Document zd8p7e90EykZ9Ox7ooj7yb2Yn

REfRIGERATION' Here are seven easy-touse formulas to find capacity of your system, be it direct or tndirect type. Study the problems solved for you so you'll be ready for some fast checks By D K CHAPMAN Jtnkt tnglnoor, Worthington Corp TYPICAL characteristic curve for o centrifugol pump In an otr-eondltioning system. Don't use this chort for figuring Quick Check List lor Alr-CondlHonlni Rtlritenlloo Systems Item Condenitng-votes Inlst temp, F Coadenilng-voter ootlot lamp. F Water-temp riu In condenser, F Freon-12 hood proisare, pslg Freon-12 section pressure, pslg Temp olr oaten cooling coll, F Temp drop through cooling coil, F Temp olr leave* cooling coil, F Air circulated per ten refrigeration, cfm Room temp for comfort cooling, F Condensing voter, gpm per ton Tomp' cbiNod-votor enters coolers, F Temp chilled-voter leaves coolers, F /tango 60-95 15-115 15-25 100.150 25-50 75-90 12-25 50-70 300-600 70-16 35 50-65 40-50 i Easy Figuring Helps You Check Out Refrigeration for Air Conditioning j .Two types of refrigeration ayatema and leaving the system'* chiller, F. Sensible-heat removal results in a 1 may be used with air-conditioning units If you don't have a flowmeter to meas change of air temperature while latent ! --direct and Indirect expansion. Direct ure chilled-water flow, multiply differ heat comes from moisture that con- hea evaporator coila in air stream. Ex ence between auction and discharge denies out of the air. Method 2 is ) pansion valvo in coil regulates refrig pressure-gage readings of circulating equally good for direct and indirect sys- erant flow. Indirect hae evaporator coils pump by 2.31. This gives pump head in terns. A that cool water or other licyuid pumped feet. Refer to characteristic curve for Totol Heat. To figure total heal re: i through the.air-conditioning unit. pump, read' gpm.corresponding to that moved from air passing over o coil use J Spot Cheeks. You can check refriger*' head. Diagrom, right, is a typical char this equation: II TM 4.S X cfm X he, .5 ating-machine performance in several acteristic curve for a centrifugal pump. where H -- total heat removed from ways. With each you must use some aim- To estimate gpm from curve you must air, Blu per hr; cfm -- cubic feet of ? pile arithmetic--but don't let that scare know both the head and horsepower of air passing over the coil per minute; you. Just follow the formulae and steps pump. Eosiest way to find pump hp is A* mm enthalpy difference between- en- S we give below. They're accurate enough to use a clamp-on wattmeter to meas tering and leaving air, Dtu per lb. | for all ordinary jobs. You can figure ure input watts. Divide reading by 746, Obtain air enthalpy from a psycliro* '} some results closer, but the extra work and multiply result by 0.80 and pump metric chart or the table, right. Reod S needed really isn't necessary unless efficiency, read from the characteristic enthalpy at the air's wet-bulb tempera- J| you're designing a plant. curve. (The value O.BO is the assumed lure, not the dry-bulb. Once you've ^ System Copoclty. On trouble shooting motor efficiency, 80%, expressed as a figured total heat removed by refriger- and other jobs you often benefit.from decimal.) am, convert this to Iona refrigeration knowing actual capacity of a system that Remember, you must use the char by dividing by 12,000 Btu per hr per 4 cools chilled water for air conditioning. acteristic curve for the pump in your ton. If wet-bulb temperatures you're ^ Most systems like this have e water system. Diagram is only an example of working with are above or below these .]jl chiller in which refrigerant expands in a typical centrifugal-pump curve. in table; see a standard reference book VI a coil. Water passes through chilter Direct-expansion systems may be fig- such as ASH&VE Guide. 4| shell. Rut you can use some method for .ured in three waya if you wish to check Moasuro air velocity with a velo-'^l a fioodecl-tjrpe chiller, too. over-all capacity of system. These are: meter, if possible. This gives velocity /jl Figure the capacity from C -- 0.042 (1) measuring totol heat removed from readings directly, Convert to cfm by-}| gpm X *, where C -- tons refrigera air poising through colt (2) measuring multiplying average velocity, obtained M tion; gpm -- gallons of chilled water best rejected to condensing water (3) from 10 or 12 readings across.the duct, | circulated per mloute; t* -- tempera separately measuring sensible and latent by duct area in square feet. You can ? ture difference between water entering heat removed from air. also use a pilot tube to obtain elr i 13} PIANT OPERATION AMO MAINTENANCE SECTION powei | velocity. It's often handy when a duct opening isn't accessible. Be sure to fol low instructions provided with these in struments when taking measurements. . Hoot rejected to condensing water Is figured from h -- SCO X 8Pm X *v where A -- best to condenser water, Btu per hr; gpm gallons of water pissing through condenser per minute; t temperature rise of water through condenser, F. Deduct 10% of the fig ured lieat to get actual heat rejected beeause this much heat in condenser water comes from what's called "heat of compression." This is heat from work that compressor must do to raise the pressure of the vapor. Divide result by. 12,000 lo convert to tons refrigera tion. . . Don't forget that if compressor is hermetic type, having motor winding cooled by suction gas, this heat is also rejected to the condenser water. So we mutt deduct the Blu equivalent of motor ' heal, or motor, hp X 2544, from heat load on condenser. Treat fan motors in ih-cool air stream the same way. < This figuring is good for ell shell-andtube condensers. Use a flowmeter or measuring barrel to find water flow in 8pm. On. large joba where It's lough to flow accurately with a barrel, install pressure gsges on inlet and out let of condenser water pump. Multiply difference in readings hy 2.31 to find . *?l*l head In feet. Read gpm from pump characteristic curve, after figuring motor "P sa shown above. Evaporative condensers are figured by method l above, exeept that we're deal- mi with an air-temperature rise instead * drop. Also, we must subtract heat Enttofa of Mold Air Wst-bulA temp, F 68 69 70 71 72 73 74 75 76 n 78 79 80 81 82 83 84 85 intholpf, Btu per lb 31.92 32.71 33.51 34.33 35.17 36.03 36.91 37.81 38.73 39.67 40.64 41.63 42.64 43.67 44.72 45.80 46.91 48.04 of compression (10%) and electrical heat, if any. Total sensible heat Is found from A* -- 1.1 X cfm X 4. where A. a totol sensible heat removed from air, Btu per hr; cfm -- cubic feet of air cooled per minute by refrigerant; t* temperature difference between air entering and leaving coil, F. Figure total latent heat from A> -- B'fSO X Bph. where Ai total latent heat removed from air, Blu per hr; gph -- gallona of moisture condensed from air per Hour. Add total'sensible and latent heat to find amount of heat the coil removea from olr. You have enough methods here to make a good spot check on most job* you run across. But since they are all subject to s little Inaccuracy, depending on conditions, your best bet is to figure the job in several waya and take the average of your results. For quick checks of refrigeration sys tems use the list shown above. Ranges given are typical, and you'll find them valid for most jobs you're called on to .trouble-shoot. .(samples. Here ere several typical problems showing how to use the equa tions above. Example: A compressor chills 100 gpm of water through s temperature range of 20 F. What is compressor, capacity at this load? Solution: Use first equation; C -- 0.024X 100X20.48 tons. Example: Air enters a coil at a wetbulb temperature of 80 F, leaves at 70 F. What Is compressor tonnage when 1000 cfm flow through coil? Solution: Difference in total heal, using values from table, is 42.64 - 33.51 ->9.13 Blu per lb. Put this in equation to find total heat removed as If -- 4.5 X 1000 X 9.13 -- 41,083 Btu per hr. This is 41,085/12,000 -- 9.43 tons of refrig eration. Example: What Is the capacity of a refrigeration unit using 30 gpm of con densing water when temperature rise through condenser Is 20 F? Solution: Using method 2, A ,500 X 30 X 20 mm 300,000 Btu per hr. Sub tracting 10% for heat of compression, quantity of heat rejected to cooling waleris 270,000 Blu per hr and tonnage is 270,000/12,000 -- 22.5 tons. Wit |Wj PLANT OPERATION ANO MAINTENANCE SECTION