Power Plant Engineering Equipment 51 | 4.2.1 Steam Turbine Figure 4.2: Steam Turbine Figure 4.2 shows a one-tenth scale, cutaway model of a steam turbine at Think Tank, the science museum in Birmingham, England. Steam enters from the left through the gray pipe at the top, then passes through the low-pressure reaction turbine in the center, which drives the electricity generator on the right ▪ Of all the heat engine and prime movers, the steam turbine is the nearest to the ideal and it is widely used in power plants and in all industries where power and/or heat is needed for processes: such as pulp mills, refineries, Petro-chemical plants, food processing plants, incinerating and heating plants. ▪ Operation principle: In principle, the impulse steam turbine consists of a casing containing stationary steam nozzles and a rotor with moving or rotating buckets. The steam passes through the stationary nozzles and is directed at high velocity against the rotor buckets causing the rotor to rotate at high speed. ▪ The following events take place in the nozzles: ▪ The steam pressure decreases ▪ The enthalpy of the steam decreases ▪ The steam velocity increases ▪ The volume of the steam increases.
Power Plant Engineering Equipment 52 | 4.2.2 Gas Turbine Figure 4.3: Gas Turbine • A gas turbine has a compressor, combustion chamber, and turbine. The turbine and the compressor are on the same shaft. • The compressor raises the pressure of atmospheric air and sends this air to the combustion chamber. • Here a fuel (oil, gas, or pulverized coal) burns, raising the temperature and increasing the heat energy. • The hot gas in the turbine expands to develop mechanical energy, as expanding steam does in a steam turbine. • The basic parts of a turbine are rotor, which has blades projecting radials from its periphery; and nozzles, through which the gas is expanded and directed. • The conversion of kinetic energy to mechanical energy occurs at the blade. • The basic distinction between the type of turbines is the manner in which the gas causes the turbine rotor to move.
Power Plant Engineering Equipment 53 | 4.2.3 Jet Engine Figure 4.4: Jet Engine • Jet engines move the airplane forward with a great force that is produced by a tremendous thrust and cause the plane to fly very fast. • All jet engines, which are also called gas turbines, work on the same principle. • The engine sucks air in at the front with a fan. • A compressor raises the pressure of the air. • The compressor is made up of fans with many blades and attached to a shaft. • The blades compress the air. • The compressed air is then sprayed with fuel and an electric spark lights the mixture. • The burning gases expand and blast out through the nozzle, at the back of the engine. • As the jets of gas shoot backward, the engine and the aircraft are thrust forward as shown above. • In a jet engine airplane, thrust is a result of hot gases (exhaust) rushing out of the engine’s nozzle. The action of the gases rapidly moving backward causes a reaction in the air. • The air puts out a force equal to the thrust, but in the opposite direction, moving the airplane forward.
Power Plant Engineering Equipment 54 | 4.2.4 Injector Figure 4.5: Injector ▪ One example of an injector is a fuel injector. ▪ It is an electronically controlled valve. ▪ It is supplied with pressurized fuel by the fuel pump in your car, and it is capable of opening and closing many times per second. ▪ When the injector is energized, an electromagnet moves a plunger that opens the valve, allowing the pressurized fuel to squirt out through a tiny nozzle. ▪ The nozzle is designed to atomize the fuel to make the fuel as fine mist as possible so that it can burn easily. ▪ The injectors are mounted in the intake manifold so that they spray fuel directly at the intake valves. ▪ A pipe called the fuel rail supplies pressurized fuel to all of the injectors.
Power Plant Engineering Equipment 55 | 4.3 Types of Nozzles Figure 4.6: Types of Nozzles Nozzles can be described as convergent (narrowing down from a wide diameter to a smaller diameter in the direction of the flow) or divergent (expanding from a smaller diameter to a larger one). A nozzle has a convergent section followed by a divergent section and is often called a convergentdivergent nozzle. 4.3.1 Convergent Nozzle When the cross-section of a nozzle decreases continuously from entrance to exit, it is called a convergent nozzle. The cross-sectional area of the nozzle is dictated by the expansion process and the condition of the steam at inlet and at exit. In convergent nozzles, the cross-sectional area diminishes from the inlet section to the outlet section. This type of nozzle is useful up to a pressure ratio of 0.58 when using saturated steam. This is known as critical pressure ratio and is different for different fluids. Convergent nozzles accelerate subsonic fluids. If the nozzle pressure ratio is high enough the flow will reach sonic velocity at the narrowest point (i.e. the nozzle throat). In this situation, the nozzle is said to be choked.
Power Plant Engineering Equipment 56 | 4.3.2 Convergent-Divergent Nozzle When the cross-section of a nozzle first decreases from its entrance to throat, and then increases from its throat to exit, it is called a convergent –divergent nozzle. A convergent-divergent nozzle or CD nozzle is a tube that is pinched in the middle, making a carefully balanced, asymmetric hourglass-shape. It is used to accelerate a hot, pressurized gas passing through it to a supersonic speed, and upon expansion, to shape the exhaust flow so that the heat energy propelling the flow is maximally converted into directed kinetic energy. Because of this, the nozzle is widely used in some types of steam turbines, and is used as a rocket engine nozzle. It also sees use in supersonic jet engines. Flow of steam through nozzle Consider a nozzle as shown below: Figure 4.7: Convergent Nozzle Applying the energy equation to section 1 and 2 of the convergent nozzles and considering a flow rate of one kg per second.
Power Plant Engineering Equipment 57 | 2 2 2 2 2 2 1 1 C W Q h C h + = + Where h = enthalpy C = velocity of steam W = work Q = heat transfer Since the expansion through a nozzle is considered as isentropic, and as there is no external work done during the flow of steam, both the heat transfer and work transfer have zero values. Q = 0 and W = 0. Using these conditions, the above equation simplifies to: 2 2 2 2 2 2 1 1 C W Q h C h + = + 2 2 2 2 2 2 1 1 C h C h + = + ( ) 2 2 2 1 1 2 2 2 C h h C = − + ( ) 2 C2 = 2 h1 − h2 +C1 Usually, the velocity of steam entering the nozzle (C1) is very small as compared to the velocity at exit, and therefore, C1 can be neglected. Otherwise the quantity, 0 2 1 1 2 h C h = + may be used where h0 is known as the enthalpy at inlet. ( ) ( ) C2 = 2 h1 − h2 = 2Cp T1 −T2 ( ) 3 C2 = 2 h1 − h2 10 where (h) is in kJ/kg ( ) 2 72 1 2 C = 44. h − h
Power Plant Engineering Equipment 58 | If the area at the section X –X is A, and the specific volume is v, then the mass flow rate will be: Mass flow, v CA m = . Area per unit mass flow, C v m A = . . Then substituting for the velocity C, Area per unit mass flow = ( ) . 2 h1 ho v − 4.4 Critical Pressure Ratio It has been stated that the velocity at the throat of a correctly designed nozzle is the velocity of sound. In the same way for a nozzle that is convergent only, then the fluid will attain sonic velocity at exit if the pressure drop across the nozzle is large enough. The ratio of the pressure at the section where sonic velocity is attained to the inlet pressure of a nozzle is called the critical pressure ratio. Critical Pressure Ratio = 1 1 1 2 − + = P Pc For a perfect gas the pressure ratio required to attain sonic velocity in a nozzle depends only on the value of for the gas. For example, for air = 1.4, therefore 1 1 1 2 − + = P Pc 0.5283 1.4 1 2 1.4 1 1.4 1 = + = − P Pc For example, for carbon dioxide = 1.3, therefore
Power Plant Engineering Equipment 59 | 1 1 1 2 − + = P Pc 0.5457 1.3 1 2 1.3 1 1.3 1 = + = − P Pc The ratio of the temperature at the section of the nozzle where the velocity is sonic to the inlet temperature is called the critical temperature ratio. Critical temperature ratio = 1 2 1 + = T Tc Critical velocity, Cc = ( ) RTc 4.5 Important Formula Entrance at Throat Critical Pressure Ratio = 1 1 1 2 − + = P Pc 0.5283 1.4 1 2 1.4 1 1.4 1 = + = − P Pc Critical temperature ratio = 1 2 1 + = T Tc Critical velocity, Cc = ( ) RTc Mass flow, v CA m = . = . m mass flow rate (kg/s) C = velocity of stream (m/s) A = Area (m2 ) V = volume (m3 )
Power Plant Engineering Equipment 60 | PCVC = mRTc 4.6 Important Formula Exit Areas 1 2 1 2 1 − = P P T T ( ) ( ) C2 = 2 h1 − h2 = 2Cp T1 −T2 P2V2 = mRT2 Mass flow, v CA m = .
Power Plant Engineering Equipment 61 | TUTORIAL: NOZZLE QUESTION 1: Air at 8.6 bar and 190C expands at the rate of 4.5 kg/s through a convergent-divergent nozzle into a space at 1.03 bars. Assuming that the inlet velocity is negligible, calculate the throat and the exit cross-sectional areas of the nozzle. QUESTION 2: A fluid at 6.9 bar and 93C enters a convergent nozzle with negligible velocity, and expands isentropically into a space at 3.6 bar. Calculate the mass flow per square meter of exit area: (i) When the fluid is helium (Cp = 5.19 kJ/kg K ) Assume helium is perfect gas and take molar masses as 4 kg/kmol. QUESTION 3: Calculate the throat and exit areas of a nozzle to expand air at the rate of 4.5 kg/s from 8.3 bar, 327 C into a space at 1.38 bar. Neglect the inlet velocity and assume isentropic flow. QUESTION 4: Fluid at 7.3 bar and 95oC enters a convergent nozzle with negligible velocity and expands isentropically into a space at 3.8 bar. Calculate outlet temperature and mass flow per m2 of exit area when the fluid is helium (Cp = 5.19 kJ/kgK) Assume helium is perfect gas and take molar masses as 4 kg/ kmol. QUESTION 5: Fluid flows through a horizontal nozzle. The fluid has an enthalpy 245 kJ/kg at entrance point and 546 kJ/kg at channel exit of nozzle with flow velocity at 90 m/s. The entrance area and specific volume
Power Plant Engineering Equipment 62 | given are 0.23m2 and 0.41 m3 /kg. The heat transfers from the nozzle is estimated at 0.02 kJ/kg. Calculate: i. Velocity at entrance nozzle ii. Mass flow rate of fluid QUESTION 6: Air at 400ºC flows through a convergent-divergent nozzle with a specific volume of 0.345 m3 /kg. The temperature decreases at throat to 314ºC. The air flow rate is 4.6 kg/minute. Given a specific gas constant at 0.287kJ/kg K, calculate: i. Ratio of specific heat ii. Ratio of critical pressure iii. Throat area QUESTION 7: The inlet condition of air to a convergent divergent nozzle is 2.2 MN/m2 and 260C. The exit pressure is 0.4 MN/m2 . Determine: i. The flow rate for a throat area of 32.2 cm2 . ii. The exit area
Power Plant Engineering Equipment 63 | CHAPTER 5: VAPOUR COMPRESSION SYSTEM 5.1 Vapour Compression System • Refrigeration may be defined as the process of removing heat from a body or enclosed space so that its temperature is first lowered and then maintained at a level below the temperature of surroundings. • The system maintained at the lower temperature is known as refrigerated system while equipment used to maintain this lower temperature is known as refrigerating equipment. • Work is required to transfer heat from lower temperature body to higher temperature body. • Amount heat removed by refrigerating equipment from refrigerated system is known as refrigerating effect. Unit kJ/s. • Effectiveness of refrigeration is given by Coefficient of Performance (COP) • COP = Refrigerating effect / Work supplied. 5.2 Application of Refrigeration System • Comfort air conditioning of auditoriums, hospitals, residents, offices, hotel. • Manufacturing and preservation of medicine • Storage and transportation of food stuffs such as dairy products, fruits, vegetables, meat and fish. • Processing of textiles, printing work and photo graphics materials. • Manufacturing of ice • Cooling of concrete for dam • Treatment of air for blast furnace • Processing of petroleum and other chemical products. • Production of rockets fuel. • Computer functioning • 5.3 Vapour Compression Process • In a simple vapour compression system fundamental processes are completed in one cycle. These is: o COMPRESSION (COMPRESSOR)
Power Plant Engineering Equipment 64 | o CONDENSATION (CONDENSER) o EXPANSION (EXPANSION VALVE) o VAPOURISATION (EVAPORATOR) 5.3.1 The Compressor • The compressor is the heart of the system. The compressor does just what its name is. It compresses the low pressure refrigerant vapor from the evaporator and compresses it into a high pressure vapor. • The inlet to the compressor is called the “Suction Line”. It brings the low pressure vapor into the compressor. • After the compressor compresses the refrigerant into a high pressure Vapor, it removes it to the outlet called the “Discharge Line”. Figure 5.1: The Compressor 5.3.2 The Condenser • The “Discharge Line” leaves the compressor and runs to the inlet of the condenser. • Because the refrigerant was compressed, it is a hot high pressure vapor (as pressure goes up – temperature goes up). • The hot vapor enters the condenser and starts to flow through the tubes. • Cool air is blown across the outside of the finned tubes of the condenser (usually by a fan or water with a pump). • Since the air is cooler than the refrigerant, heat jumps from the tubing to the cooler air (energy goes from hot to cold – “latent heat”).
Power Plant Engineering Equipment 65 | • As the heat is removed from the refrigerant, it reaches its “saturated temperature” and starts to “flash” (change states), into a high-pressure liquid. • The high-pressure liquid leaves the condenser through the “liquid line” and travels to the “metering device”. Sometimes running through a filter dryer first, to remove any dirt or foreign particles. Figure 5.2: The Condenser Placement in Refrigeration Unit 5.3.3 Thermal Expansion Valves • A very common type of metering device is called a TX Valve (Thermostatic Expansion Valve). This valve has the capability of controlling the refrigerant flow. If the load on the evaporator changes, the valve can respond to the change and increase or decrease the flow accordingly. • The TXV has a sensing bulb attached to the outlet of the evaporator. This bulb senses the suction line temperature and sends a signal to the TXV allowing it to adjust the flow rate. This is important because, if not all, the refrigerant in the evaporator changes state into a gas, there could be liquid refrigerant content returning to the compressor. This can be fatal to the compressor. Liquid cannot be compressed and when a compressor tries to compress a liquid, mechanical failing can happen. The compressor can suffer mechanical damage in the valves and bearings. This is called” liquid slugging”. • Normally TXV's are set to maintain 10 degrees of superheat. That means that the gas returning to the compressor is at least 10 degrees away from the risk of having any liquid.
Power Plant Engineering Equipment 66 | Figure 5.3: The Thermal Expansion Valve Unit 5.3.4 The Evaporator • The evaporator is where the heat is removed from your house, business or refrigeration box. • Low pressure liquid leaves the metering device and enters the evaporator. • Usually, a fan will move warm air from the conditioned space across the evaporator finned coils. • The cooler refrigerant in the evaporator tubes, absorb the warm room air. The change of temperature causes the refrigerant to “flash” or “boil”, and changes from a low-pressure liquid to a low pressure cold vapor. • The low pressure vapor is pulled into the compressor and the cycle starts over. • The amount of heat added to the liquid to make it saturated and change states is called “Super Heat”. • One way to charge a system with refrigerant is by super heat. Figure 5.4: The Evaporator Unit
Power Plant Engineering Equipment 67 | 5.4 Basic Refrigeration Cycle • Starting at the compressor; • Low pressure vapor refrigerant is compressed and discharged out of the compressor. • The refrigerant at this point is a high temperature, high pressure, “superheated” vapor. • The high-pressure refrigerant flows to the condenser by way of the "Discharge Line". • The condenser changes the high-pressure refrigerant from a high temperature vapor to a low temperature, high pressure liquid and leaves through the "Liquid Line". • The high-pressure refrigerant then flows through a filter dryer to the Thermal Expansion valve or TXV. • The TXV meters the correct amount of liquid refrigerant into the evaporator. • As the TXV meters the refrigerant, the high-pressure liquid changes to a low pressure, low temperature, saturated liquid/vapor. • This saturated liquid/vapor enters the evaporator and is changed to a low pressure, dry vapor. • The low pressure, dry vapor is then returned to the compressor in the "Suction line". • The cycle then starts over. Figure 5.5: The Refrigeration Cycle 5.5 The Refrigerant Cycle: What Is It? • An air conditioner works similar to a refrigerator. The refrigerant flows through the system, and changes in state or condition. There are four processes in the 'refrigeration cycle'.
Power Plant Engineering Equipment 68 | o Processes 1: The compressor which pumps the refrigerant around the system, is the heart of the air conditioner. Before the compressor, the refrigerant is a gas at low pressure. Because of the compressor, the gas becomes high pressure, gets heated and flows towards the condenser. o Process 2: At the condenser, the high temperature, high pressure gas releases its heat to the outdoor air and becomes sub cooled high pressure liquid. o Process 3: The high-pressure liquid goes through the expansion valve, which reduces the pressure, and thus temperature goes below the temperature of the refrigerated space. This results in cold, low pressure refrigerant liquid. o Process 4: The low-pressure refrigerant flows to the evaporator where it absorbs heat from the indoor air through evaporation and becomes low pressure gas. The gas flows back to the compressor where the cycle starts all over again. In case of a heat pump the cycle can be reversed. Figure 5.5: The Air Conditioner Cycle
Power Plant Engineering Equipment 69 | 5.6 Vapour Compression System Figure 5.6: Figure 5.7: Condenser Evaporator High Pressure Side Low Pressure Side Compressor Expansion Device 3 4 1 2 Condenser Evaporator High Pressure Side Low Pressure Side Compressor Expansion Device 3 4 1 2 The superheated vapor enters the compressor where its pressure is raised
Power Plant Engineering Equipment 70 | Figure 5.8: Figure 5.9: Condenser Evaporator High Pressure Side Low Pressure Side Expansion Device Compressor 3 4 1 2 Condenser Evaporator High Pressure Side Low Pressure Side Compressor Expansion Device 3 4 1 2 The high pressure superheated gas is cooled in several stages in the condenser Liquid passes through expansion device, which reduces its pressure and controls the flow into the evaporator
Power Plant Engineering Equipment 71 | Figure 5.10: 5.7 Refrigerants Defination: • A refrigerant is defined as any substance that absorbs heat through expansion or vaporisation and loses it through condensation in a refrigeration system. Figure 5.11: Condenser Evaporator High Pressure Side Low Pressure Side Compressor Expansion Device 3 4 1 2 Low pressure liquid refrigerant in evaporator absorbs heat and changes to a gas
Power Plant Engineering Equipment 72 | 5.7.1 Classification of Refrigerants a) Azeotrope Refrigerants • The term ‘azeotrope’ refers to a stable mixture of refrigerants whose vapour and liquid phase retain identical compositions over a wide range of temperature. o R-500 o R-502 o R-503 o R-504 b) Halo-carbon refrigerants o R-11 o R-12 o R-13 o R-14 o R-21 o R-22 o R-30 o R-40 o R-100 o R-113 o R-114 c) Inorganic Refrigerants • The inorganic refrigerants were exclusively used before the introduction of halocarbon refrigerants. • These refrigerants are still in due to their inherent thermodynamic and physical properties. o R-717 (Ammonia) o R-729 (Air) o R-744 (Carbon dioxide) o R-764 (Sulphur dioxide) o R-118 (Water) d) Hydro-Carbon
Power Plant Engineering Equipment 73 | • Most of the hydrocarbon refrigerants are successfully used in industrial and commercial installations. • They possess satisfactory thermodynamic properties but are highly flammable and explosive. o R-170 (Ethane) o R-290 (Propane) o R-600 (Butane) o R-1120 o R-1130 o R-1150 o R-1270 5.7.2 Desirable Properties of An Ideal Refrigerant • Low boiling point • High critical temperature • High latent heat of vaporisation • Low specific heat of liquid • Low specific volume of vapour • Non-corrosive to metal • Non-flammable and non-explosive • Non-toxic • Low cost • Easy to liquify at moderate pressure and temperature • Easy of locating leaks by odour or suitable indicator • Mixes well with oil 5.8 Reversed Heat Engine System Operating on The Carnot Cycle The best COP will be given by a cycle which is a Carnot cycle operating between the given temperature conditions. Such a cycle using a wet vapor as the working substance is shown diagrammatically in Figure below: Wet vapor is used as the example, since the process of constant-pressure heat supply and heat rejection are made at constant temperature, a necessary requirement of the Carnot cycle and one which is not fulfilled by using a superheated vapor.
Power Plant Engineering Equipment 74 | Figure 5.12 (a) and (b): The changes in the thermodynamic properties of the refrigerant throughout the cycle are indicated on the T-s diagram of Fig. 5.10.2 (b). The cycle events are as follows: 1 – 2 Isentropic compression process (Work required) Wet vapor at stage 1 enters the compressor and is compressed isentropic ally to state 2. The work input for this process is represented by W1-2 2 – 3 Condensation process (Heat rejected) The vapor enters the condenser at stage 2 and is condensed at constant pressure and temperature to state 3 when it is completely liquid. The heat rejected by the refrigerant is Q2-Q3 3 – 4 Isentropic expansion process (work produced) The saturated liquid expands isentropic ally where the temperature drops from T2 to T1 and its state changes to wet vapor. 4 – 1 Evaporation process (Heat supplied) At the lower pressure and temperature of state 4 the refrigerant enters the evaporator where the heat necessary for evaporation, Q4-1, is supplied from the cold source. The boundaries of the system are as shown in Figure 5.10.2 (b) and therefore • The Net Work input to the system W = W1-2 + W3-4 • The Net Heat supplied to the system Q = Q2-3 + Q4-1 • Refrigerating effect Q: (Maximum) Q = T (s1 – s4) = T (s2 – s3) • Refrigerating effect Q: (Actual) Q = h1 – h4
Power Plant Engineering Equipment 75 | • Coefficient of performance (COPref): (Actual) • Coefficient of performance (COPref): (Maximum) Example 1: A refrigerator has working temperatures in the evaporator and condenser coils of -30 and 32C respectively. What is the possible maximum COP? If the actual refrigerator has a COP of 0.75 of the maximums calculate the required power input for a refrigerating effect of 5 kW. SOLUTION: T1 = -30 +273 = 243 K and T2 = 32 + 273 = 305 K Actual COPref =0.75 x 3.92 =2.94 Required power input = 1.7 kW. 5.9 Vapour Compression Cycles • The most widely used refrigerators and heat pumps are those which use a liquefiable vapor as the refrigerant. − − = = 2 1 1 4 . 21 14 h h h h W Q COPref ( ) ( )( ) 2 1 1 . 2 1 1 4 1 1 4 T T T COP T T s s T s s COP ref ref − = − − − = W kW COP Q W h h h h W Q COP ref ref 1.7 2.94 5 14 2 1 1 4 . 21 14 = = = − − = = 3.92 305 243 243 2 1 1 = − = − = ref ref COP T T T COP
Power Plant Engineering Equipment 76 | • The evaporation and condensation process take place when the fluid is receiving and rejecting the specific enthalpy of vapor of vaporization, and these are constant –temperature and constantpressure process. • The cycle used in this process is reverse Carnot cycle but due to practical consideration have led to several modifications to the ideal Carnot cycle which transform to Rankine cycle. 5.10 Modification 5.10.1 Replacement of The Expansion Engine By A Throttle Valve • The plant is simplified by replacing the expansion cylinder with a simple throttle valve. • The throttling process was shown to occur such that initial enthalpy equals the final enthalpy. • The process is highly irreversible so that the whole cycle becomes irreversible. • The process is represented by the dotted line 3-4 on Figure 5.13 below. • Figure 5.13 shows that refrigerating effect Q1=(s1-s4), is reduced by using a throttle valve instead of the expansion cylinder. Figure 5.13: Reversed cycle using a throttle valve 5.10.2 Condition at The Compressor Inlet • In practice it is very difficult to determine condition 1 (figure 5.13) that is wet steam. • In a practical unit this process is extended to give the vapor a definite amount of superheated as it leaves the evaporator. • This is really undesirable, since the work to be done by the compressor is increased, as will be shown later.
Power Plant Engineering Equipment 77 | • It is a practical necessity to allow the refrigerant to become superheated in this way in order to prevent the carry-over of liquid refrigerant into the compressor, where it interferes with the lubrication. Figure 5.14: T-s diagram for a reversed cycle with superheated vapor in the compressor 5.10.3 Undercooling of The Condensed Vapor • The condensed vapor can be cooled at constant pressure to a temperature below that of the saturation temperature corresponding to the condenser pressure. • This effect is shown in Figure 5.12.3, in which the constant-pressure line is shown further from the liquid line than it would actually appear, in order to illustrate the point. • The effect of undercooling is to move the 3-4, representing the throttling process, to the left on the diagram. • The result of this is that the refrigerating effect in process 4-1 is increased. Figure 5.15: T-s diagram for a reversed cycle with undercooling in the condenser
Power Plant Engineering Equipment 78 | Figure above includes all the modification of this section, and this can be taken as showing the practical cycle. For process 4-1: h4 + Q1 = h1 + 0 Q1 = (h1 – h4) Refrigerating effect Q1 = (h1 – h4) For process 1-2: h1 + 0 = h2 + W W = (h2 – h1) Work done on the refrigerant = (h2 – h1) For process 2-3: h2 + Q2 = h3 + 0 Q2 = (h2 – h3) Heat rejected by the refrigerant = (h2 – h3) For process 3-4: h3 + 0 = h4 + 0 h3 = h4 In throttling process = (h3 – h4) The solution to numerical problems depends on the means of obtaining the enthalpies h1, h2, and h3. Example 2: A refrigeration machine operates between a condenser temperature of 45C and a refrigerant temperature of -15C using R12 refrigerant. Determine the actual COPref for the refrigerant: Modification replaces the expansion device by using a throttle valve h1 = f fg h + xh pada -15C From the diagram above s1 =s2 and s1 is in wet condition
Power Plant Engineering Equipment 79 | S2 = sf1 + x1sfg pada -15C 0.681 = 0.0906 + x1 (0.6145) = 0.6145 0.6811 − 0.00906 = 0.96 h1 = hf + xhfg at -15C h1 = 22.33 + 0.96 (180.97 -22.33) =174.62 kJ/kg h2 = hg at 45C h2= 204.87 kJ/kg. h3 = hf at 45C h3 = 79.71 kJ/kg. =h4 = hf at 45C h4 = 79.71 kJ/kg. (i) COPref = ( ) ( ) 2 1 1 4 h h h h W Q − − = = 30.25 94.91 204.87 174.62 174.62 79.71 = − − = 3.14 Example 3: A refrigeration machine operates between a condenser temperature of 45C and a refrigerant temperature of -15C using R12 refrigerant. Determine the actual COPref for the refrigerant: The modification continues the evaporation process until point 1 becomes dry saturation h1 = hg at -15C h1 = 180.97 kJ/kg h3 = h4 = hf at 45C h3 = h4 = 79.71 kJ/kg
Power Plant Engineering Equipment 80 | From the diagram, s1=s2 and s2 is in the superheated zone. S1 = Sg at -15C S1 = 0.7051 kJ/kgK S2 = Sg at 45C is at superheated vapor between 0 to 15K. From the interpolation: h2 = (216.74 204.87) 0.7175 0.6811 0.7051 0.6811 204.87 − − − + h2 = 212.70 kJ/kg. (i) COPref = ( ) ( ) 2 1 1 4 h h h h W Q − − = = 31.73 101.26 212.70 180.97 180.97 79.71 = − − = 3.191 Example 4: A refrigeration machine operates between a condenser temperature of 45C and a refrigerant temperature of -15C using R12 refrigerant. Determine the actual COPref for the refrigerant: Modifications whereby the liquid at condition 3 is allowed to undergo sub-cooling to 5C before entering the expansion valve h1 = hg at -15C h1 = 180.97 kJ/kg h3 = h4 = hf at 40C h3 = h4 = 74.59 kJ/kg From the diagram, s1=s2 and s2 is in the superheated zone. S1 = Sg at -15C S1 = 0.7051 kJ/kgK
Power Plant Engineering Equipment 81 | S2 = Sg at 45C is at superheated vapor between 0 to 15K. From the interpolation: h2 = (216.74 204.87) 0.7175 0.6811 0.7051 0.6811 204.87 − − − + = 212.70 kJ/kg. (i) COPref = ( ) ( ) 2 1 1 4 h h h h W Q − − = = 31.73 106.38 212.70 180.97 180.97 74.59 = − − = 3.35
Power Plant Engineering Equipment 82 | TUTORIAL: REFRIGERANT SYSTEM Question 1: An ammonia vapor – compression refrigerator operates between an evaporator pressure of 2.077 bars and a condenser pressure of 12.37 bars. The vapor is dry saturated at entry to the compressor. There is no under cooling in the condenser and isentropic compression may be assumed. Calculate: (i) COPref (ii) Refrigerating effect per unit mass Question 2: A heat pump uses ammonia as the refrigerant operates between saturation temperature of 6 and 38C. The refrigerant is compressed isentropically from dry saturation and there is 6K of under cooling in the condenser. Calculate: (i) COPref (ii) The mass flow of refrigerant per kilowatt power input (iii) The heat available per kilowatt power input Question 3: An ammonia refrigerator operates between evaporating and condensing temperature of -16 and 50C. The vapor is dry saturated at the compressor inlet, and there is no under cooling of the condensate. Calculate: (i) The refrigerating effects (ii) The mass flow rate per kilowatt of refrigeration capacity. (iii) The power input per kilowatt of refrigerating capacity (iv) COPref Question 4: In a refrigerating plant using R12 the vapor leaves the evaporator dry saturated at 1.826 bars and is compressed to 7.449 bars. The temperature of the vapor leaving the compressor is 45C. The liquid leaves the condenser at 25C and is throttled to the evaporator pressure. Calculate: (i) The refrigerating effects (ii) The specific work input (iii) The COPref
Power Plant Engineering Equipment 83 | Question 5: A vapor compression refrigerator using R 12 works between the temperature limits of -20C and 25C. The refrigerant leaves the compressor in dry saturated condition. If the liquid refrigeration is undercooled to 20C before entering the throttle valve, determine: (i) Work required to drive the compressor (ii) Refrigerating effect (iii) COPref Question 6: In a vapor compression refrigerating cycle using Freon 12 as refrigerant, the temperature in the evaporator is -10C and the pressure in the condenser is 13.66 bars. The vapor is dry saturated at the inlet to the compressor and the liquid which is under cooled by 10C enters the expansion valve of the cycle. (i) Sketch the cycle with the aid of temperature – entropy (T-s) (ii) The rate of flow of refrigerant in kg/s if 15kW of heat to be rejected in the cycle. (iii) The power input to the compressor (iv) The coefficient of performance
Power Plant Engineering Equipment 84 | CHAPTER 6: PSYCHROMETRY 6.0 Introduction • It is often necessary to provide a controlled atmosphere in buildings where industrial process is to be carried out. • The atmospheric air generally consists of water vapor. • The amount of water vapor plays important role in psychrometry. • If it exceeds or lowers certain limit it will create discomfort to us. • It is very important to keep the moisture content in the air within the specified limit in case of processing industries and air-conditioned buildings. • The science which deals with the study of behaviour of moist air (mixture of dry air and water vapor) is known as psychrometry. 6.1 Air Conditioning Terminology a) Dry Air • The dry air is nothing but the air without moisture or water vapor. • The pure dry air is a mixture of number of gasses such as nitrogen, oxygen, carbon dioxide, hydrogen etc. • The volumetric composition of dry air is 77% of nitrogen and 23% of oxygen. b) Moist Air • It is a mixture of dry air and water vapor. The amount of water vapor present in the most air varies with temperature. c) Moisture • The water vapor present in the air is known as moisture. d) Dry Bulb Temperature (DBT) • The temperature measured by an ordinary thermometer is known as dry bulb temperature.
Power Plant Engineering Equipment 85 | e) Wet Bulb Temperature (WBT) • It is the temperature of air measured by a thermometer when its bulb is covered with wet cotton and is exposed to a current rapidly moving air. f) Wet Bulb Depression (WBD) • It is the difference between dry bulb temperature and wet bulb temperature. g) WBD = DBT – WBT • The value of wet bulb depression is zero when the air becomes saturated. h) Dew Point Temperature • It is the temperature at which the water vapor present in air begins to condense when the air is cooled. • For saturated air, the dry bulb, wet bulb and dew point temperature are all same. i) Dew Point Depression (DPD) • It is the difference between dry bulb temperature and dew point temperature. Figure 6.1: Thermostat Wet Bulb
Power Plant Engineering Equipment 86 | 6.2 Specific Humidity (Or) Moisture Content (Or) Humidity Ratio • It is defined as the mass of water vapor present in one kg of dry air. • The specific humidity is the ratio of the mass of water vapor to the mass of dry air in a given volume of the mixture. = ms / ma • Where the subscript s denotes the vapor, and the subscript a denote the dry air. • From the equation: = 0.622 x (Ps / P – Ps) Where: P = Barometric Pressure (1.013 bar) Ps = Partial pressure of the water vapor Pa = Partial pressure of the dry air Where P = Pa + Ps • Specific humidity, = Mass of water vapor / Mass of dry air (water vapor), Where: Ps = Partial pressure of water vapor V = Volume of mixture R = Gas content for water vapor T = Temperature of mixture • Specific humidity, = = Mass of water vapor / Mass of dry air (dry air), Where: Pa = Partial pressure of dry air V = Volume of mixture R = Gas content for dry air T = Temperature of mixture
Power Plant Engineering Equipment 87 | 6.3 Relative Humidity Relative humidity plays a major role in the comfort air conditioning and industrial air conditioning to compare with specific humidity. It is defined as the actual mass of water vapor in a given volume to saturated mass of water in same volume and temperature. Relative humidity ( ) = ms / (ms) sat or ( ) = Ps / Pg Where Pg is the saturation pressure at the temperature of the mixture. 6.4 Percentage Saturation The term percentage saturation is also used, defined as the ratio of the specific humidity of a mixture to the specific humidity of the mixture when saturated at the same temperature, expressed as a percentage. = 100(p – pg) / 0.622pg Or pg corresponds to the dry bulb temperature. 6.5 Dalton’s Law of Partial Pressure The total pressure exerted by air and water vapor mixture is equal to the barometric pressure. Where P = Pa + Ps Where: P = Barometric Pressure (1.013 bar) Ps = Partial pressure of the superheated vapor Pa = Partial pressure of the dry air a) Partial pressure of water vapor is given by: − − = b s b g g s p p p p p p ( )( ) w b sw d w s sw t P P t t P P 1527.4−1.3 − − = −
Power Plant Engineering Equipment 88 | Psw = saturation pressure corresponding wet bulb temperature td = dry bulb temperature tw = wet bulb temperature Total enthalpy of moist air is the sum of the enthalpy of dry air and the enthalpy of water vapor associated with the dry air. b) Total enthalpy of moist air is given by: H = Cp.td + .hg Where: Cp = Specific heat at constant pressure = 1.005 kJ/kg K td = Dry bulb temperature = specific humidity hg = specific enthalpy air corresponding to dry bulb temperature.
Power Plant Engineering Equipment 89 | Tutorial: Psychrometry & Air Conditioning Question 1: Air at 32C is saturated with water vapor at a barometric pressure of 1.013 bars. Calculate: (i) The partial pressures of the vapor and dry air. (ii) The volume of the mixture per kilogram of vapor. (iii) The mass of dry air per kilogram of vapor. (iv) The specific humidity of the mixture (v) The relative humidity of the mixture Question 2: The pressure of the water vapor in an atmosphere which is at 32C and 1.013 bars is 0.02063 bars. Calculate: (i) The degree of superheat of the water vapor. (ii) The specific humidity (iii) The relative humidity (iv) The temperature to which the air must be cooled for it to become just saturated. Question 3: An air and water vapor at 1.013 bar and 26.7C have a specific humidity of 0.0085. Calculate the percentage saturation. Question 4: A mixture of air and water vapor at 1.013 bar and 16C has a dew point of 5C. Calculate: (i) Relative humidity (ii) Specific humidity Question 5: The readings taken in a room from a sling psychrometer give a dry and wet bulb temperature of 25C and 19.9C. using a psychometric chart, taking the atmospheric pressure as 1.01325 bars, calculate: (i) specific humidity (ii) relative humidity
Power Plant Engineering Equipment 90 | (iii) specific volume of the mixture (iv) specific enthalpy of the mixture Question 6: The air supplied to a room of a building in winter is to be at 17C and have a percentage relative humidity of 60%. If the barometric pressure is 1.01326 bars, calculate the specific humidity. What would be the dew point under these conditions? Question 7: The pressure and temperature of the air in a room are 1 bar and 28C. If the relative humidity is found to be 30 per cent, find: (i) the partial pressure of the water vapor (ii) the dew points (iii) the specific humidity (iv) density Question 8: A 5m x 5m x 3m room contains air at 25C and 100 kPa at a relative humidity of 75%. Determine: (i) the partial pressure of dry air (ii) the specific humidity of the air (iii) The masses of the dry air and water vapor in the room. Question 9: The atmospheric condition 1.0132 bar are 20C and specific humidity of 0.0095 kg/kg of dry air. Calculate the following: (i) Partial pressure of vapor (ii) Relative humidity (iii) Dew point temperature
Power Plant Engineering Equipment 91 | Question 10: Atmospheric air with barometric pressure of 1.013 bar has 40C dry bulb temperature and 26C wet bulb temperature. Without the aid of psychrometric chart, determine: (i) Relative humidity (ii) Saturation percentage
Power Plant Engineering Equipment 92 | Summary POWER PLANT ENGINEERING EQUIPMENT guides students to understand and comprehend in detail the working principle of the components’ basic equipment and the additional equipment contained in a power plant and processing plant. The learning components cover the functions and classification of the pump, compressed air plant, fuel and combustion, nozzle, refrigeration and air conditioning. Pump covers understand of different pumps, calculate the parameter for pump head losses and solve the problem related to different types of pumps. Compressed air plant covers the understand of their layout and components in compressor. Compressed air plant also covers the problem that need to be solved related to reciprocating compressed air. Fuel and Combustion covers the basic understanding and combustion equations. Students are required to solve the problem related to stoichiometric Air Fuel Ratio. Nozzle covers the application and how different types of nozzle and solve problem related to nozzle. Refrigeration describe the method used to increase refrigeration performance and solve the problem related to vapor compressed refrigeration cycle. And lastly Air conditioning cover the function of air conditioning and different type of air conditioning used in industry and in workplace. Students are to explain the used of psychometric chart and calculate the value for air conditioning.
Power Plant Engineering Equipment 93 | REFERENCES Thomas C. E, Kao Chen and Swanekamp, Robert C. (2005).Standard Handbook OfPower Plant Engineering. 2nd Edition. Tata McGraw-Hill. New Delhi. ISBN: 0070194351 Rajput,R.K(2010). ThermalEngineering in S.I Units. 8thEdition. LaxmiPublication, New Delhi Eastop,T.D and McConkey. (2009) Applied Thermodynamics for Engineering Technologists. 5th Edition. Pearson Education Ltd. New Delhi. Nag P.K. (2005). Power Plant Engineering.2nd Edition. Tata McGraw-Hill. New DelhiRajput, R.K. (2006). A Textbook of Power System Engineering.1st Edition. Laxmi Publication, New Delhi. RogerG.F.CandY.R.Mayhew(2005).Engineering Thermodynamics,Work and Heat Transfer. S.I Units, 3rd Edition, Longman Sharma,P,C (2018). A Text book of Power Plant Engineering. S.K Kataria & Sons. ISBN-13: 978-93-5014-384-1 Yunus A. Cengel, Micheal A. Boles. (2015). Thermodynamics: An Engineering Approach, 8th Edition. McGraw-Hill Education. New York. ISBN-13: 978- 0073398174 Richardson, Duncan C (2014). Plant Equipment and Maintenance Engineering Handbook. McGraw Hill Professional. ISBN 9780071809900 Afzairizal, M (2018). Pengenalan kepada Teknologi Kejuruteraan Loji. Pelangi Engineering Series. Pelangi Professional Publishing Sdn Bhd Selangor Nag P.K (2017). Power Plant Engineering. 4th Kindle Edition. Kumar, D.S (2017), Refrigeration and Air Conditioning. S.K Kataria & Sons. ISBN:978-93-5014-634-7