T.Y. Diploma : Sem. VI Refrigeration and Air Conditioning [ME/MH/MI/PG/PT]

T.Y. Diploma : Sem. VI Refrigeration and Air Conditioning [ME/MH/MI/PG/PT] Time: 3 Hrs.] Prelim Question Paper Solution [Marks : 100 Q.1(a) Attempt any THREE of the following : [12] Q.1(a) (i) Define C.O.P and unit of refrigeration also explain heat engine. [4] (A) The practical unit of refrigeration is expressed in terms of tonne of refrigeration (written as TR) A tonne of refrigeration is defined as the amount of refrigeration effect produced by the uniform melting of one tonne (1000 kg) of ice from and at 0C in 24 hours. Taking latent heat of ice = 335 KJ / Kg 1 TR = 1000 335 KJ in 24 hours = 1000 335 = 232.6 KJ / min 24 60 Coefficient of performance (COP) is the ratio of heat extracted in the refrigerator to the work done on the refrigerant. Q C.O.P = W Heat engine : In a heat engine, the heat supplied to the engine is converted into useful work. If Q 2 is the heat supplied to the engine and Q 1 is the heat rejected from the engine then the net work done by the engine is given by W E = Q 2 Q 1 T 2 Hot body (High Q Heat 2 E W Engine E Efficiency = Networkoutput Heatadded Q 1 Q = 1 2 Q 1 T 1 T 1 > T a Q Q Q = 2 1 Thus the efficiency of heat engine is always less than unity. 2 Cold body (Low temp) Q.1(a) (ii) Sketch Bell column cycle on P-V and T-S. List process involved. [4] (A) P V and T S diagram Process 1 2 The air is compressed isentropically Process 2 3 Heat rejected to atmosphere at constant pressure. Process 3 4 The air is expanded isentropically Process 4 1 Heat is absorbed at constant pressure. 1

: T.Y. Diploma RAC Q.1(a) (iii) Draw a neat labelled symmetric diagram of a simple vapour compression refrigeration system? (A) [4] Fig. :Flow diagram of simple vapour compression refrigeration cycle Q.1(a) (iv) Draw a neat labeled sketch and explain working of pulse tube refrigeration. [4] (A) Pulse tube refrigeration uses the idea of sudden pressurization and depressurization are employed to get refrigeration effect. Pulse Tube Refrigeration Working of pulse tube refrigeration The main components are pulse tube refrigerator, valve mechanism and cooling system. A high pressure gas from gas source is supplied at temperature equal to ambient temperature to the base of the pulse tube. Q.1(b) Attempt any ONE of the following : [6] Q.1(b) (i) Explain and draw Electrolux refrigeration system. [6] (A) 2 Fig. : Domestic electrolux type refrigerator

Prelim Question Paper Solution Operation or Working of Electrolux system : The strong aqua-ammonia solution is heated in the generator by external source usually a gas burner. During heating process aqua-ammonia vapour rises up through analyser where dehydration of aqua-takes places then it passes through a rectifier where complete water particles are removed and reback to generator and the strong ammonia vapour are sent to condenser. In condenser the ammonia vapour changes its phase from vapour to liquid. Condensation of ammonia vapour is done by external cooling source. The liquid refrigerant leaving the condenser flours under gravity to the evaporator where it meets with the hydrogen gas. The hydrogen gas which is in the evaporator increase the rate of evaporation. During the process of evaporation it absorbs latent heat from the refrigerated space and thus produces cooling effect. The mixture of ammonia vapour and hydrogen from evaporator is passed to the absorber where ammonia vapour is absorbed in water while the hydrogen rises to the top and flows back to the evaporator. Thus cycle is complete Heatabsorbedintheevaporator C.O.P. = Heatsup pliedinthegenerator Q.1(b) (ii) What is the ultimate effect of under cooling (sub-cooling) on C.O.P.? Show on P-h [6] and T-s diagram? (A) (1) Subcooling reduces flashing of the liquid during expansion and increases the refrigerating effect. (2) Thus the piston displacement and power per ton are reduced for all refrigerant. (3) The ultimate effect of undercooling or subcooling is to increase the value of co-efficient of performance under the same set of condition. Refrigerating effect (R.E) = h 1 h 4 = h 1 h 3 W C = Work done = h 2 h 1 C.O.P. = R.E. W = h1 h4 h h C 2 1 = h1 h3 h h 2 1 Fig. : Theoretical vapour compression cycle with subcooling or undercooling of the refrigerant Q.2 Attempt any TWO of the following : [16] Q.2(a) A refrigerating machine using R-12 as refrigerant operates between the pressures 2.5 [8] bar and 9.0 bar. The compression is isentropic and there is not under cooling in the condenser. The vapour is dry and saturated condition at the beginning of the compression. Estimate the theoretical COP. If the actual COP is 0.65 of theoretical COP, calculate the net cooling produced per hour. The refrigerant flow is 5 Kg/min. The properties of refrigerant are : 3

: T.Y. Diploma RAC (A) Pressure Satu. temp. Enthalpy (kj/kg) Entropy (kj/kg K) (Bar) (C) Liquid Vapour Vapour 9.0 36 70.55 201.8 0.6836 2.5 7 29.62 184.5 0.7001 Take specific heat of superheated vapour at 9 bar as 0.64 kj/kg K. Compressor pressure = 9 bar Evaporator pressure = 2.5 bar From the properties of Refrigerant, h 1 = 184.5 kj/kg, S 1 = 0.7001 kj/kg K, h ' 2 = 201.8 kj/kg, S 2 = 0.6836 kj/kg K, ' h 3 = 70.55 kj/kg, h 4 29.62 kj/kg. Consider the process 34; Throttling process. Enthalpy before throttling = Enthalpy after throttling h 3 = h 4 = 70.55 kj/kg Enthalpy at compressor inlet = h 1 = 184.5 kj/kg Consider the process 12; PV = C Entropy before compressor = Entropy after compressor S 1 = S 2 ' T 2 S 1 = S 2 + C p log e ' T 2 T2 0.7001 = 0.6836 + 0.64 log e 36 T 2 = 317 K Enthalpy at compressor exit, h 2 = h2cpt2 T2 = 201.8 + 0.64 + (317 309) = 206.92 kj/kg h1h4 (COP) th = h2h1 184.570.55 = 206.92184.5 = 5.083 Actual COP = 0.65 (COP) th = 0.65 5.083 = 3.304 Compressor power = m (h 2 h 1 ) = 5 (206.92 184.5) 60 = 1.868 kw 4

Prelim Question Paper Solution Actual COP = NetRE Comporessorpower NetRE 3.304 = 1.868 Net RE = 6.17 kw Net cooling produced per hour = 6.17 60 60 = 22.212 kj Q.2(b) Refrigerating machine working between the temperature limits of 13C and 37C and has 90% relative COP. It consumes 4.8 kw power. Determine TR capacity. For same TR capacity, how much power will be consumed by carnot refrigerator? Also for the same power consumption, determine TR capacity of carnot refrigerator operating between same temperature limits. (A) Temperature = T 1 = 13 + 273 = 260k T 2 = 37 + 273 = 310k Relative COP = 0.90 Power consume = P = 4.8 kw T COP = 1 260 5.2 T2 T1 310 260 ActualCOP Relative COP = CarnotCOP Actual COP = Relative COP Carnot COP = 0.90 5.2 = 4.68 Refrigeration effect (R COP = E) Work of compressor (W C) 4.68 = RE 4.8 Actual RE = 4.68 4.8 = 22.464 kj / Sec = 1347.84 kj / min Capacity in TR = 1347.84 6.387 tonnes 211 RE Carnot cop = WC For same capacity i.e. R E = 6.387 ton or = 1347.84 R E = Carnot COP W C 1347.84 = 5.2 W C W C = 259.2 kj / min = 4.32 kw (kj / sec.) For same capacity power consume will be 4.32 kw For same power (P = 4.8 kw given) TR capacity of carnot refrigerator RE COP = WC 5.2 = RE 4.8 Carnot RE = 5.2 4.8 = 24.96 kj / Sec = 1497.6 kj / min T.R. capacity = 1497.6 = 7.097 tonnes 211 [8] 5

: T.Y. Diploma RAC Q.2(c) With Neat sketch. Explain Li-Br refrigeration system in brief? [8] (A) Fig. : Lithium-bromide system without LiBr refrigeration system In this system water acts as a refrigerant and lithium bromide (Li Br) acts as a absorbent. The LiBr solution has a strong affinity for water vapour because of its very low vapour pressure. Lithium chromate is often used as a corrosion inhabitor because LiBr solution is corrosive. This system is very popular for airconditioning in which low refrigeration temperatures are required. In this system the absorber and the evaporator are placed in one shell which operates at the same low pressure of the system. The generator and condenser are placed in another shell which operates at the high pressure of the system. The pressure inside the system is below atmosphere. Since the pressure inside the evaporator is maintained very low, the refrigerant water evaporates. The water vapours thus farmed will be absorbed by the strong lithium bromide solution which is sprayed in the absorber. In absorbing the water vapour the lithium-bromide solution helps in maintaining very low pressure needed in the evaporator and the solution becomes weak. This weak solution is pumped by a pump to the generator where it is heated up by using steam or hot water in the heating coils. A portion of water is evaporated by the heat and solution becomes more strong this strong solution is passed through heat exchanger and then sprayed in the absorber. The weak solution of LiBr from the absorber to the generator is also passed through the heat exchanger. This weak solution gets heat from the strong solution in the heat exchanger, thus reducing the quantity of steam required to heat the weak solution in the generator. Q.3 Attempt any FOUR of the following : [16] Q.3 (a) Explain with a neat sketch thermostatic expansion valve? [4] (A) The operation of this valve is based on the principle of constant degree of superheat for the vaporator exist i.e. by controlling the flow of liquid refrigerant through the evaporator. The thermostatic expansion valve consists of a needle valve and a seat, a metallic diaphragm, spring and adjusting screw. 6

Prelim Question Paper Solution In addition to this it has a feeder or thermal bulb which is mounted on the suction line of compressor near the outlet of the evaporator coil. The filler bulb is partly filled with the same liquid refrigerant as used in refrigeration system. The opening or closing of valve is depended upon the force on the diagram. Thermostatic expansion valve Operation : The remote bulb is charged with fluid which is open on one side of the diaphragm through capillary tube is firmly to evaporate outlet. The pressure (P b ) of the fluid in the bulb tends to open the valve. This pressure is balanced by pressure due to spring (P S ) and in the evaporator (P e ). If the evaporator temperature is high or the load on the evaporator increase, more fluid from feeler bulb will be vapourised and bulb pressure will rises which exert this force on diaphragm. This will widen the valve opening and the refrigerant flow will increase to meet load demand and if load on evaporator decreases reverse action takes place. Q.3 (b) Explain with neat sketch the working of ice-plant? [4] (A) The ice plant of can system uses a simple vapour compression cycle in its primary circuit using ammonia (NH 3 ) as refrigerant and brine in secondary circuit. The ice can in ice plant contains water which rejects heat to brine which is been circulated in secondary circuit. This brine is pumped through the evaporator and again return to the brine tank. In evaporator the heat of brine is transferred to the refrigerant NH 3 in primary circuit. This cooled brine is again returned to insulated brine tank. Ice Plant (can type) 7

: T.Y. Diploma RAC The vapour refrigerant (ammonia) is sucked by compressor from evaporator and then compressed to a high pressure which is then condensed in a condenser with the help of water by cooling water circuit. The high pressure liquid ammonia is collected in the receiver and then it is passed through expansion valve. The throttle liquid ammonia form at lower pressure and lower temperature enter in an evaporator which is coil dipped in brine. So brine is cooled and ammonia absorbs heat and forms vapour and cycle is repeated. Q.3 (c) State the effects of suction pressure and discharge pressure on performance of vapour compression system? (A) Following are the effect of reduced suction pressure (i) Increases the mass flow rate of the refrigerant. (ii) C.O.P. of the system reduce. (iii) Reduce in net refrigerating effect by = [(h 2 h 1 ) (h 2 h 1 ) = (h 2 h 2 )] kj/kg (iv) Compressor work increase by [(h 3 h 2 ) (h 3 h 2 )] (v) Compressor capacity increase due to increased specific work. Effect of suction pressure Effect of change in discharge pressure (i) The mass of refrigerant to be circulated per ton of refrigeration increases. (ii) C.O.P. of the system decrease. (iii) Heat rejected in the condenser per ton of refrigeration increase due to increase mass flow of refrigeration therefore size of condenser increases. (iv) As mass flow rate increases, compressor capacity increases per ton of refrigeration through specific volume of refrigerant at entry to compressor remains the same. (v) Refrigeration effect reduces by an amount equal to [(h 2 h 1 ) (h 2 h 1 )] by (h 1 h 1 ) KJ/kg. [4] Q.3 (d) Explain the concept of green house effect and global warning? [4] (A) All the living organism except trees releases CO 2 which at is absorbed by trees and releases O 2. The balance of O 2 and CO 2 in the atmosphere is responsible for affecting atmospheric conditions. 8

Prelim Question Paper Solution Due to rapid industrialization large forest of were cut which lead to increase in CO 2 in atmosphere. The excess of CO 2 destroys the stratospheric ozone (O 3 ) layer which protects the earth from excess heat coming from the sun. As the ozone layer thickness is reduced it creates a void which allows solar rays to directly fall on earth which increase the earth temperature and hence the earth atmosphere. This effect of reducing O 3 layer is called green house effect. Global warming The part of the solar energy absorbed by the atmosphere and earths surface is re-emitted at infrared radiation by them. These radiation are absorbed by greenhouse gases like CFC refrigerants, CO 2, O 3, NO 3 and CH 4 present in the lower atmosphere. This increases the temperature on the surface of earth which is called as global warming due to green house effect. Q.3 (e) State the desirable properties of ideal refrigerants. [4] (A) Low boiling and freezing point. ii) High critical pressure and temperature. iii) High latent heat of vapourisation. iv) Low specific heat of liquid and high specific heat of vapour. v) Low specific volume of vapour. vi) High thermal conductivity. vii) Non-corrosive to metal. viii) Non-flammable and non-explosive. ix) Non-toxic. x) Should be low in cost. xi) Easily and regularly available. xii) Easy to liquefy at moderate pressure and temperature. xiii) Easy of locating leaks by odour or suitable indicator. xiv) Ozone friendly. xv) High co-efficient of performance. Q.4 (a) Attempt any THREE of the following : [12] Q.4 (a)(i) State the classification of condenser used in refrigeration system? [4] (A) Q.4 (a)(ii) Explain working principle of evaporative condenser with neat sketch? [4] (A) The evaporative condenser perform both the combined function of a water-cooled condenser and a cooling tower. 9

: T.Y. Diploma RAC In its operation the water is pumped from sump to spray header and sprayed through nozzles over the condenser coil through which hot refrigerant from compressor is passing. Heat is transferred from refrigerant in the condenser into the water that is outside the surface of tuber. A fan is also used which draws air from the bottom side of condenser and discharges out at the top of condenser. The air causes the water from the surface of the condenser coils to evaporate and absorb the latent heat of evaporation from the remaining water to cool it. Since heat for vaporizing the water is taken from the refrigerant, therefore the vapour refrigerant condenses into liquid refrigerant. The cold water that drops down into a sump is recirculated. A float valve keeps a check of water level. The eliminator is provided above the spray header to stop particles of water escaping along with the discharge air. Note : For exam student must draw any one figure. Q.4 (a)(iii)explain the concept of sensible heat factor and bypass factor with suitable [4] sketches? (A) By-pass factor : When air passes over a coil, some of it say x just by-passes unaffected while the remaining (1 x) kg comes in direct contact with the coil. This by-pass process of air is measured in terms of by-pass factor. Balancing the enthalpies, we get xc pm td 1 + (1 x) C pm td 3 = 1 C pm td 2 (C pm = specific humid heat) x (td 3 td 1 ) = td 3 td 2 td3 td 2 x =, where x is by pass air td td 3 1 Evaporative Condensers 10

Prelim Question Paper Solution Sensible Heat Factor : The ratio of sensible heat to total heat added is known as sensible heat factor (SHF) process of sensible heating on the psychromatic chart is shown by a horizontal lin 1 2 extending from left to right. Sensibleheat Sensible heat factor = SensibleheatLatentheat Q.4 (a)(iv) List different types of dehumidifiers. Describe most commonly used type with sketch? (A) Different types of dehumidifiers (i) Spray type dehumidifier ii) Cooling coil type dehumidifier (i) Spray type dehumidifier : This type of dehumidifier is similar with humidifier with an exception that the water used for spraying purpose is having temperature below the dew point temperature of air. The outside fresh air enters through filter which removes the dust particle. This air is then passed through cold water having temperature below dew point temperature of the room air. By this the outside air is cooled below its dew point temperature (DPT) and thus excess moisture is removed. However, the dry bulb temperature of air leaving the eliminator is much below the DBT of desired condition space. Hence an heater is used to maintain the temperature through which air is passed to condition space. A refrigeration system is used to cool the water supply the and an overflow pipe is required to drain the added water. [4] Spray type Dehumidifier Q.4 (b) Attempt any ONE of the following : [6] Q.4 (b)(i) Explain the thermal exchange mechanism of human body with environment? [6] (A) The human body works best at a certain temperature, like any other machine, but it cannot tolerate wide range of variation in their environmental temperature like machines. 11

: T.Y. Diploma RAC Human body tries to maintain its thermal equilibrium with the environment by means of three modes of heat transfer i.e. Evaporation, radiation and convection. A human body feels comfortable when the heat produced by metabolism is equal to sum of heat dissipated to surrounding and stored in human body by increasing the temperature of body tissues. Q M W = Q E ± Q R ± Q C ± Q S Q M = Metabolic heat produced within the body W = Useful rate of working Q R, Q C is heat test or gained by radiation and convection respectively. + when heat is lost to surrounding. when heat is gained from surrounding. In summer the energy stored Q S increases and temperature of body increases. This leads to increase in blood flow rate through the extremities and body starts perspiring and thus reduces body temperature. This condition is called vasodilation. In winter stored energy may become negative because of decrease in temperature. Thus reducing the blood flow rate through extremities which results in shivering. Such a condition is called as vasoconstriction. For comfort feeling thermal neutrality is required and in order to achieve that there should be no stored energy and hence no change in temperature. Any variation in the body temperature acts as a stress to the brain which ultimately results in either perspiration or shivering. Q.4 (b)(ii) Explain factors affecting human comfort? [6] (A) Following are the factors affecting human comfort : i) Effective temperature ii) Moisture content of air iii) Heat production and regulation in human body iv) Heat and Moisture losses from the human body v) Quality and Quantity of air vi) Air motion vii) Air stratification viii)hot and Cold surfaces Q.5 Attempt any TWO of the following : [16] Q.5 (a) Draw a labeled sketch and explain working of window air conditioning system? [8] (A) Working : First the low pressure, low temperature refrigerant vapour is sucked by hermitically sealed compressor and compressed to high pressure, high temperature and it is then discharged to condenser to reject the latent heat. The liquid refrigerant passes through the filter into the capillary tube where it is throttled and the flows to the evaporator coil at lower pressure. This liquid refrigerant than rapidly boils at low pressure and picks up evaporation enthalpy from the evaporator surface. 12

Prelim Question Paper Solution A fan or blower is used to drive the air from room through air filter from the lower part of the unit and forces it to flow over the evaporator coil. The temperature of the cooling coil absorbs the heat from the air and is circulated back into the conditioned room. Due to this the temperature of room air is reduced hence air becomes chilled and circulated back into the conditioned room. But due to reduction in the temperature of the air dew is formed on the surface of the cooling coil. For this purpose the temperature of the cooling coil is lower than then the dew point temperature of the air. This moisture present in circulating air is removed and flows from coil surface and drips in the tray at the bottom. This moisture in the tray (pan) evaporates to some extent which helps in cooling the compressor and condensor. This type of air conditioning is used for office, bed room, drawing office etc. Q.5 (b) Explain with neat sketch the various losses in the duct? [8] (A) (i) Surface Frictional Loss: The frictional resistance of a duct of any cross-section is given by Darey s equation, flv h f = 2 2gD D = Diameter of circular duct. V = Velocity of the fluid flowing in m/sec. f = Friction factor. L = Length of the duct in meters. The friction factor f is depend on Reynold number, 64 f = R when R e < 2300 e (ii) Dynamic Losses in Ducts: Whenever there is change in direction or velocity in the flow through duct, the pressure loss is inevitable. The additional loss called dynamic loss. The change in magnitude of velocity occurs when the area of duct changes. The change in magnitude or direction which cause accelerating and decelerating force which may be internal or external. The pressure loss due to the change of direction of velocity at elbow is known as velocity pressure head. 13

: T.Y. Diploma RAC (iii)loss due to Enlargement: When the area of duct changes, the velocity of air flowing through the duct changes, when area increase, the velocity decrease with rise in pressure which form eddies at the corner thus sudden or abrupt change is neglected. iv) Loss due to Sudden Contraction: When the air is flowing and having a sudden or abrupt contraction, the eddies are formed at the shoulders of the large section and beyond the entry at the smaller section forming a venacontracta. (a) Sudden contraction (b) Gradual contraction The loss of pressure due to sudden contraction is not due to contraction itself but it is due to sudden enlargement of flow area from vena-contracta to the section of smaller duct. v) Pressure Losses in Elbow and Bend: L e = Equivalent length of duct. K d = Dynamic loss coefficient. The value of (L e / K d ) is different for different elbow. The value of (L e /K d ) is mostly affected by the geometry of elbow and surface roughness of duct wall and remains unaffected by the air velocity. To minimize the pressure loss in the bend, the splitters are generally used, aspect ratio is small. 14

Prelim Question Paper Solution vi) Losses at Suction and Discharged Opening: (a) Abrupt entrance (b) Formed entrance When an abrupt suction opening is provided the air is accelerated at the opening, forming a vena-contracta inside the duct. In this case sudden changes from infinity to duct area and dynamic loss coefficient. Q.5 (c) A cold storage room has walls made of 0.23 m of brick on the outside, 0.08 m of plastic foam and finally 15 mm of wood on the inside. The outside and inside temperature are 22C and 2C respectively. If the inside and outside heat transfer coefficient are 29 and 12 W/m 2 K respectively the thermal conductivities of bricks, foam and wood are 0.98, 0.02 and 0.17 W/m K respectively. Determine rate of heat removal by refrigeration per unit area of wall. (A) Inside heat transfer coefficient = h i = 29 W/m 2 K Outside heat transfer coefficient = h o = 12 W/m 2 K U = Overall heat transfer coefficient 1 U = 1 X1 X2 X3 1 h0 K1 K2 K3 hi 1 15 0.08 0.23 1 = 12 10000.17 0.02 0.98 29 1 U = 4.4407 U = 0.225 W/m 2 K The rate of heat removed by refrigeration per unit area of wall, Q = UA (To Ti) Q A. = U(T o T i ) = 0.225 (295 271) = 5.40 kj/m 2 Q.6 Attempt any FOUR of the following : [16] Q.6(a) State application of refrigeration from domestic, commercial and industrial area? [4] (A) Domestic application : The household refrigerator used for storing food, vegetables, milk etc. is a common example of refrigeration from domestic area. [8] 15

: T.Y. Diploma RAC Commercial application : 1) They are used to store frozen food, fish, cold drinks, Ice creams, milk for longer duration. 2) Commercial application involves the deep freezers. 3) Air conditioning for restaurants and hotels 4) Air conditioning used in hospitals. Industrial application : 1) Pharmaceutical industries uses refrigeration system to produce granular product which requires dry and low temperature of air for its preservation. 2) Rubber industries. 3) Oil refinery : Use of refrigeration system is done in various stages such as during the removal wax from oil. 4) Textile industries are also using refrigeration methods to maintain dry condition. Q.6 (b) What are the factors which need to be considered during design of car air [4] conditioning? (A) The automobile compartment is designed by considering several factors to achieve the required conditions for all weather such as : (i) Higher outside temperature (ii) Dissipation of heat by engine. (iii) Colour of vehicle is also considered (iv) Infiltration of air and ventilation (v) Solar radiations through glass, window and roof depending upon the position of sun and intensity of solar radiation, cleanness of sky. (vi) Vehicle speed and wind direction and velocity of air. Q.6 (c) Draw and Explain a neat diagram for closed perimeter system for duct management? (A) This type of conditioner is usually placed in the basement and it is located near the geometric centre of all the outlet. The ducts run through the basement foundation slab, the flour and connect the air conditioner to supply to outlet grill. The return grill are generally located on the bottom side of the inside wall. This arrangement is commonly used for residential system. Q.6 (d) What is sensible heat gain and latent heat gain 2 list the sources of sensible and [4] latent heat gain in a restaurant? (A) Sensible heat gain : When there is direct addition of heat to the enclosed space, a gain in the sensible heat is said to occur. [4] 16

Prelim Question Paper Solution Latent heat gain : When there is addition of water vapour to the air in enclosed space, a gain of heat is latent heat. The sensible heat gain may occur due to following sources of heat transfer : (a) Heat flowing into the building by conduction through exterior walls, floors, ceilings, doors and windows due to temperature difference on their two sides. (b) Heat load due to solar radiation are divided into two forms. (i) Heat transmitted directly by radiation through glass of windows and ventilators. (ii) Heat from sun will be absorbed by the walls and roof and transferred to room by conduction. (c) Heat received from the occupants. (d) Heat received from different equipments which are commonly used in the air-conditioned space. (e) Miscellaneous heat sources which include the following : (i) Heat gain through the walls of ducts carrying conditioned air through unconditioned space in the building, leakage of ducts. (ii) The heat gain from infiltrated air. When there is addition of water vapour to the air in enclosed space, a gain of heat is latent heat. The latent heat gain occur due to following factors. The latent heat gain occur due to following factors : (a) The heat gain due to moisture in the outside air entering by infiltration. (b) Heat gain from occupant (person). (c) Heat gain from cooking, due to condensation of moisture from any process. (d) Heat gain from products or material brought into the space. Q.6 (e) Air at 25C WBT 25% RH is to be conditioned to 22C. DBT and 11 gm / kg d.a. specific humidity. Determine heat transfer per kg of dry air referring the psychrometric chart. Represent the process on chart by sketch. (A) Given : t w = 25C WBT, 25% RH, t d = 22C DBT, W = 11 gm / kg d.a h 2 h 1 t w 22 25 DBT 25% RH From Chart W 1 = 0.0128 kg of d.a h 1 = 77.5 KJ / Kg h 2 = 49 KJ / Kg W 2 = 0.011 Kg / Kg of d.a. Heat transfer per kg of d.a. = h 1 h 2 = 77.5 49 = 28.5 kj / Kg W W W = 0.11 kg/kg of d. [4] 17