HEAT TRANSFER ME 323
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This 42 page Class Notes was uploaded by Darian Metz on Wednesday September 2, 2015. The Class Notes belongs to ME 323 at Portland State University taught by Graig Spolek in Fall. Since its upload, it has received 34 views. For similar materials see /class/168305/me-323-portland-state-university in Mechanical Engineering at Portland State University.
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Date Created: 09/02/15
Groundwater Heat Pump Plate amp Frame Heat Exchanger in Engineering Building Graham Corporation Model 6P667 WaterWater Surface area 740 ft2 For more information see httpwwwgrahammfgcompheoverviewhtml Measurements performed yielded the owrates and temperatures given in the diagram below Tchiller out F Tweuin F Tchiller in F V 260 GPM Twell out F Perform the following calculations for this heat pump A 03 0 U Heat balance on well water solve for Q Heat balance on chilled water solve for Q Overall heat transfer coefficient solve for U using the average Q determined for parts A and B above Compare this result to the typical values for waterwater heat exchangers as listed in Table 112 of Incropera and DeWitt Assume that the plate heat exchanger is approximated by counter ow Effectiveness solve for the effectiveness of the heat exchanger YEEASE 50 SHEETS EVEEASE 5 SQUARES 235 4232 I00 SHEETS EYEEASE 5 SQUARES 42469 200 SHEETS E 39139 Marina Dram ME SZCoaql 24X 90349 V 51Kva 76 quotquot7 g by 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Q F e vC SKT gces 0 14 u 9 AT 0 HA K 1 6 gt fig heron mm A Use L r l M4 J g 2 1 Aaho 753 14 h 373 4 quotJ L34 momlt3 zmA ozA WW3 A W C 5 45 M 194 2 QQJCW 30m 398ch C Lamak MI H I A Z I T Tuc GA L I 3 comp acanxc 10 quotva 622 i z Lgt 639 6 M56 26 005quot 170455 T 5 2 L 904 T1 393 706 M an fm a I66 1912 EVEEASE 5 SQUARES 5 SQUARES 382 3900 SHEETS EVE EASE39 5 SQUARES EYEEAS A38 POO SHEEN 6238 50 SHEETS Ilatianal rand 23 A 3APV 04px 7 TR w I r 1 f 61 V BTW Aghw w WI T361 H S YER h mw m ugb fGMM 41n nvn m4 Q N w samp M Ne wt w n R v a weak in D1 mw b mm Jr 3 amp 84 N A H N N mu Q W xw gxow Solutions to Practical Problems 1 N E 4 V39 0 gt1 Greenhouse 7 The greenhouse works due to selective transmissivity of glass Short wave radiation from sun is transmitted through the glass long wave radiation from plants is blocked Internal temperature rises until heat gain by radiation balances shell heat loss Insulation 7 Insulation inhibits air movement which eliminates convection Hence heat transfer occurs by conduction through air glass ber blend and by radiation between bers For radiation color matters But pink and yellow have about the same emissivity so little difference could be detected Parallel plates on heaters 7 Plates act as f1ns which increase the area for convection from the heater to the surrounding air Q h A AT For a Q set by heater power as AT the ATL for the same h value Reducing AT lowers the operating temperature of the heating element increases useful life and reduces surface temperature of all parts safety Thermos bottle 7 The thermos bottle ask is a sealed double wall container Air is evacuated between the two walls so only radiation can transmit thermal energy which greatly inhibits heat loss Coating the surfaces with a shiny layer silver reduces the emissivity to reduce radiation losses further Chill factor 7 The human body internally generates thermal energy which is lost continuously by convection at the surface as well as respiration and perspiration The heat loss in still air can be calculated by Q h A AT where AT Tsurface Too When the wind blows the convective heat transfer coefficient hwind is much larger so the heat loss is the same as it would be for still air with the original h but a much lower temperature Twind Chm Q h A Tsurface 39 Twind chill hwind A Tsurface 39 Too Thermal touch 7 When objects initially at a temperature lower than that of the skin surface is touched heat will be lost to that object The initial rate of heat lost affects the sensation of warm or cool That rate is predicted by factor 1lkpCp for the material being touched the larger this factor the cooler an object will feel See section 57 of Incropera and DeWitt Car roof frost 7 A car s roof will reach an equilibrium temperature which balances the convection from surrounding air and radiation loss to the clear sky Since deep space is approximately 40 F the roof s surface can cool to below the freezing point During day indirect radiation changes the balance The side of the car does not freeze because of reduced shape factor with the cold sky 9 0 Window pane frost 7 As air is cooled its density increases and it drops Air cooled by contact with a cold window surfaces cools to the ice point near the bottom of the window pane on its drop from the top down Frost forms after adequate cooling Warm water ice cubes 7 In an all things equal experiment warm water will not freeze faster than cold water because the amount of heat to be removed is greater However open trays of warm water will often freeze faster than cold ones because warm water evaporates faster especially into a frostfree freezer compartment The more water that evaporates the less remains to freeze Since latent heat rather than sensible hear dominates this entire coolingfreezing process freezing less water will appear to produce ice cubes faster Cooling effect of open refrigerator 7 As you stand before an open refrigerator your skin will begin to lose heat faster due to radiation exchange with the cold interior refrigerator surfaces Greater heat loss is sensed as feeling cold Microwave cooking 7 Yes a microwave oven does indeed cook from the inside out Microwaves cause water molecules in the food to vibrate which is equivalent to higher thermal energy Microwaves are absorbed moreorless uniformly by food representing uniform internal generation of heat Thus the core of the food is warmer than the surface Potato baking This can be solved using transient heat transfer assuming the same dimensionless temperature for both potatoes Only the Bi391 changes due to the size change so the F0 number for the larger potato can be found Typical numbers reveals about 32 longer cooking for a potato twice the weight Hot tub operation 7 The energy required to heat the hot tub is equal to the energy loss from the hot tub integrated over time Hot tub heat loss is dictated by AT between the water and the surroundings A reduction in AT reduces energy costs Allowing the hot tub to cool down between uses reduces AT and therefore energy use When to add cream to coffee 7 The cream will cool down the coffee when added and by approximately the same amount whether added when the coffee is hotter or cooler So by adding the cream when the coffee is first served the coffee cream mixture has a reduced temperature for longer prior to drinking the heat loss is smaller and the drinking temperature is higher So add the cream initially to keep the coffee warmer for drinking Beverage cooling 7 The cooling rate is driven by AT but also by the convective coefficient h Q h A AT The h for water is likely to be 10100 times as great as that for air while the temperature difference AT for the two cases is not that large So leave the beverage in the air at least until the beverage reaches water temperature MESZS0409 I Dlwowsfam apg7 7911 WJ JZ My m uopJam caWM Wm who my ample 46mm 4 3an on vb Memggtow I 6116 MNk 494 JM uC39Qrofer ero wu dpes food2954 Epuu avq CGWKJI39JIIUW 4 Solve Av 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07quotquot J Mk Twig AX E CQCWJ K 14Aij ksz 61M n 41 rm 39 Tquot a aT airN quot b quT H al 77 39f 023T 3 02 4233 50 SHEFI S EYEEASEquot 5 SQUARES 3912 382 00 SHEETS EYEEASE sms A ZJOG 200 SHEETS EVEgtEASE A 5 SQUARES Mariana Brand J I V MAJ VMC ctorM angE E Design a heating system for a small factory in Portland OR This problem will be a multipart problem which will require making reasonable assumptions making engineering decisions and then checking the reasonableness of the design Part 1 Determine the heating load on the building In the rst part you were asked to estimate the maximum amount of heat required for a small factory building suf cient maintain the inside temperature at 20 C in the winter In order to accomplish that task you were forced to employ a rough estimate of the convective coef cients for building surfaces Part2 Refine the heating load estimate on the building Repeat the calculation for energy losses for the same building with the following changes 1 Heating Employ convection theory to refine your estimates for the heat transfer coef cient h I Outside conditions 7 assume that the prevailing winds in Portland are 15 mph I Inside conditions fuse h is 10 W m2 K for all surfaces 2 Lighting 7 Estimate the effect that lighting for the building will have on the heating load if electric lighting requires 20 incandescent bulbs of 150 watts each Discuss the effect of this lighting during summer months Deliverables 1 Total energy input required 2 Total energy input required when lighting effects are included 3 Table of unit R values for major building components walls windows etc I Note Rvalues are in hr oF Btu 4266 50 SHEETS EVEEASE 5 SQUARES 382 00 EHSEYEEASP5WARES latinnaIQBrand 22 I 5 200 SHEETS EYEEASE 5 SQUARES MEBZE HOQ EK LWV J Convoup om Flt1 M Q cL aau c S EOMV JIZRO Layer EF leo l gum u r j v f N EGMMamp0V7 39BL La7ampr reessv vm u vxaPPeJrj Baawdavq La rows 67 as K4 A 5 AtA CA6 BOWMLM a er gm l AT 0 as Jqu deVJoF J fW 4 0 le xc al X0 391ij xct m 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