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Unit6 Unit6 Document Transcript

  • FLOW PROCESS J2006/6/1 UNIT 6FLOW PROCESS OBJECTIVESGeneral Objective: To understand the concept of flow process and its application in steady flow energy equation.Specific Objectives : At the end of the unit you will be able to:  derive the meaning and interpret the steady-flow energy equation  apply the steady-flow energy equation to : • boiler • condenser
  • FLOW PROCESS J2006/6/2 INPUT6.0 STEADY FLOW PROCESSES Do you know the conditions for Steady Flow Processes ?In heat engine it is the steady flow processes which are generally of most interest. Theconditions which must be satisfied by all of these processes are :i. The mass of fluid flowing past any section in the system must be constant with respect to time.ii. The properties of the fluid at any particular section in the system must be constant with respect to time.iii. All transfer of work energy and heat which takes place must be done at a uniform rate.A typical example of a steady flow process is a steam boiler, operating under a constant loadas shown diagrammatically in Fig. 6.0. In order to maintain the water level in the boiler, the STEAM OUTfeed pump supplies water at exactly the same rate as that at which steam is drawn off fromthe boiler. To maintain the production of steam at this rate at a steady pressure, the furnace WATERwill need to supply heat energy at a steady rate. Under these conditions, the properties of the LEVELworking fluid at any section within the system must be constant with respect to time. BOUNDRY WATER IN FURNACE
  • FLOW PROCESS J2006/6/3 Figure 6.0 Steam Boiler6.1 STEADY FLOW ENERGY EQUATION This equation is a mathematical statement on the principle of Conservation of Energy as applied to the flow of a fluid through a thermodynamic system. The various forms of energy which the fluid can have are as follows: a) Potential energy If the fluid is at some height Z above a given datum level, then as a result of its mass it possesses potential energy with respect to that datum. Thus, for unit mass of fluid, in the close vicinity of the earth, Potential energy = g Z ≈ 9.81 Z
  • FLOW PROCESS J2006/6/4 b) Kinetic energy A fluid in motion possesses kinetic energy. If the fluid flows with velocity C, then, for unit mass of fluid, C2 Kinetic energy = 2 c) Internal energy All fluids store energy. The store of energy within any fluid can be increased or decreased as a result of various processes carried out on or by the fluid. The energy stored within a fluid which results from the internal motion of its atoms and molecules is called its internal energy and it is usually designated by the letter U. If the internal energy of the unit mass of fluid is discussed this is then called the specific internal energy and is designated by u. d) Flow or displacement energy In order to enter or leave the system, any entering or leaving volume of fluid must be displaced with an equal volume ahead of itself. The displacing mass must do work on the mass being displaced, since the movement of any mass can only be achieved at the expense of work. Thus, if the volume of unit mass of liquid (its specific volume) at entry is v1 and its pressure is P1, then in order to enter a system it must displace an equal specific volume v1 inside the system. Thus work to the value P1v1 is done on the specific volume inside the system by the specific volume outside the system. This work is called flow or displacement work and at entry it is energy received by the system. Similarly, at exit, in order to leave, the flow work must be done by the fluid inside the system on the fluid outside the system. Thus, if the pressure at exit, is P2 and the specific volume is v2 the equation is then, Flow or displacement work rejected = P2v2 e) Heat received or rejected During its passage through the system the fluid can have direct reception or rejection of heat energy through the system boundary. This is designated by Q. This must be taken in its algebraic sense. Thus,
  • FLOW PROCESS J2006/6/5 Q is positive when heat is received. Q is negative when heat is rejected. Q = 0 if heat is neither received nor rejected. f) External work done During its passage through the system the fluid can do external work or have external work done on it. This is usually designated by W. This also must be taken in its algebraic sense. Thus if, External work is done by the fluid then W is positive. External work is done on the fluid then W is negative. If no external work is done on or by the fluid then W = 0. Figure 6.1 illustrates some thermodynamic system into which is flowing a fluid with pressure P1, specific volume v1, specific internal energy u1 and velocity C1. The entry is at height Z1 above some datum level. In its passage through the system, external work W may be done on or by the fluid and also heat energy Q may be received or rejected by the fluid from or to the surroundings. The fluid then leaves the system with pressure P2, specific volume v2, specific internal energy u2 and velocity C2. The exit is at height Z2 above some datum level. P1 v1 W U1C1 SYSTEM ENTRY (OR CONTROL VOLUME) Z1 P2 v2 u2 C2 EXIT Z2 Q Figure 6.1 Thermodynamic system
  • FLOW PROCESS J2006/6/6 The application of the principle of energy conservation to the system is, Total energy entering the system = Total energy leaving the system or, for unit mass of substance, C12 C2 gZ 1 + u1 + P1v1 + + Q = gZ 2 + u 2 + P2 v 2 + 2 + W (6.1) 2 2 This is called the steady flow energy equation. This equation is not applicable to all energy forms. In such cases, the energy forms concerned are omitted from the energy equation. In equation 6.1, it was stated that the particular combination of properties of the form, u + Pv is called specific enthalpy and is designated as h. Thus, the steady flow energy equation is written as C12 C2 gZ 1 + h1 + + Q = gZ 2 + h2 + 2 + W (6.2) 2 2 Steady flow energy equation Potential energy + Kinetic energy + Internal energy +Flow or Displacement energy+ Heat or Work6.2 APPLICATION OF STEADY FLOW EQUATION
  • FLOW PROCESS J2006/6/7 The steady flow energy equation may be applied to any apparatus through which a fluid is flowing, provided that the conditions stated previously are applicable. Some of the most common cases found in engineering practise are dealt with in detail as below. 6.2.1 Boilers In a boiler operating under steady conditions, water is pumped into the boiler along the feed line at the same rate as which steam leaves the boiler along the steam main, and heat energy is supplied from the furnace at a steady rate. 2 STEAM OUT 2 BOUNDRY SYSTEM 1 WATER IN 1 Q FURNACE Figure 6.2.1 Steam Boiler The steady flow energy equation gives C12 C2 gZ 1 + h1 + + Q = gZ 2 + h2 + 2 + W 2 2 and with the flow rate, m (kg/s) the equation may be written as   C 2 − C12 2   Q − W = m ( h2 − h1 ) +     + ( gZ 2 − gZ 1 )   (6.3)    2    In applying this equation to the boiler, the following points should be noted :
  • FLOW PROCESS J2006/6/8 i. Q is the amount of heat energy passing into the fluid per second ii. W is zero since a boiler has no moving parts capable of affecting a work transfer iii. The kinetic energy is small as compared to the other terms and may usually be neglected iv. The potential energy is generally small enough to be neglected. v. m (kg/s) is the rate of the flow of fluid. Hence the equation is reduced to Q = m( h2 − h1 )  (6.4) FOR BOILER UNDER A STEADY CONDITION, WORK = 0 KINETIC ENERGY = NEGLECTED POTENTIAL ENERGY = NEGLECTED
  • FLOW PROCESS J2006/6/9 6.2.2 Condensers In principle, a condenser is a boiler reverse. In a boiler, heat energy is supplied to convert the liquid into vapour whereas in a condenser heat energy is removed in order to condense the vapour into a liquid. If the condenser is in a steady state then the amount of liquid, usually called condensate, leaving the condenser must be equal to the amount of vapour entering the condenser. VAPOUR WATER OUT SYSTEM WATER BOUNDARY IN CONDENSATE Figure 6.2.2 Condenser The steady flow energy equation gives   C 2 − C12   Q − W = m ( h2 − h1 ) +  2    + ( gZ 2 − gZ 1 )      2    Points to note, i. Q is the amount of heat energy per second transferred from the system ii. W is zero in the boiler iii. The kinetic energy term may be neglected as in the boiler iv. The potential energy is generally small enough to be neglected v. m is the rate of the flow of fluid.  Thus the equation is reduced to Q = m( h2 − h1 )  (6.5)
  • FLOW PROCESS J2006/6/10 Example 6.1 A boiler operates at a constant pressure of 15 bar, and evaporates fluid at the rate of 1000 kg/h. At entry to the boiler, the fluid has an enthalpy of 165 kJ/kg and on leaving the boiler the enthalpy of the fluid is 2200 kJ/kg. Determine the heat energy supplied to the boiler. Solution to Example 6.1 2 STEAM OUT 2 SYSTEM 1 BOUNDARY WATER IN 1 Q The steady flow energy equation gives   C 2 − C12   Q − W = m ( h2 − h1 ) +  2    + ( gZ 2 − gZ 1 )      2    Q = heat energy per hour entering system W = work energy per hour leaving system = 0 m = fluid flow rate = 1000 kg/h h2 = 2200 kJ/kg h1 = 165 kJ/kg C1& C2 = neglected Z1& Z2 = neglected Thus, the steady flow energy equation becomes kJ kg kJ Q = 1000 ( 2200 − 165) h h kg kJ Q = 2.035 x10 6 h
  • FLOW PROCESS J2006/6/11 Example 6.2 If 65 % of the heat energy supplied to the boiler in example 6.1 is used in evaporating the fluid, determine the rate of fuel consumption required to maintain this rate of evaporation, if 1 kg of fuel produces 32000 kJ of heat energy. Solution to Example 6.2 2.035 x10 6 Heat energy from fuel required per hour = 0.65 = 3.13 x 106 kJ/h Heat energy obtained from the fuel = 32 000 kJ/kg 3.13 x10 6 kJ kg ∴ Fuel required = x 32000 h kJ = 97.8 kg/h
  • FLOW PROCESS J2006/6/12 Example 6.3 Fluid enters a condenser at the rate of 35 kg/min with a specific enthalpy of 2200 kJ/kg, and leaves with a specific enthalpy of 255 kJ/kg. Determine the rate of heat energy loss from the system. Solution to Example 6.2 The steady flow energy equation gives   C 2 − C12   Q − W = m ( h2 − h1 ) +  2    + ( gZ 2 − gZ 1 )      2    For a condenser, W = 0, and the term representing the change in kinetic and potential energy may be neglected. Therefore the equation is reduced to Q = m( h2 − h1 )  From the above equation kg kJ Q = 35 (255 − 2200) min kg = - 68 000 kJ/min
  • FLOW PROCESS J2006/6/13 Activity 6 TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 6.1 In an air conditioning system, air is cooled by passing it over a chilled water coil condenser. Water enters the coil with an enthalpy of 42 kJ/kg and leaves the coil with an enthalpy of 80 kJ/kg. The water flow rate is 200 kg/h. Find the rate of heat absorption by the water in kilowatts. 6.2 In a steady flow system, a substance flows at the rate of 4 kg/s. It enters at a pressure of 620 kN/m2, a velocity of 300 m/s, internal energy 2100 kJ/kg and specific volume 0.37 m3/kg. It leaves the system at a pressure of 130 kN/m2, a velocity of 150 m/s, internal energy 1500 kJ/kg and specific volume 1.2 m3/kg. During its passage through the system the substance has a loss by heat transfer of 30 kJ/kg to the surroundings. Determine the power of the system in kilowatts, stating whether it is from or to the system. Neglect any change in potential energy.
  • FLOW PROCESS J2006/6/14 Feedback to Activity 6 200 6.1 Data: m = 200 kg/h =  = 0.056 kg/s 3600 h1 = 42 kJ/kg; h2 = 80 kJ/kg Q =? The diameter of the water tube in a cooler is normally constant. Therefore, there is no change in water velocity and kinetic energy. In general the change in potential energy is also negligible. The equation of steady flow is therefore reduced to Q = m( h2 − h1 )  = 0.056(80 – 42) = 2.13 kJ/s or kW The rate of heat absorption by the water is 2.13 kW
  • FLOW PROCESS J2006/6/15 6.2 By neglecting the change in potential energy, for a unit mass of substance, the steady flow energy equation becomes: C12 C 22 u1 + P1v1 + + Q = u 2 + P2 v 2 + +W (1) 2 2 Q is written negative since 30 kJ/kg are lost to the surroundings. From equation (1) C12 − C 2 2 Specific W = (u1 − u1 ) + ( P1v1 − P2 v 2 ) + ( )−Q 2 Working in kilojoules (kJ) 300 2 − 150 2 Specific W = (2100-1500)+(620x0.37-130x1.2)+( )-30 2 x10 3 = 676.75 kJ/kg. The substance flows at the rate of 4 kg/s ∴Output (since W is positive) = 676.75 x 4 = 2707 kJ/s or kW
  • FLOW PROCESS J2006/6/16 SELF-ASSESSMENT You are approaching success. Try all the questions in this self-assessment section and check your answers with those given in the Feedback to Self-Assessment on the next page. If you face any problem, discuss it with your lecturer. Good luck. 1. A boiler uses coal at the rate of 3000 kg/h in producing steam with a specific enthalpy of 2700 kJ/kg from feed water with a specific enthalpy of 280 kJ/kg. The combustion of 1 kg of coal produces 28000 kJ, of which 80% is useful in producing steam. Calculate the rate at which steam is produced. 2. Fluid with specific enthalpy of 2280 kJ/kg enters a condenser at the rate of 4500 kg/h, and leaves with a specific enthalpy of 163 kJ/kg. If the enthalpy of the cooling water circulating through the condenser tubes increases at the rate of 148 000 kJ/min, determine the rate at which heat energy flows from the condenser to the atmosphere.
  • FLOW PROCESS J2006/6/17 Feedback to Self-Assessment Have you tried the questions????? If “YES”, check your answers now. 1. 27800 kg/h 2. 646000 kJ/h CONGRATULATIONS!!!!….. May success be with you always….