SOLUTION The chilling room of a meat plant with a capacity of 450 beef carcasses is considered. The cooling load, the airflow rate, and the heat transfer area of the evaporator are to be determined.
Assumptions 1 Water evaporates at a rate 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^{\circ} C are 333.7 kJ/kg and 2501 kJ/kg (Table A-9). The density and specific heat of air at 0^{\circ} C are 1.292 kg / m ^{3} \text { and } 1.006 kJ / kg \cdot{ }^{\circ} C (Table A-15). Also, the specific heat of beef carcass is determined from the relation in Table A-7b to be
C_{p} = 1.68 + 2.51 \times(\text { water content })
= 1.68+2.51 \times 0.58 = 3.14 kJ / kg \cdot{ }^{\circ} C
Analysis (a) A sketch of the chilling room is given in Figure. The amount of beef mass that needs to be cooled per unit time is
m_{\text {beef }}=(\text { Total beef mass cooled }) /(\text { Cooling time })
= (450 \text {carcasses })(285 kg / \text {carcass}) /(10 \times 3600 s )= 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^{\circ} C at a rate of 3.56 kg/s and is determined to be Then the total refrigeration load of the chilling room becomes
\dot{Q}_{\text {total chillroom }}=\dot{Q}_{\text {beef }}+\dot{Q}_{\text {fan }}+\dot{Q}_{\text {lights }}+\dot{Q}_{\text {heat gain }}
= 235 + 26 + 3 + 13 = 277 kW
The amount of carcass cooling due to evaporative cooling of water is
\dot{Q}_{\text {beef, evaporative }} = \left(m h_{f g}\right)_{\text {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^{\circ} C \text { to } 0.7^{\circ} C as a result. Therefore, the mass flow rate of air is
\dot{m}_{\text {air }}=\frac{\dot{Q}_{\text {air }}}{\left(C_{p} \Delta T_{\text {air }}\right)}=\frac{277 kW }{\left(1.006 kJ / kg \cdot{ }^{\circ} C \right)\left[0.7-(-2){ }^{\circ} C \right]} = 102.0 kg / s
Then the volume flow rate of air becomes
\dot{V}_{\text {air }}=\frac{\dot{m}_{\text {air }}}{\rho_{\text {air }}} = \frac{102 kg / s }{1.292 kg / m ^{3}} = 78.9 m ^{3} / 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 frozen as the temperature drops to -2^{\circ} C , and the evaporator must also remove the latent heat of freezing, which is determined from
\dot{Q}_{\text {freezing }}=\left(\dot{m} h_{\text {latent }}\right)_{\text {water }}
= (0.080 kg / s )(334 kJ / kg ) = 27 kW
Therefore, the total rate of heat removal at the evaporator is
\dot{Q}_{\text {evaporator }}=\dot{Q}_{\text {total, chill room }}+\dot{Q}_{\text {freezing }}=277+27 = 304 kW
Then the heat transfer surface area of the evaporator on the air side is determined from \dot{Q}_{\text {evaporator }}=(U A)_{\text {airside }} \Delta T,
A=\frac{\dot{Q}_{\text {evaporator }}}{U \Delta T}=\frac{304,000 W }{\left(20 W / m ^{2} \cdot{ }^{\circ} C \right)\left(5.5^{\circ} C \right)} = 2764 m ^{2}
Obviously, a finned surface must be used to provide such a large surface area on the air side.
