the figure below is used to measure the pressure difference between two points A and B which has water flowing. The difference in level h = 0.3 m, a = 0.25 m and b = 0.15 m. Calculate the pressure difference pB – pA if the top of the manometer is filled with: (a) air (b) oil of relative density 0.8. 5 4
T H E E N D O F C H A P T E R T H R E E
CHAPTER FOUR Fluid Dynamics Objectives: At the end of this chapter student will able to: 1. Describe the different types of flow 2. Describe flow rate 3. Apply continuity equation law 4. Apply Bernoulli theorem 56
INTRODUCTION Fluid dynamics is "the branch of applied science that is concerned with the movement of liquids and gases, " according to the American Heritage Dictionary. Fluid dynamics is one of two branches of fluid mechanics, which is the study of fluids and how forces affect them. Fluid dynamics provides methods for studying the evolution of stars, ocean currents, weather patterns, plate tectonics and even blood circulation. Some important technological applications of fluid dynamics include rocket engines, wind turbines, oil pipelines and air conditioning systems. 57
TYPES OF FLOW Steady flow The cross-sectional area and velocity of the stream of fluid are the same at each successive cross-section. Example: flow through a pipe of uniform bore running completely full. Uniform flow The cross-sectional area and velocity of the stream may vary from cross-section, but for each cross-section they do not change with time. Example: a wave travelling along a channel. Laminar flow Also known as streamline or viscous flow, in which the particles of the fluid move in an orderly manner and retain the same relative positions in successive cross-sections. Transition flow The process of a laminar flow becoming turbulent Turbulent flow Turbulent flow is a non steady flow in which the particles of fluid move in a disorderly manner, occupying different relative positions in successive cross-sections. 58
TYPES OF FLOW 5 9
Discharge and Mass Flowrate Discharge The volume of liquid passing through a given cross-section in unit time is called the discharge. It is measured in cubic meter per second, or similar units and denoted by Q. = Example 1: If the diameter d = 15 cm and the mean velocity, v = 3 m/s, calculate the actual discharge in the pipe. 60
Discharge and Mass Flowrate Mass Flowrate The mass of fluid passing through a given cross section in unit time is called the mass flow rate. It is measured in kilogram per second, or similar units and denoted by . = Example 2: Oil flows through a pipe at a velocity of 1.6 m/s. The diameter of the pipe is 8 cm. Calculate discharge and mass flowrate of oil. Take into consideration soil = 0.85. Example 3: The weight of an empty bucket is 2.0 kg. After 7 seconds of collecting water the weight of the bucket is 8.0 kg. Calculate the mass flowrate of the fluid. 61
Discharge and Mass Flowrate Example 2: Oil flows through a pipe at a velocity of 1.6 m/s. The diameter of the pipe is 8 cm. Calculate discharge and mass flowrate of oil. Take into consideration soil = 0.85. 62
Discharge and Mass Flowrate Example 3: The weight of an empty bucket is 2.0 kg. After 7 seconds of collecting water the weight of the bucket is 8.0 kg. Calculate the mass flowrate of the fluid. 63
Continuity Equation For continuity of flow in any system of fluid flow, the total amount of fluid entering the system must equal the amount leaving the system. This occurs in the case of uniform flow and steady flow 64
1 2 Continuity Equation Example 4: If the area A = 10 10 -3 m2 and A = 3 10 -3 m2 and the upstream mean velocity, v1=2.1 m/s, calculate the downstream mean velocity. Example 5: Referring to the Figure the diameter at section 1 is d1 = 30 mm and at section 2 is d2=40 mm and the mean velocity at section 2 is v2 = 3.0 m/s. Calculate the velocity entering the diffuser. 6 5
1 2 Continuity Equation Example 4: If the area A = 10 10 -3 m2 and A = 3 10 -3 m2 and the upstream mean velocity, v1=2.1 m/s, calculate the downstream mean velocity. 6 6
Continuity Equation Example 5: Referring to the Figure the diameter at section 1 is d1 = 30 mm and at section 2 is d2=40 mm and the mean velocity at section 2 is v2 = 3.0 m/s. Calculate the velocity entering the diffuser. 6 7
Continuity Equation Test your understanding: 68
Solution: 69
= Energy of a flowing fluid A liquid may possess three forms of energy: • Potential energy Potential energy = Wz Potential energy per unit weight = z The potential energy per unit weight has dimensions of Nm/N and is measured as a length or head z and can be called the potential head. • Pressure energy When a fluid flows in a continuous stream under pressure it can do work. If the area of cross-section of the stream of fluid is a, then force due to pressure p on cross-section is pa. If a weight W of liquid passes the cross-section Volume passing cross-section = Distance moved by liquid = x Work done = force distance = pressure energy per unit weight = Similarly the pressure energy per unit weight p/W is equivalent to a head and is referred to as the pressure head. 70
= Energy of a flowing fluid • Kinetic energy A liquid may possess three forms of energy: If a weight W of liquid has a velocity v, Kinetic energy = 1 2 2 Kinetic energy per unit weight 2 2 The kinetic energy per unit weight measured as a length and referred velocity head. is also to as the The total energy of the liquid is the sum of these three forms of energy Total head = potential head + pressure head + velocity head Total energy per unit weight = z+ + 2 2 71
Bernoulli’s Equation Definition; Bernoulli’s Theorem states that the total energy of each particle of a body of fluid is the same provided that no energy enters or leaves the system at any point. The division of this energy between potential, pressure and kinetic energy may vary, but the total remains constant. In symbols: 72
Bernoulli’s Equation By Bernoulli’s Theorem, Total energy per unit Total energy per unit weight at weight at section 1 = section 2 Bernoulli’s Equation has some restrictions in its applicability, they are : • the flow is steady • the density is constant (which also means the fluid is compressible) • friction losses are negligible • the equation relates the state at two points along a single streamline (not conditions on two different streamlines). 73
Application of Bernoulli’s Equation ➢Horizontal Pipe ➢Inclined Pipe ➢Horizontal Venturi Meter ➢Inclined Venturi Meter ➢Orifice Meter ➢Simple Pitot Tube 74
Application of Bernoulli’s Equation Horizontal Pipe Example 1: Water flows through a pipe 36 m from the sea level as shown in figure 5.2. Pressure in the pipe is 410 kN/m2 and the velocity is 4.8 m/s. Calculate total energy of every weight of unit water above the sea level. 75
Application of Bernoulli’s Equation Inclined Pipe Example 2: A bent pipe labeled MN measures 5 m and 3 m respectively above the datum line. The diameter M and N are both 20 cm and 5 cm. The water pressure at M is 5 kg/cm2 . If the velocity at M is 1 m/s, determine the pressure at N in kg/cm2 . 76
p1 = pressure of section 1 v1 = velocity of section 1 A1 = area of section 1 p2 = pressure of section 2 v1 = velocity of section 1 A1 = area of section 1 ω1 = liquid in pipeline (specific weight, spec.wg) ωg = liquid in the gauge (specific weight, spec.wg) g = gravity (9.81 m/s 2 ) z = height above datum Application of Bernoulli’s Equation Horizontal Venturi Meter Venturi meter: It is a device used for measuring the rate of flow of a non- viscous, incompressible fluid in non- rotational and steady-stream lined flow. Although venturi meters can be applied to the measurement of gas, they are most commonly used for liquids. The following treatment is limited to incompressible fluids. 77
Application of Bernoulli’s Equation Horizontal Venturi Meter 7 8
Application of Bernoulli’s Equation Horizontal Venturi Meter Example 3: A venture tube tapers from 300 mm in diameter at the entrance to 100 mm in diameter at the throat; the discharge coefficient is 0.98. A differential mercury U- tube gauge is connected between pressure tapping at the entrance at throat. If the meter is used to measure the flow of water and the water fills the leads to the U-tube and is in contact with the mercury(133.7 kN/m3), calculate the discharge when the difference of level in the U-tube is 55 mm. 79
Application of Bernoulli’s Equation Horizontal Venturi Meter Example 4: A horizontal venturi meter measures the flow of oil of specific gravity 0.9 in a 75 mm diameter pipe line. If the difference of pressure between the full bore and the throat tapping is 34.5 kN/m2 and the area ratio, m is 4, calculate the rate of flow, assuming a coefficient of discharge is 0.97. 80
Application of Bernoulli’s Equation Inclined Venturi Meter 81
Application of Bernoulli’s Equation Inclined Venturi Meter Example 5: A vertical venturi meter measures the flow of oil of specific gravity 0.82 and has an entrance of 125 mm diameter and throat of 50 mm diameter. There are pressure gauges at the entrance and at the throat, which is 300 mm above the entrance. If the coefficient for the meter is 0.97 and pressure difference is 27.5 kN/m2 , calculate the actual discharge in m3 /s. 82
Application of Bernoulli’s Equation Inclined Venturi Meter Example 6 The water supply to a gas water heater contracts from 10mm in diameter at A (Figure 5.6) to 7 mm in diameter at B. If the pipe is horizontal, calculate the difference in pressure between A and B when the velocity of water at A is 4.5 m/s. The pressure difference operates the gas control through connections which is taken to a horizontal cylinder in which a piston of 20 mm diameter moves. Ignoring friction and the area of the piston connecting rod, what is the force on the piston? 83
Application of Bernoulli’s Equation Orifice Meter The principle of the orifice meter is identical with that of the venturi meter. The reduction at the cross section of the flowing stream in passing through the orifice increases the velocity head at the expense of the pressure head, and the reduction in pressure between the taps is measured by a manometer. 8 4
Application of Bernoulli’s Equation Orifice Meter Example 7: A meter orifice has a 100 mm diameter rectangular hole in the pipe. Diameter of the pipe is 250 mm. Coefficient of discharge, Cd = 0.65 and specific gravity of oil in the pipe is 0.9. The pressure difference that is measured by the manometer is 750 mm. Calculate the flow rate of the oil through the pipe. 85
Application of Bernoulli’s Equation Small Orifice 86
Application of Bernoulli’s Equation Small Orifice Example 8: A tank 1.8 m high, standing on the ground, is kept full of water. There is an orifice in its vertical site at depth, h m below the surface. Find the value of h in order the jet may strike the ground at a maximum distance from the tank. 87
Application of Bernoulli’s Equation Orifice Meter Example 9: An orifice meter consists of a 100 mm diameter in a 250 mm diameter pipe (Figure 5.9), and has a coefficient discharge of 0.65. The pipe conveys oil of specific gravity 0.9. The pressure difference between the two sides of the orifice plate is measured by a mercury manometer, that leads to the gauge being filled with oil. If the difference in mercury levels in the gauge is 760 mm, calculate the flowrate of oil in the pipeline. 88
Application of Bernoulli’s Equation Pitot tube 89
Application of Bernoulli’s Equation Pitot tube Example10: A Pitot Tube is used to measure air velocity in a pipe attached to a mercury manometer. Head difference of that manometer is 6 mm water. The weight density of air is 1.25 kg/m3 . Calculate the air velocity if coefficient of the pitot tube, C = 0.94. Do you understand? Test you understanding by trying all the questions in the self- assessment section. All the best 90
T H E E N D O F C H A P T E R FOUR
CHAPTER FIVE Energy Loss In Pipelines Objectives: At the end of this chapter student will able to: 1. Explain the round pipe system 2. Apply the types of head loss 3. Apply head loss in pipeline systems 92
INTRODUCTION Losses of energy in a pipeline cannot be ignored. When the shock losses and friction loss have been determined, they are inserted in Bernoulli’s equation in the usual way p1 g v1 2 2g z1 p2 g v2 2 2g z2 h 93
ROUND PIPE SYSTEM Velocity profile in circular pipe system 94
ROUND PIPE SYSTEM Velocity profile in circular pipe system Because of the shear force near the pipe wall, a boundary layer forms on the inside surface and occupies a large portion of the flow area as the distance downstream from the pipe entrance increase. At some value of this distance the boundary layer fills the flow area. The velocity profile becomes independent of the axis in the direction of flow, and the flow is said to be fully developed. Pipe Entrance 95
ROUND PIPE SYSTEM Types of head loss Sudden Enlargement Friction Sudden Contraction Sharp Inlet &Outlet 96
9 8 − 1 ROUND PIPE SYSTEM 1 Sudden = enlargement − HEAD 2 Sudden LOSS c = contraction 3 Friction = Nomenclature: 4 d=diameter of pipe CC = Contraction coefficient f= friction coefficient g=gravity (9.81m2/s) L= pipe length in m h = head loss v= velocity (m/s) Inlet & outlet , i = Outlet, o = 97
ROUND PIPE SYSTEM Example 1: A pipe carrying 1800 l/min of water increases suddenly from 10 cm to 15 cm diameter. Find a) the head loss due to the sudden enlargement b) the difference in pressure in kN/m2 in the two pipes Example 2: A pipe carrying 0.06 m3 /s suddenly contracts from 200 mm to 150 mm diameter. Assuming that the vena contracta is formed in the smaller pipe, calculate the coefficient of contraction if the pressure head at a point upstream of the contraction is 0.655 m greater than at a point just downstream of the vena contracta Example 3: Determine the loss of head due to friction in a pipe 14 m long and 2 m diameter which carries 1.5 m/s oil. Take into consideration f = 0.05. 98
1 1 2 HEAD LOSS IN PIPELINE SYSTEM Pipeline Systems The basic approach to all piping systems is to write the Bernoulli equation between two points, connected by a streamline, where the conditions are known. For example, between the surface of a reservoir and a pipe outlet. Thus: + 1 2 2 + = + 2 2 2 + 2+shock loss + friction loss 99