CSU MECH 344 - HW1 - Solutions (5 pages)

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HW1 - Solutions



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HW1 - Solutions

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Pages:
5
School:
Colorado State University- Fort Collins
Course:
Mech 344 - Heat and Mass Transfer
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PROBLEM 1 4 KNOWN Dimensions thermal conductivity and surface temperatures of a concrete slab Efficiency of gas furnace and cost of natural gas FIND Daily cost of heat loss SCHEMATIC ASSUMPTIONS 1 Steady state 2 One dimensional conduction 3 Constant properties ANALYSIS The rate of heat loss by conduction through the slab is q k LW T1 T2 t 1 4 W m K 11m u 8 m 7qC 0 20 m 4312 W The daily cost of natural gas that must be combusted to compensate for the heat loss is Cd q Cg Kf t 4312 W u 0 02 MJ 0 9 u 106 J MJ 24 h d u 3600s h 8 28 d COMMENTS The loss could be reduced by installing a floor covering with a layer of insulation between it and the concrete PROBLEM 1 24 KNOWN Dimensions of a cartridge heater Heater power Convection coefficients in air and water at a prescribed temperature FIND Heater surface temperatures in water and air SCHEMATIC ASSUMPTIONS 1 Steady state conditions 2 All of the electrical power is transferred to the fluid by convection 3 Negligible heat transfer from ends ANALYSIS With P qconv Newton s law of cooling yields P hA Ts Tf hS DL Ts Tf P Ts Tf hS DL In water 2000 W 5000 W m K u S u 0 02 m u 0 200 m Ts 20 C Ts 20 C 31 8 C 51 8 C Ts 20 C Ts 20 C 3183 C 3203 C 2 In air 2000 W 50 W m K u S u 0 02 m u 0 200 m 2 COMMENTS 1 Air is much less effective than water as a heat transfer fluid Hence the cartridge temperature is much higher in air so high in fact that the cartridge would melt 2 In air the high cartridge temperature would render radiation significant PROBLEM 1 34 KNOWN Vacuum enclosure maintained at 77 K by liquid nitrogen shroud while baseplate is maintained at 300 K by an electrical heater FIND a Electrical power required to maintain baseplate b Liquid nitrogen consumption rate c Effect on consumption rate if aluminum foil Hp 0 09 is bonded to baseplate surface SCHEMATIC ASSUMPTIONS 1 Steady state conditions 2 No heat losses from backside of heater or sides of plate 3 Vacuum enclosure large compared to baseplate 4 Enclosure is evacuated with negligible convection 5 Liquid nitrogen LN2 is heated only by heat transfer to the shroud and 6 Foil is intimately bonded to baseplate PROPERTIES Heat of vaporization of liquid nitrogen given 125 kJ kg ANALYSIS a From an energy balance on the baseplate E in E out 0 q elec q rad 0 and using Eq 1 7 for radiative exchange between the baseplate and shroud 4 q elec H p A pV Tp4 Tsh Substituting numerical values with A p S D 2p 4 find q elec 0 25 S 0 3 m 2 4 5 67 u 10 8 W m 2 K 4 3004 77 4 K 4 8 1 W b From an energy balance on the enclosure radiative transfer heats the liquid nitrogen stream causing evaporation E in E out 0 LN2 h fg 0 q rad m LN2 is the liquid nitrogen consumption rate Hence where m LN2 q rad h fg 8 1 W 125 kJ kg 6 48 u 10 5 kg s 0 23 kg h m c If aluminum foil Hp 0 09 were bonded to the upper surface of the baseplate q rad foil q rad H f H p 8 1 W 0 09 0 25 2 9 W and the liquid nitrogen consumption rate would be reduced by 0 25 0 09 0 25 64 to 0 083 kg h PROBLEM 1 52 KNOWN Power consumption diameter and inlet and discharge temperatures of a hair dryer FIND a Volumetric flow rate and discharge velocity of heated air b Heat loss from case SCHEMATIC ASSUMPTIONS 1 Steady state 2 Constant air properties 3 Negligible potential and kinetic energy changes of air flow 4 Negligible work done by fan 5 Negligible heat transfer from casing of dryer to ambient air Part a 6 Radiation exchange between a small surface and a large enclosure Part b ANALYSIS a For a control surface about the air flow passage through the dryer conservation of energy for an open system reduces to u pv m u pv m i o q 0 ii i o where u pv i and q Pelec Hence with m p To Ti mc Pelec m Pelec c p To Ti m Vo U Ac p Ti To mc 500 W 1007 J kg K 25 C 0 0199 kg s 1 10 kg m 4 S D2 3 0 0199 kg s 0 0181 m3 s 4 u 0 0181 m3 s S 0 07 m 2 4 7 m s b Heat transfer from the casing is by convection and radiation and from Equation 1 10 q where q As 4 hAs Ts Tf H AsV Ts4 Tsur S DL S 0 07 m u 0 15 m 0 033 m 2 Hence 4W m 2 K 0 033 m 2 20 C 0 8 u 0 033 m 2 u 5 67 u 10 8 W m 2 K 4 3134 2934 K 4 q 2 64 W 3 33 W 5 97 W The heat loss is much less than the electrical power and the assumption of negligible heat loss is justified COMMENTS Although the mass flow rate is invariant the volumetric flow rate increases because the air is heated in its passage through the dryer causing a reduction in the density However for the is small prescribed temperature rise the change in U and hence the effect on Extra Problem HW1


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