0
COPYRIGHT Published by : Politeknik Sultan Azlan Shah Behrang Stesen, Behrang 35950 Perak Tel : 05-4544431 Faks : 05-4544993 Email : http://www.psas.edu.my First Published 2022 All right reserved. No parts of this publication may be reproduced stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying or otherwise without permission of Sultan Azlan Shah Polytechnics. PERPUSTAKAAN NEGARA MALAYSIA DJM20053 THERMOFLUIDS
AN INTRODUCTION OF DJM20053 THERMOFLUIDS Thermofluids was developed in order to provide students with a clear and through presentation of theory and application of subject. In this edition, visualization aids such as picture and figure are provided to enable students to better understand concept taught and problems posed. The problem provided in this book are arranged in increasing level of difficulty to build students’ analytical and problem-solving skills as well as give them the practice they require. This course introduces the subject areas of fluid mechanics and thermodynamics from an engineering perspective and concerned with the application of these topics to analysis and design in an engineering context. The aims of this course to teach the basic analytical methods, that is, the fundamental concepts and techniques of thermodynamics and fluid mechanics. Topic includes in this e-book are conceptual principles in Thermofluids, fluid application, properties of pure substances, first law of thermodynamics and second law of thermodynamics. Thank you.
CONTENTS Copyright Declaration Contents An Introduction of DJJM20053 Thermofluids Topic 1 Conceptual Principles in Thermofluids Introduction of Thermal-Fluid 1 Characteristic of Solid, Liquid, and Gas 3 Concepts of Dimension, SI and Imperial Unit 4 System, Boundary and Surrounding 6 Process and Cycle 7 Properties of a System 7 State and Equilibrium 8 Energy Conversion 8 Types of Pressure 9 Tutorial 10 Topic 2 Fluid Application Relationship Between Pressure and Depth 11 Pascal Law and Hydraulic Jack 12 Tutorial 1 15 Manometer 16 Tutorial 2 20 Piezometer and Barometer 22 Tutorial 3 23 Buoyancy 24 Types of Flow 25 Discharge and Mass Flowrate 28
Continuity Equation Law 29 Tutorial 4 31 Bernoulli Theorem 32 Tutorial 5 40 Topic 3 Properties of Pure Substances Importance of steam 42 Pure Substances 43 Phase Change Process 43 T-V and P-V Diagram 44 Steam Table 45 Tutorial 1 48 Interpolation 49 Tutorial 2 51 Ideal Gas 52 Specific Heat 53 Analysis of Process 54 Tutorial 3 54 Topic 4 First Law of Thermodynamics Concept of The First Law of Thermodynamics 55 Internal Energy 56 Tutorial 1 58 Closed System/ Non-Flow Process 59 Tutorial 2 61 Open System/ Flow Process 63 The Steady Flow Energy Equation 66 Equation of Continuity 67 Tutorial 3 67
Topic 5 Second Law of Thermodynamics Second Law of Thermodynamics 69 Heat Engine 70 Tutorial 1 71 Reversed Heat Engine 72 Tutorial 2 73 Tutorial 3 74 Carnot & Rankine Cycle 75 Tutorial 4 75 Tutorial 5 77
1 TOPIC 1: CONCEPTUAL PRINCIPLES IN THERMOFLUIDS Introduction of Thermal-Fluid The design and analysis of most thermal systems such as power plants, automotive engines, and refrigerators involve all categories of thermal-fluid sciences as well as other sciences. For example, designing the radiator of car involves the determination of the amount of energy transfer from the knowledge of the properties of the coolant using thermodynamics, the determination of the size and shape of their inner tubes and the outer fins using heat transfer, and the determination of the size and type of the water pump using fluid mechanics. Basic Concept of Fluid Mechanics and Thermodynamics Definition of Fluid Mechanics Understanding Fluids • any substance which can flow is known as a fluid. • Example: gases and liquids
2 • Mechanics – branch of science in which we study about Force and its Effects. • For example, when you kick the ball, you exert force on it and the force produce acceleration. • Fluid mechanics- the branch of science in which we study about behaviour of fluids under the action force. • Fluid mechanics divided into three parts- fluids statics, fluids kinematics and fluid dynamics. Definition of Thermodynamics • Thermodynamics is the branch of physics dealing with the relationships between heat, work and other forms of energy. • From the Greek:
3 Characteristic of Solid, Liquid, and Gas Solid Liquid Gas Have strong intermolecular force. Weak intermolecular force. Very weak intermolecular force. Very less intermolecular space. Large intermolecular space. Very large intermolecular space. Have definite shape and volume. Do not have definite shape but have definite volume. No definite shape and volume. Have high density. Density is low Very low density Solids cannot be compressed. Liquids can be compressed Gases can be highly compressed.
4 Concepts of Dimension, SI and Imperial Unit • A dimension is a measure of a physical variable (without numerical values), while a unit is a way to assign a number or measurement to that dimension. For example, length is a dimension, but it is measured in units of feet (ft) or meters (m). • The SI system (International System of Units) is the modern metric system of measurement and the dominant system of international commerce and trade. • The Imperial Unit is also called The British Imperial because it came from the British Empire that ruled many parts of the world from the 16th to the 19th century. After the U.S gained independence from Britain, the new American government decided to keep this type of measurement, even though the metric system was gaining in popularity at the time. Fundamental/Primary units: Fundamental/Primary (sometimes called basic) dimensions are defined as independent or fundamental dimensions, from which other dimensions can be obtained. The fundamental units are listed in Table 1‐1. Table 1-1: Fundamental units
5 Derived/Secondary Units: All other dimensions can be derived as combinations of these seven primary dimensions. These are called secondary dimensions, with their corresponding secondary units. The examples of derived units are listed in Table 1-2. Table 1-2: Example of Derived Units Unit Conversion 1 kg = 1000 g 1 m = 100 cm = 1000 mm 1 km = 1000 m = (100 000 cm @ 105 cm) = (1 000 000 mm @ 106 mm) 1 hour = 60 minutes = 3600 seconds 1 m3 = 1000 litre, or 1 litre = 1 x 10-3 m3 1 bar = 1 x 105 N/m2 = 1 x 102 kN/m2 0°C = 273 K Watt = J/s Pascal = N/m2
6 System, Boundary and Surrounding • A system is defined as a quantity of matter or a region in space chosen for study. • The mass or region outside the system is called the surroundings. • The boundary is surface that separates the system from its surroundings.
7 Process and Cycle • Process = any change that a system undergoes from one equilibrium state to another. There are five processes: – isothermal process, (T1 = T2) – isobaric process, (P1 = P2) – isochoric (or isometric), (V1 = V2) – isentropic / adiabatic process, (s1 = s2) – polytrophic process, PV = C • A system is said to have undergone a cycle if it returns to its initial state at the end of the process. Reversible & Irreversible Process Properties of a System • Property - Any characteristic of a system – pressure P, temperature T, volume V, and mass m. – considered to be either intensive or extensive. • Intensive - properties are those that are independent of the size of a system, such as temperature, pressure, and density. • Extensive - values depend on the size of the system, such as Total mass, total volume, and total momentum.
8 State and Equilibrium Energy Conversion • Energy can change from one form to another form, but the total amount of energy remains constant. That is energy cannot be created or destroyed. • For example, chemical energy from fossil fuels (coal, oil, and natural gas) can be converted into heat energy when burned. The heat energy may be converted into kinetic energy in a gas turbine and finally into electrical energy by a generator. A rock falling off the cliff, picks up speed as a result of its potential energy being converted to kinetic energy.
9 Types of Pressure Atmospheric Pressure (patm): The pressure due to atmosphere at the surface of the earth depends upon the head of the air above the surface. Gauge Pressure ( pG ): It is the pressure, measured with the help of a pressure measuring instrument, in which the atmospheric pressure is taken as datum. In other words, the atmospheric pressure at the gauge scale is marked zero. Absolute Pressure ( pA ) It the pressure that equals to the algebraic sum of the atmospheric and gauge pressures. Absolute pressure= Gauge pressure+ Atmospheric pressure A G atm p = p + p Vacuum ( pv ) A completely empty space where the pressure is zero. Fluid Properties PROPERTIES FORMULA SYMBOL UNIT Mass density = = 3 Specific weight Sp ℎ = ℎ = × = = = 3 Specific gravity, relative density = ℎ ℎ = = - Specific volume = = = 1 3 Viscosity = = 2
10 Tutorial 1. A bourdon gauge is attached to a boiler which is located at sea level with a reading pressure of 7 bar. If atmospheric pressure is 1.013 bar, calculate the absolute pressure in that boiler in kN/m2 . 2. What is the pressure gauge of air in the cylinder if the atmospheric gauge is 101.3 kN/m2 and absolute pressure is 350 kN/m2 ? 3. What is the volume in m3 of this syrup if it has a density of 0.63 g/mL and a mass of 78 g? 4. A piece of cooper has a mass of 57.54 g. It is 9.36 cm long, 7.23 cm wide and 9.5 mm tick. Calculate density (kg/m3 ) and specific volume. 5. Determine the specific weight, (in kN/m3 ) and specific gravity, S of fluid if the weight of fluid is 10 N and the volume is 500 cm3 . 6. Calculate the specific weight, density, specific volume, and specific gravity of 1 liter of petrol weights 7 N. 7. Given the mass of fluid is 500 g and the volume is 200 cm3. calculate: i. Mass density of fluid ii. Specific weight of fluid iii. Specific volume of fluid iv. Specific gravity of fluid 8. Given the mass of fluid is 500 g and the volume is 200 cm3. calculate: i. Mass density of fluid ii. Specific weight of fluid iii. Specific volume of fluid iv. Specific gravity of fluid
11 TOPIC 2: FLUID APPLICATION Relationship Between Pressure and Depth Pressure defines as a force per unit area. Thus, in SI systems the pressure unit is measured as Newton per square meter (N/m2 ). It is also expressed as Pascal (Pa). The basic formula for pressure: = A fluid is a substance that flows easily. Gases and liquids are fluids, although sometimes the dividing line between liquids and solids is not always clear. Because of their ability to flow, fluids can exert buoyant forces, multiply forces in a hydraulic system, allow aircraft to fly and ships to float. If a fluid is within a container then the depth of an object placed in that fluid can be measured. The deeper the object is placed in the fluid, the more pressure it experiences. This is because of the weight of the fluid above it. The denser the fluid above it, the more pressure is exerted on the object that is submerged, due to the weight of the fluid. The formula that gives the pressure, P on an object submerged in a fluid is: = ℎ = ℎ Where, ▪ is the density of the fluid, ▪ g is the acceleration of gravity ▪ h is the height of the fluid above the object ▪ ω is specific weight of fluid If the container is open to the atmosphere above, the added pressure must be included if one is to find the total pressure on an object. The total pressure is the same as absolute pressure on pressure gauge readings, while the gauge pressure is the same as the fluid pressure alone, not including atmospheric pressure. Ptotal = Patmosphere + Pfluid Ptotal = Patmosphere + (ρgh)
12 Pascal Law and Hydraulic Jack a) Pascal Law at a point in a liquid b) Pascal Law at any point Hydraulic system uses an incompressible fluid, such as oil or water, to transmit forces from one location to another within the fluid. Most aircraft use hydraulics in the braking systems and landing gear. Pneumatic systems use compressible fluid, such as air, in their operation. Some aircraft utilize pneumatic systems for their brakes, landing gear and movement of flaps. Pascal's law states that when there is an increase in pressure at any point in a confined fluid, there is an equal increase at every other point in the co Pressure at any point is the same in all directions. This is known as Pascal Law and applies to fluids at rest
13 Hydraulic Jack A Hydraulic Jack is used to lift a heavy load with the help of a light force. F1 W ①SMALL PISTON/CYLINDER A1 = Area (m2 ) F1 = Action Force (N) ②LARGE PISTON/CYLINDER A2 = Area (m2 ) W = Load Lifted (N)
14 The Situation of Hydraulic Jack a) The small and the larger piston are at the same level b) The small piston is below the larger piston. c) The large piston is below the smaller piston. W F Area , a p1 p2 Area, A Area , a Area, A W F p2 p1 h Area , a Area, A W F p2 p1 h Equation for the piston at the same level P1 = P2 1 = 2 Equation for the small piston is below the larger piston P1 = P2 + ℎ 1 = 2 + ℎ Equation for the large piston is below the smaller piston P1 + ℎ = P2 1 + ℎ = 2
15 Tutorial 1 1. Calculate the pressure in kN/m2 due to a column of 0.4 m of; a. Oil (oil = 900 kg/m3 ) b. Brine (Sbrine = 1.1) c. Mercury (SHg = 13.6) 2. A force of 800 N is applied on the small cylinder which has an area of 20 cm2 . Determine the load that can be lifted by the hydraulic jack if the area of the large cylinder is 200 cm2 . 3. A force, P of 650 N is applied to the smaller cylinder of a hydraulic jack. The area, A1 of a small piston is 15 cm2 and the area A2 of a larger piston is 150 cm2 . What load, W can be lifted on the larger piston if: a. the pistons are at the same level. b. the large piston is 0.65 m below the smaller piston. c. the small piston is 0.40 m below the larger piston. Consider the mass density ρ of the liquid in the jack is 103 kg/m3 . 4. A hydraulic jack consists of a small and a large cylinder with diameters of 7 cm and 20 cm respectively. The required force, F to lift up a load, W is 400 N. If the large piston is 15 cm higher than the small one, determine the weight, W that can be lifted if the specific weight of oil is 8730 N/m3 . 5. A force, F of 900 N is applied to the smaller cylinder of a hydraulic jack. The area, A1 of a small piston is 22 cm2 and the area A2 of a larger piston is 250 cm2 . What load, W can be lifted on the larger piston if: a. The pistons are at the same level. b. The large piston is 0.65 m below the smaller piston. 6. A hydraulic jack has a diameter ratio between the two pistons of 8:1. The diameter of large piston is 600 mm, and it is required to support a mass of 3500 kg. The jack is filled with hydraulic fluid of specific gravity 0.8. Calculate the force required on the small piston: a. When the two pistons are at the same level. b. When the small piston is 2.6m below than the large piston.
16 Manometer A manometer is one of the most accurate devices for measuring pressure in the lower ranges. Since manometers are so accurate, they are often used as calibration standards. All manometers operate on the principle that changes in pressure will cause a liquid to rise or fall in a tube. Typical manometer liquids are mercury, water, and light oils. The following are a few types of manometers: a. Simple manometer, b. Differential manometer and c. Inverted differential manometer.
17 a. Simple Manometer ① Divide into 2 parts - LEFT and RIGHT ③ ② Determine the datum line – B and C Write the equation for LEFT B p = Pressure, A p at A + Pressure due to depth, 1 h of fluid P = + ℎ1 = 1 p gh A + P ④Write the equation for RIGHT pC = Pressure D p at D + Pressure due to depth 2 h of liquid Q But, D p = Patm = 0 (Zero gauge pressure) And so, pC = ℎ2 = ℎ2 ωP SQ ⑤ PB = PC (Datum line at same level) ⑥ Substitute Equation LEFT and RIGHT into PB = PC PB = PC + ℎ1 = ℎ2 = ℎ2 – ℎ1
18 b. Differential Manometer ① Divide into 2 parts - LEFT and RIGHT ② Determine the datum line – B and C ③ Write the equation for LEFT = Pressure, A p at A + Pressure due to depth, h of fluid P = + ℎ = + ℎ ④Write the equation for RIGHT PD = Pressure PB at B + Pressure due to depth h1 of fluid Q + Pressure due to depth 2 h of liquid P PD = PB + ℎ1 + ℎ2 = PB + ℎ1 + ℎ2 ⑤ PC = PD (Datum line at same level) ⑥ Substitute Equation LEFT and RIGHT into PC = PD PC = PD + ℎ = PB + ℎ1 + ℎ2 PA – PB = ℎ1 + ℎ2 - ℎ
19 c. Inverted Differential Manometer ωP ωQ ① Divide into 2 parts - LEFT and RIGHT ② Determine the datum line – B and C ③ Write the equation for LEFT = − ℎ1 − ℎ = − ℎ1 − ℎ ④Write the equation for RIGHT PD = PB - ℎ2 = PB - ℎ2 ⑤ PC = PD (Datum line at same level) ⑥ Substitute Equation LEFT and RIGHT into PC = PD PC = PD − ℎ1 − ℎ = PB - ℎ2 PA – PB = - ℎ2 + ℎ1 + ℎ
20 Tutorial 2 1. A U-tube manometer shown in figure below is used to measure the gauge pressure of water (mass density ρ = 1000 kg /m3 ). If the density of mercury is 13.6 × 103 kg/m3 , what will be the gauge pressure at A if h1 = 0.45 m and D is 0.7 m above BC. 2. A U-tube manometer shown in figure below is used to measure the gauge pressure of a fluid P of density ρ = 1000 kg/m3 . If the density of the liquid Q is 13.6 × 103 kg/m3 , what will be the gauge pressure at A if h1 = 0.15 m and h2 = 0.25 m above BC. Take into consideration patm = 101.3 kN/m2 . 3. U-tube manometer as shown in figure below, measures the pressure difference at points A and B. Fluid P is oil (S = 0.85 ) and Fluid Q is mercury (S = 13.6 ). Calculate the pressure difference if h = 2.0 m, h2 = 0.35 m and h1 = 0.5 m.
21 4. Figure shows a differential manometer connected at two points A and B. At A, air pressure is 10 kN/m2 . Find the absolute pressure at B. Given ρwater = 1000 kg/m3 . 5. 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 oil of relative density 0.8. 6. Figure below shows a u-tube manometer that used to measure the pressure difference between pipe P and pipe Q that contains water. If the fluid in u-tube is oil with specific gravity of 0.9, calculate the pressure difference between these two pipes in kN/m2 . Given M=80 cm and N=25 cm. water oil
22 Piezometer Piezometer is one of the simplest forms of manometers. It can be used for measuring moderate pressures of liquids. The setup of piezometer consists of a glass tube, inserted in the wall of a vessel or of a pipe. The tube extends vertically upward to such a height that liquid can freely rise in it without overflowing. The pressure at any point in the liquid is indicated by the height of the liquid in the tube above that point. Pressure at point A can be computed by measuring the height to which the liquid rises in the glass tube. The pressure at point A is given by P = ωh, where ω is the specific weight of the liquid. Barometer A barometer is a device for measuring atmospheric pressure. The barometer, invented by the Italian mathematician and physicist Evangelista Torricelli (1608–1647) in 1643, is constructed from a glass tube closed at one end and filled with mercury. The tube is then inverted and placed in a pool of mercury. This device measures atmospheric pressure, rather than gauge pressure, because there is a nearly pure vacuum above the mercury in the tube. The height of the mercury is such that Patm = gh. When atmospheric pressure varies, the mercury rises or falls.
23 Tutorial 3 1. A pressure tube is used to measure the pressure of oil (mass density,640 kg/m3 ) in a pipeline. If the oil rises to a height of 1.2 above the centre of the pipe, what is the gauge pressure in N/m2 at that point? (gravity = 9.81 m/s2 ). 2. What is the atmospheric pressure in N/m2 if the level of mercury in a Barometer tube is 760 mm above the level of the mercury in the bowl? Given the specific gravity of mercury is 13.6 and specific weight of water is 9.81×103 N/m3 .
24 Buoyancy Principle of Archimedes Upthrust on body = weight of fluid displaced by the body If the body is immersed so that part of its volume, v1 is immersed in a fluid of density, 1 and the rest of its volume, v2 in another immiscible fluid of mass density, 2. The upthrust will act through the centre of gravity of the displaced fluid, which is called the centre of buoyancy. The positions of G1 and G2 are not necessarily on the same vertical line, and the centre of buoyancy of the whole body is, therefore, not bound to pass through the centroid of the whole body. Archimedes Principle states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object. Upthrust on upper part, R1 = 1gv1 acting through G1, the centroid of v1, Upthrust on lower part, R2 = 2gv2 acting through G2, the centroid of v2, Total upthrust = 1gv1 + 2 gv2
25 Types of Flow 1) Steady and Unsteady Flow: i) Steady Flow: A flow is said to be steady flow if different fluid properties and fluid velocity does not change with respect to time. ii) Unsteady Flow: A flow is said to be unsteady flow if different fluid properties and fluid velocity changes with respect to time.
26 2) Uniform and non-uniform flow: i) Uniform flow: In this type of flow, different fluid properties and fluid velocity does not vary with respect to space (or directions). E.g.: - Constant discharge through a constant diameter pipe. ii) Non-uniform flow: In this type of flow, different fluid properties and fluid velocity varies with respect to space (or directions). E.g.: - Constant discharge through a variable diameter pipe.
27 3) Laminar and Turbulent flow: i) Laminar flow: In a laminar flow, the flow of fluid is smooth and highly ordered. In this flow, the fluid layers move parallel to each other and do not cross each other. ii) Turbulent flow: In a turbulent flow, the flow of fluid is chaotic and not in any order. In this flow, the fluid layers cross each other and do not move parallel to each other.
28 Discharge and Mass Flowrate i) Discharge, Q 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 (m3 /s), or similar units and denoted by Q. ii) Mass Flowrate, ṁ 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 (kg/s), or similar units and denoted by ṁ. ṁ1 = ṁ2 1 x A1 x v1 = 2 x A2 x v2 1Q1 = 2Q2 Q = A v A = Area, m2 v = Velocity, m/s ṁ = x A x v = Q = mass density, kg/m3 A = Area, m2 v = Velocity, m/s
29 Continuity Equation Law 1) Single pipe with same diameter Qin = Qout Q1 = Q2 A1v1 = A2v2 2) Single pipe with different diameter Qin = Qout Q1 = Q2 A1v1 = A2v2 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.
30 3) Single pipe on a diffuser Qin = Qout Q1 = Q2 A1v1 = A2v2 4) Branch pipe Qin = Qout Q1 = Q2 + Q3 A1v1 = A2v2 + A3v3
31 Tutorial 4 1. For the pipe in figure below, the following data are given: Find: a. The discharge, Q b. The fluid velocity at station 2, v2 2. Water flows through a pipe with a diameter of 50 mm. then the pipe split into two, one of the pipes has a diameter 25 mm with the velocity of flow 0.4 m/s and the other one has a diameter 15 mm with the velocity 0.6 m/s. Calculate the velocity in the main pipe. 3. A pipe is split into 2 pipes which are BC and BD as shown in the figure below. The following information is given: Calculate: a. discharge at section A if vA = 2 m/s b. velocity at section B and section D if velocity at section C = 4 m/s 4. Oil is flow in a 20mm pipe. The pipe is split into two pipe which is the first pipe is 10mm diameter with velocity 0.3m/s and the other pipe is 15mm diameter with velocity 0.6m/s. Calculate the flow rate for 20mm pipe. 5. Water is flow in 50cm pipe which is connected from two pipes A and B, which is the diameter for both pipe are 25cm and 30cm. The velocity for both pipes A and B are 0.3m/s and 0.2m/s. Calculate the flow rate of pipe A, B and C and also find the velocity for pipe C. D1 = 10 cm V1 = 4 m/s D2 = 5 cm diameter pipe AB at A = 0.45 m diameter pipe AB at B = 0.3 m diameter pipe BC = 0.2 m diameter pipe BD = 0.15 m
32 Bernoulli Theorem 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. = + + 2 2 = According to Bernoulli’s equation; The energy in section 1 = energy in section 2 1 + 1 + 1 2 2 = 2 + 2 + 2 2 Total Energy/ Total Head Potential Energy/ Potential Head Pressure Energy/ Pressure Head Kinetic Energy/ Velocity Head
33 Assumptions The following are the assumptions made in the derivation of Bernoulli’s equation: ▪ The fluid is ideal or perfect, that is viscosity is zero. ▪ The flow is steady (The velocity of every liquid particle is uniform). ▪ There is no energy loss while flowing. ▪ The flow is incompressible. ▪ The flow is Irrotational. ▪ There is no external force, except the gravity force, is acting on the liquid. Appication of Bernouli's Theorem PIPE -Horizontal Pipe -Inclined Pipe PITOT TUBE ORIFICE METER VENTURI METER - Horizontal -Inclined
34 Horizontal Pipe = + + 2 2 Inclined Pipe 1 + 1 + 1 2 2 = 2 + 2 + 2 2
35 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. Where: Cd = coefficient of Discharge A1 = Area of entrance v1 = Velocity at entrance g = gravity (9.81 m/s2 ) m = Area ratio H = Pressure difference expressed as a head of the liquid flowing in venturi meter (m) x = Difference of level in U-tube manometer ωHg = Specific weight of Mercury ωsub = Specific weight of fluid entering the venturi meter P1 – P2 = Difference of pressure in entrance and throat
36 Basic step to calculate the Actual Discharge of Venturi meter: 1. Find the Value of A1 2. Find the Value of Area Ratio, m = 1 2 = 1 2 2 2 3. Find the Value of H using: = 1 − 2 = − 1 4. Insert in QActual Equation ℎ = 1 2 2 − 1 Qactual = Cd x A1 x v1 Qactual = Cd X Qtheory = Cd 1 2 2−1
37 Inclined Venturi Meter Where: Cd = coefficient of Discharge A1 = Area of entrance v1 = Velocity at entrance g = gravity (9.81 m/s2 ) m = Area ratio H = Pressure difference expressed as a head of the liquid flowing in venturi meter (m) x = Difference of level in U-tube manometer ωHg = Specific weight of Mercury ωsub = Specific weight of fluid entering the venturi meter P1 – P2 = Difference of pressure in entrance and throat Z1 – Z2 = Difference of height between entrance and throat
38 Basic step to calculate the Actual Discharge of Inclined Venturi meter: 1. Find the Value of A1 2. Find the Value of Area Ratio, m = 1 2 = 1 2 2 2 3. Find the Value of H using: = 1 − 2 + (1 − 2) = − 1 + (1 − 2) 4. Insert in QActual Equation ℎ = 1 2 2 − 1 Qactual = Cd x A1 x v1 Qactual = Cd X Qtheory = Cd 1 2 2−1
39 Orifice meter The Orifice Meter consists of a flat orifice plate with a circular hole drilled in it. There is a pressure tap upstream from the orifice plate and another just downstream. There are three recognized methods of placing the taps and the coefficient of the meter will depend upon the position of the taps. Where is, A = area of section v = velocity of section P = pressure of section Pitot Tube The Pitot Tube is a device used to measure the local velocity along a streamline. The pitot tube has two tubes: one is a static tube (b), and another is an impact tube(a). Theoretically = √(2ℎ) , pitot tubes may require calibration. The actual velocity is then given by = √(2ℎ) , where C is the coefficient of the instrument.
40 Tutorial 5 1. Water flows through a pipe 36 m from the sea level as shown in figure below. Pressure in the pipe is 410 kN/m2 and the velocity is 4.8 m/s. Calculate total energy of water above the sea level. 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 inlet is 5 kg/cm2 . If the velocity at M is 1 m/s, determine the pressure at N in kN/m2 . 3. A horizontal venturi meter has a diameter of entrance and throat 0.25 m and 0.18 m respectively. The pressure at entrance is 280 kN/m2 and throat is 100 kN/m2 . Determine the actual discharge of oil if Soil = 0.9 and coefficient of discharge is 0.92. 4. 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 (SHg = 13.6), calculate the actual discharge when the difference of level in the U-tube is 55 mm. 5. A horizontal venturi meter has a diameter of throat 150 mm and 250 mm for entrance. This venturi meter is used to measure the flow rate of fluid that has mass density 900 kg/m3 . A Utube mercury (SHg = 13.6) is use and show the difference of level 0.15 m. Calculate the Qactual if Cd = 0.98. 6. A vertical venturi meter measured the flow of water 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 250 mm above the entrance. If the coefficient for the meter is 0.97 and the pressure difference is 50 kN/m2 , calculate the actual discharge in m3 /s.
41 7. 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. 8. A meter orifice has a 100 mm diameter circular 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 difference of level is measured by manometer is 750 mm. Calculate the actual flow rate of the oil through the pipe.
42 TOPIC 3: PROPERTIES OF PURE SUBSTANCES Importance of steam Steam, odorless, invisible gas consisting of vaporized water. In nature, steam is produced by the heating of underground water by volcanic processes and is emitted from hot springs, geysers, fumaroles, and certain types of volcanoes. Steam also can be generated on a large scale by technological systems, as, for example, those employing fossil-fuel-burning boilers, coal-fired power plants, nuclear reactors, or even by sunlight in a solar thermal power plant. Steam power constitutes an important power source for industrial society. Water is heated to steam in power plants, and the pressurized steam drives turbines that produce electrical current. The thermal energy of steam is thus converted to mechanical energy, which in turn is converted into electricity. Example of a steam power plant layout Steam is also widely employed in such industrial processes as the manufacture of steel, aluminum, copper, and nickel; the production of chemicals; and the refining of petroleum. In the home, steam has long been used for cooking and heating. Steam used in domestic household appliances.
43 PURE SUBSTANCES • A pure substance is a material that has constant composition (is homogeneous) and has consistent properties throughout the sample. It may exist in more than one phase. • There are 3 principal phases: Phase Change Process Solid Liquid Gas
44 T-V and P-V Diagram The variations of properties during phase‐change processes are best studied and understood with the help of property diagrams. There are T‐v, P‐v, and P‐T diagrams for pure substances. T-V diagram P-V diagram