编辑: kr9梯 | 2019-07-12 |
Htwe Tin Abstract:-Chiller and city-multi systems are being used for large buildings. Although chiller system occupies large space, city - multi system is a compact air conditioning system. It does not need large space to locate outdoor and indoor units, liquid and gas pipe lines as other systems. City-multi system with R410A was chosen to arrange air conditioning system for Technological University (Kyaukse). R410A has several advantages than R22 and its operating pressure is 1.6 times higher than R22'
S at the same temperature. System compact size design can be obtained by using R410A. Cooling loads for seminar rooms of this university were calculated to compare the system performance and power consumption for R410A and R22. Key-Words:-air conditioning , city-multi, R 410A, cooling load, performance, power consumption I. INTRODUCTION Refrigerant R410A is one of the substitutes currently accepted as a replacement for the commonly used HCFCS (hydrochlorofluorocarbons). Refrigerants such as R-12 and R-22 are family of chemicals that contain chlorine, fluorine, and carbon. The chlorine content in these compound causes the depletion of the ozone layer. R
410 A is a type of HFCS and it does not contain chlorine. It has an Ozone Depletion Potential rating of 0.00 verus 0.05 for R 22. Its boiling point (-51.4°C) is lower than R
22 (-40.8°C) and the quantity of heat received by one kilogram of refrigerant from the space being cooled is larger than R 22. And again its operating pressure is 1.6 times higher than R22. So its specific volume is less than R22, and smaller pipe diameters and small system size per KW are obtained by using R410 A. Although installation and service procedures are similar for R
410 A as the methods for R22, there are several critical differences. Systems using R410 A must be designed for components are not interchangeable and cannot be matched with R22 components. So, equipment and piping must be designed for that increased pressure. Components used on R410 A air conditioners use thicker metals to withstand the higher operating pressures [1]. Manuscript received January 4, 2009. This work was supported in part by the Ministry of Science and Technology, Union of Myanmar. Ms. Htwe Tin is with the Technological University (Kyaukse). Contact Phone: 095-066-50418 Mechanical Engineering Department, e-mail: [email protected] II. DESIGN CONSIDERATION OF COOLING LOADS Technological University (Kyakse) is located in Mandalay division of Myanmar Country. Design condition is first selected before considering the cooling load. Dry bulb temperature 95°F (DBT), relative humidity 65% (RH), and daily range 22.6°F were chosen for outside design condition (ODT) of Technological University (Kyaukse).Dry bulb temperature 78°F and 50% RH were chosen for inside design conditions [2]. A. Heat gain through exterior structure Solar radiation forms the greatest single factor of cooling load in buildings. Radiant energy from the sun is absorbed by the room materials, both the structure and furnishings. Cooling load temperature difference accounts for the heat storage effect [3]. B. Conduction heat gain through exterior structure The conduction heat gain through the exterior roof, walls and glass are each determined by the following equations For walls and glass Q = U x A x CLTDC (1) Q = net room conduction heat gain through roof, wall or glass, BTU/hr U = overall heat transfer coefficient for roof, wall or glass, BTU/hr ft2 ?F A = area of wall, roof or glass, ft2 CLTDC = corrected value of cooling load temperature difference, ?F that accounts for heat storage effect. For wall CLTDC = [(CLTD +LM) K+(78 - tR)+(t0 - 85)] f (2) For glass CLTDC = (CLTD) + (78 - tR) + (t0 - 85) (3) C. Heat gain by solar radiation through glass The net heat gain by solar radiation through glass can be calculated by the following equation. Q = SHGF x A x SC x CLF (4) SHGF = maximum solar heat gain factor, BTU/hr ft2 A = area of glass, ft2 SC = shading coefficient CLF = cooling load factor for glass D. Transmission gain through interior structure The heat flows from interior unconditioned spaces to the conditioned spaces can be calculated by the following equation. Q = U x A x TD (5) TD = temperature different between conditioned and unconditioned space, °F E. Infiltration and outside air The equations for determining the sensible and latent loads for ventilation and outside air are as follow. QS = 1.08 x CFM x TC (6) QL = 0.68 x CFM x (ωHi- ωHo) (7) QS, QL = sensible and latent cooling load from ventilation and infiltration air, BTU/hr CFM = air ventilation and infiltration rate, ft3 /min TC = temperature change between outside and inside air, ?F ωHi,o = specific humidity for outside and inside conditions, gr.wv/lb-da If (CFM)venti is greater than (CFM)infil, (CFM)infil can be neglected. For ventilation, QS = 1.08 [((CFM) venti x (B.F)] ?T (8) QL = 0.68 [(CFM) venti x (B.F)] x ? ωH (9) ?T = temperature difference between outside and inside condition, ?F B.F = By pass factor of cooling coil If the ventilation air is less than the infiltration air (cfm), sensible and latent heat can be calculated as follows. QS = QS (Infil) + QS (venti) = [ 1.08 x (CFM) ........