Hw1 Solutions - bookHW2aSolution - altPROBLEM 1.4KNOWN: 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  12TT 7Cq k LW 1.4 W / m K 11m 8 m 4312 Wt0.20mq u <The daily cost of natural gas that must be combusted to compensate for the heat loss is  gd6fqC4312 W $0.02 / MJC t 24 h / d 3600s / h $8.28 / d0.9 10 J / MJKu ' u u<COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation between it and the concrete.PROBLEM 1.24KNOWN: Dimensions of a cartridge heater. Heater power. Convection coefficients in air and water at a prescribed temperature. FIND: Heater surface temperatures in water and air. SCHEMATIC:ASSUMPTIONS: (1) Steady-state conditions, (2) All of the electrical power is transferred to the fluid by convection, (3) Negligible heat transfer from ends. ANALYSIS: With P = qconv, Newton’s law of cooling yields  P=hA T T h DL T TPTT .hDLsssSS  fffIn water, TC+2000 W5000 W/mK0.02 m 0.200 ms2 uu u20$S T C + 31.8 C = 51.8 C.s 20$$ $ <In air, TC+2000 W50 W/mK0.02 m 0.200 ms2 uu u20$S T C + 3183 C = 3203 C.s 20$$ $ <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 1.34KNOWN: 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 - E = 0 q - q = 0inoutelec radand using Eq. 1.7 for radiative exchange between the baseplate and shroud, p44ppelec shq = A T - T .HVSubstituting numerical values, with 2ppA = D/4,Sfind2824444elecq = 0.25 0.3 m / 4 5.67 10 W/m K 300 - 77 K 8.1 W.u S<(b) From an energy balance on the enclosure, radiative transfer heats the liquid nitrogen stream causing evaporation, E - E = 0 q - m h = 0inoutradLN2fgwheremLN2is the liquid nitrogen consumption rate. Hence,/mLN2 = qradhfg = 8.1 W / 125 kJ / kg = 6.48 10-5 kg / s = 0.23 kg / h.u<(c) If aluminum foil (Hp= 0.09) were bonded to the upper surface of the baseplate,prad,foil radq = q / = 8.1 W 0.09/0.25 = 2.9 WfHHand 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 iomu pv mu pv q 0  where u + pv = i and q = Pelec. Hence, with io pi omi i mc T T , po i elecmc T T P elecpo iP500 Wm 0.0199 kg/scT T1007 J/kg K 25 C $33m 0.0199 kg/s0.0181 m / s1.10 kg/m U <3o22c4 4 0.0181 m / sV 4.7 m/sAD0.07 mu SS <(b) Heat transfer from the casing is by convection and radiation, and from Equation (1.10) 44ss s s surqhAT T A T Tf  HVwhere2sA DL 0.07 m 0.15 m 0.033 m . u SS Hence,   22 2824444q 4W/m K 0.033 m 20 C 0.8 0.033 m 5.67 10 W/m K 313 293 K  u uu  $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 prescribed temperature rise, the change in U, and hence the effect on , is small.Extra Problem -