Properties the heat of fusion and the heat of

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Unformatted text preview: of 0.080 kg/s. 2 All the moisture in the air freezes in the evaporator. Properties The heat of fusion and the heat of vaporization of water at 0°C are 333.7 kJ/kg and 2501 kJ/kg (Table A-9). The density and specific heat of air at 0°C are 1.292 kg/m3 and 1.006 kJ/kg · °C (Table A-15). Also, the specific heat of beef carcass is determined from the relation in Table A-7b to be Cp 1.68 2.51 (water content) 1.68 2.51 0.58 3.14 kJ/kg · °C Analysis (a) A sketch of the chilling room is given in Figure 4–46. The amount of beef mass that needs to be cooled per unit time is Lights, 3 kW 13 kW mbeef (Total beef mass cooled)/(Cooling time) (450 carcasses)(285 kg/carcass)/(10 3600 s) Beef carcass 3.56 kg/s The product refrigeration load can be viewed as the energy that needs to be removed from the beef carcass as it is cooled from 36 to 15°C at a rate of 3.56 kg/s and is determined to be · Q beef (mC T)beef (3.56 kg/s)(3.14 kJ/kg · °C)(36 15)°C Evaporation 0.080 kg/s 36°C 285 kg Refrigerated air 235 kW Fans, 26 kW Then the total refrigeration load of the chilling room becomes · Q total, chillroom · Q beef · Q fan · Q lights · Q heat gain 235 26 3 13 277 kW 0.7°C Evaporator –2°C The amount of carcass cooling due to evaporative cooling of water is · Q beef, evaporative (mhfg)water (0.080 kg/s)(2490 kJ/kg) 199 kW which is 199/235 85 percent of the total product cooling load. The remaining 15 percent of the heat is transferred by convection and radiation. (b) Heat is transferred to air at the rate determined above, and the temperature of the air rises from 2°C to 0.7°C as a result. Therefore, the mass flow rate of air is · m air · Q air (Cp Tair) 277 kW (1.006 kJ/kg · °C)[0.7 ( 2)°C] 102.0 kg/s Then the volume flow rate of air becomes · Vair · m air air 102 kg/s 1.292 kg/m3 78.9 m3/s (c) Normally the heat transfer load of the evaporator is the same as the refrigeration load. But in this case the water that enters the evaporator as a liquid is · Qevap FIGURE 4–46 Schematic for Example 4–5. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 250 250 HEAT TRANSFER frozen as the temperature drops to 2°C, and the evaporator must also remove the latent heat of freezing, which is determined from · Q freezing · (m hlatent)water (0.080 kg/s)(334 kJ/kg) 27 kW Therefore, the total rate of heat removal at the evaporator is · Q evaporator · Q total, chill room · Q freezing 277 27 304 kW Then the heat transfer surface area of the evaporator on the air side is deter· mined from Q evaporator (UA)airside T, A · Q evaporator UT 304,000 W (20 W/m2 · °C)(5.5°C) 2764 m2 Obviously, a finned surface must be used to provide such a large surface area on the air side. SUMMARY In this chapter we considered the variation of temperature with time as well as position in one- or multidimensional systems. We first considered the lumped systems in which the temperature varies with time but remains uniform throughout the system at any time. The temperature of a lumped body of arbitrary shape of mass m, volume V, surface area As, density , and specific heat Cp initially at a uniform temperature Ti that is exposed to convection at time t 0 in a medium at temperature T with a heat transfer coefficient h is expressed as T(t) T Ti T e Q mCp[T(t) The amount of heat transfer reaches its upper limit when the body reaches the surrounding temperature T . Therefore, the maximum heat transfer between the body and its surroundings is Qmax mCp (T Bi hAs Cp V h Cp Lc (kJ) Ti ) (kJ) The error involved in lumped system analysis is negligible when bt where b Ti] (1/s) is a positive quantity whose dimension is (time) 1. This relation can be used to determine the temperature T(t) of a body at time t or, alternately, the time t required for the temperature to reach a specified value T(t). Once the temperature T(t) at time t is available, the rate of convection heat transfer between the body and its environment at that time can be determined from Newton’s law of cooling as · (W) Q (t) hAs [T(t) T ] The total amount of heat transfer between the body and the surrounding medium over the time interval t 0 to t is simply the change in the energy content of the body, hLc k 0.1 where Bi is the Biot number and Lc V/As is the characteristic length. When the lumped system analysis is not applicable, the variation of temperature with position as well as time can be determined using the transient temperature charts given in Figs. 4–13, 4–14, 4–15, and 4–23 for a large plane wall, a long cylinder, a sphere, and a semi-infinite medium, respectively. These charts are applicable for one-dimensional heat transfer in those geometries. Therefore, their use is limited to situations in which the body is initially at a uniform temperature, all surfaces are subjected to the same thermal conditions, and the body does not involve any heat generation. These charts can also be used to determine the total heat transfer from the body up to a specified time t. cen58933_ch04.qxd 9/10/2002 9:13 AM Page 251 251 CHAPTER 4 Using a one-term approximation, the solutions of onedimensional transient heat conduction problems are expressed analytically as Plane wall: Cylinder: Sphere: (x, t)wall (r, t)cyl (r, t)sph T(x, t) T Ti T 2 A1e 1 cos ( 1x/L), T(r, t)...
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