Example5 1 Psy chometric Chart Examples Nomenclature S Sensible heat gain (Btu/second) Ma Mass flow rate (a), it could be water of air (Lb\s) Cpma Specific heat capacity of (air) (kJ/Lb) T1 Temperature (at 1) (oF) L Latent heat gain (Btu/second) hg Enthalpy (of gas) (Btu/pound of dry air) W1 Absolute humidity of air (at point 1) (Lb of moisture /Lb of dry air) Q Rate of Energy Exchange (Btu/Second) Useful Equations Sensible Heat Gain: S =MaCpma(T1-T2) Latent Heat Gain: L= Mahg(W1-W2) Energy Exchange: Q = Ma(h1-h2) or MaCpma(T1-T2) for sensible heat exchangers (heating coils.) Example 1 A terminal reheat system maintains two zones of a building at 21oC (70oF) throughout the year and 55% relative humidity for the summer design case. Zone A has a sensible heat gain of 90kW (85 Btu/second) in the summer and a sensible heat loss of 40kW (38 Btu/second) in the winter. Zone B has a sensible heat gain of 60kW (57 Btu/second) in the summer and a sensible heat loss of 30kW (28 Btu/second) in the winter. Both zones have the same constant latent heat gain in the summer and winter. The external design states are 27oC (81oF) dry bulb and 21oC (70oF) wet bulb in the summer and -5oC (23oF) saturated in the winter. Recirculated air is mixed with fresh air in the ratio 3:1 by mass before entering the central plant which has a cooler and fan. The cooling coil has an ADP of 10oC (50oF) and a contact factor of 0.85. If the temp gain across the fan is negligible, calculate; I. the latent heat gains II. the cooling coil and reheater loads for the summer design case III. the room supply temperatures in the winter IV. the total heating load in the winter V. the room relative humidity in winter Cpma = 1.02 kJ/kgK (0.24 BTU/lb/degree F ) hg = 2540 kJ/kg (1092 Btu/pound of dry air)Example5 2Solution for Example 1 Figure 1: Air system containing an economizer, cooling coil, fan and a reheater per zone Outside Air and ADP are fixed. Return state is also given for summer design case. Calculate Mixed Point M based on the proportional relationship given in the question (3 parts return air: 1 part fresh air). Plot M ¾ of length O-R from R. Temperature = 54 F Enthalpy = 28.4 Btu/ pound dry air Absolute humidity = .0099 lb/lb The off-coil condition is determined by plotting the Apparatus Dew Point (ADP) and drawing a line from M to ADP. The coil has a stated contact factor of 0.85 which denoting 85% efficiency. Point W, the off coil condition, is .85 length of M-ADP starting at M. The question states to ignore the rise across the fan. Therefore W= S or the main supply state. Temperature = 54 oF Enthalpy = 21.6 Btu/ pound dry air. Absolute humidity = .00795 lb/lb The maximum temperature differential should be used for the case where load is largest, i.e. Zone A at 85Btu/second. In this case the reheater should not be used so (TRa-TSa) = (70-54)= 16oF S =MACpma(TRa-TSa) MA = S/ Cpma(TRa-TSa) MA = 85/ (.24*(70-54)) MaA= 22 lb/secondExample5 3 Now calculate the Latent heat gains from the data available. Take Wr from your chart. L= Mahg(WRa-WSa) LA =22*1092*(.00865-.00795) LA= 16.5 Btu/second N.B. We know that the latent gains to both spaces are the same. Therefore the supply air absolute humidity must be the same for both spaces. Therefore the Mass Flow Rates to both spaces must be the same = 22lb/second The supply temperature to Zone B can now be calculated. S =MbCpma(TRb-TSb) 57= 22(.24)(70- TSb) TSb = 70- (57/(22*.24)) TSb = 59.2 oF ii) Cooling Coil (CC) and Reheater loads Qcc = MA+B(hM-hW), same mass flow rates for both zones Qcc = (22+22)(28.4-21.6) Qcc = 299.2 Btu/second Zone A reheater is not used at design case Zone B reheater QrhB =MACpma(TSb-TS) QrhB = 22(.24) (59.2-54) QrhB = 27.456 Btu/second iii) Solve for homework iv) Solve for homework v) Solve for homework Plot out winter case on Psychometric chart for homework.Example5 4 Figure 2: Psychometric chart solution for example 1 Example5 5Example 2 The summer sensible heat gains in two zones of a building on the design day together with the outside air dry bulb temperature are given at four times of the day in the following table Time (Hours) 10:00 12:00 14:00 16:00 Zone A (Btu/second) 28 47 8 6 Zone B (Btu/second) 2 4 30 52 OA Temp db/F 66 70 73 75 Both zones are to be maintained at a temperature of 21ºC (70ºF)db by a dual-duct plant which handles a mixture of outside air and re-circulated room air at a ratio 1:3. If the cold duct temperature remains at 16ºC (61ºF) throughout the day, calculate the: i. mass of air supplied in each zone; ii. mass of air flowing in the hot duct at 12:00 and 14:00 hours (assumes the hot duct heater battery is turned off); iii. dry bulb temperature of the air in the hot duct at 10:00 and 16:00 hours. Solution for Example 2 Figure 3: Dual Duct System containing fan, a hot deck, a cold deck, two constant volume terminal units and two zones i) Summer case We base the air flow rate on the maximum sensible heat gains to each zone. Sensible Heat Gain: S =MaCpma(T1-T2)Example5 6 MA = S/ Cpma(TRa-TSa) MA = 47/ (.24*(70-61)) MA = 21.56 lb/second (Based on the assumption that only the cold duct supplies air in the summer design case) S =MbCpma(TRb-TSb) MB = S/ Cpma(TRb-TSb) MB = 52/ (.24*(70-61)) MB = 23.71 lb/second Total Mass Flow Rate = MA + MB = 45.27 lb/second At 12:00 TOA = 70oF and TR = 70oF Therefore, mixed temperature TM = (TR (3) + TOA (1))/4 TM = (70(3) + 70(1))/4 = 70oF Heater is off in the summer scenario S =MACpma(TR-TSa) TSa = 70- (28.4/(21.56*.24)) TSa = 64.5 oF TSb = 70- 4/(23.71*.24) TSb = 69.3 oF Energy Balance Hot Deck + Cold Deck = Supply A + Supply B MhTh + McTc = MATSa + MBTSb MhTh + (Mt-Mh)Tc = MATSa + MBTSb Mh70 + (45.27-Mh)61 = 21.56(64.5) + 23.71(69.3) Mh70 + 2761 -Mh61 = 1390 + 1643 Mh = (272.1)/9 Mh = 30.23 lb/s At 14:00 Solve for Homework iii) Assume heater is off.Example5 7 Temperature of air in hot duct = Temperature of Air at M At 10:00 TOA = 66 and TR= 70 TM = (70(3) + 66(1))/4 = 69oF Th = 69oF At 16:00 – Try to solve on your own Plot out one case on a Psychometric