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Question 13.9: Sensible Heating Moist air enters a steam-heating coil at 40......

Sensible Heating

Moist air enters a steam-heating coil at 40°F (4.4°C) dry-bulb temperature and 36°F (2.2°C) wet-bulb temperature at a rate of 2000 ft³/min (943.8 L/s). The air leaves the coil at a dry-bulb temperature of 90°F (32.2°C). Determine

(a) The heat transfer rate that occurs at the coil

(b) The amount of steam needed if saturated steam at 220°F (100°C) enters the coil and leaves as a condensate at 220°F
Given: V˙1=2000 cfm,Tdb1=40°F,Twb1=36°F\dot{V}_{1} = 2000  cfm, T_{db1} = 40°F, T_{wb1} = 36°F, and Tdb2=90°FT_{db2} = 90°F.
Figure: See Figure 13.11, process 1–2.
Find: Q˙h\dot{Q}_{h}.

13.11
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Assumptions: Thermodynamic equilibrium exists. The local atmospheric pressure is 1 atm, or 14.696 psia.

Lookup values: Let states 1 and 2 represent the entering and exiting airstream conditions, respectively. From the psychrometric chart, v1=13.66 ft3/lba,h1=13.47 Btu/lbav_{1} = 13.66  ft^{3}/lb_{a}, h_{1} = 13.47  Btu/lb_{a}, and h2=26.5 Btu/lbah_{2} = 26.5  Btu/lb_{a}.
From the steam tables, h3=1153.4 Btu/lbwh_{3} = 1153.4  Btu/lb_{w}, and h4=168.1 Btu/lbwh_{4} = 168.1  Btu/lb_{w}.
The mass flow rate of moist air is

m˙a=V˙1v1=2000 ft3/min12.66 ft3/lba=160 lba/min= 9479 lba/h\dot{m}_{a} = \frac{\dot{V}_{1}}{v_{1}} = \frac{2000  ft^{3}/min}{12.66  ft^{3}/lb_{a}} = 160  lb_{a}/min =  9479  lb_{a}/h

Finally, from Equation 13.36,

Q˙sen=m˙a(h2 – h1)\dot{Q}_{sen} = \dot{m}_{a} (h_{2}  –  h_{1} )             (13.36)

Q˙h=9,479 lba/h×(26.5 – 13.47)Btu/lba=\dot{Q}_{h} = 9,479  lb_{a}/h \times (26.5  –  13.47) Btu/lb_{a} =

123,511 Btu/h (36.2 kW).

The amount of steam required is easily determined from an energy balance on the heating coil:

m˙steam=m˙a(h2 – h1)(h3 – h4)=9479×⟮26.5 – 13.471153.4 – 168.1⟯\dot{m}_{steam} = \dot{m}_{a} \frac{(h_{2}  –  h_{1})}{(h_{3}  –  h_{4})} = 9479 \times \left\lgroup \frac{26.5  –  13.47 }{1153.4  –  168.1} \right\rgroup

 

=125.3 lbw/h(56.5 kgw/h) = 125.3  lb_{w}/h (56.5  kg_{w}/h)

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