Question 3.9: A given space is to be maintained at 78 F db and 65 F wb. Th......

A simplified schematic is shown in Fig. 3-12. The given quantities are shown and stations are numbered for reference. By Eq. 3-47 the sensible heat factor for the conditioned space is

SHF  =  \frac{\dot{q}_{s}}{\dot{q}_{s}  +  \dot{q}_{l}}  =  \frac{\dot{q}_{s}}{\dot{q}}                                      (3-47)

SHF  =  \frac{42,000}{60,000}  =  0.7

State 3 is located as shown in Fig. 3-13, where a line is drawn from point 3 and parallel to the SHF = 0.7 line on the protractor. State 2, which may be  any point on that line, fixes the quantity of air supplied to the space. Its location is determined by the operating characteristics of the equipment, desired indoor air quality, and what will be comfortable for the occupants. These aspects of the problem will be developed later. For now assume that the dry bulb temperature of the entering air t_{2} is 20 F less than the space temperature t_{3}.  Then  t_{2} = 58 F, which fixes state 2. The air quantity requiredmay now be found from an energy balance on the space:

\dot{m}_{a2} i_{2}  +  \dot{q}  =  \dot{m}_{a3} i_{3}

or

\dot{q}  =  \dot{m}_{a2}  (i_{3}  –  i_{2})

and

\dot{m}_{a2}  =  \frac{\dot{q}}{i_{3}  –  i_{2}}

From Chart 1a, i_{3}  =  30  Btu/lbma,  i_{2}  =  23  Btu/lbma, and

\dot{m}_{a2}  =  \dot{m}_{a3}  =  \frac{60,000}{30  –  23}  =  8570  lbma/hr

Also from Chart 1a, v_{2}  =  13.21  ft^{3}/lbma and the air volume flow rate required is

\dot{Q}_{2}  =  \dot{m}_{a2} v_{2}  =  \frac{8570(13.21)}{60}  =  1885  or  1890  cfm

Before attention is directed to the cooling and dehumidifying process, state 1 must be determined. A mass balance on the mixing section yields

\dot{m}_{a0}  +  \dot{m}_{a4}  =  \dot{m}_{a1}  =  \dot{m}_{a2}

\dot{m}_{a0}  =  \frac{\dot{Q}_{0}}{v_{0}}  ,  v_{0}  =  14.23   ft^{3}/lbma

\dot{m}_{a0}  =  \frac{500(60)}{14.23}  =  2108  lbma/hr

Then the recirculated air is

\dot{m}_{a4}  =  \dot{m}_{a2}  –  \dot{m}_{a0}  =  8570  –  2108  =  6462  lbma/hr

By using the graphical technique discussed in Example 3-7 and referring to Fig. 3-13, we see that

\frac{\bar{31}}{\bar{30}}  =  \frac{\dot{m}_{a0}}{\dot{m}_{a1}}  =  \frac{2108}{8570}  =  0.246

\bar{31}  =  0.246 (\bar{30})

State 1 is located at 81 F db and 68 F wb. A line constructed from state 1 to state 2 on Chart 1a then represents the process for the cooling coil. An energy balance gives

\dot{m}_{a1} i_{1}  =  \dot{q}_{c}  +  \dot{m}_{a2} i_{2}

Solving for the rate at which energy is removed in the cooling coil

\dot{q}_{c}  =  \dot{m}_{a1} (i_{1}  –  i_{2})

From Chart 1a, i_{1}  =  32.4 Btu/lbma and

\dot{q}_{c}  =  8570 (32.4  –  23)  =  80,600  Btu/hr  =  6.7  tons

The SHF for the cooling coil is found to be 0.6 using the protractor of Chart 1a (Fig. 3-13). Then

\dot{q}_{cs}  =  0.6 (80,600)  =  48,400  Btu/hr

and

\dot{q}_{cl}  =  (80,600)  –  48,400  =  32,200  Btu/hr

The sum of \dot{q}_{cs}  and  \dot{q}_{c1} is known as the coil refrigeration load. Notice that because of outdoor air cooling the coil refrigeration load it is different from the space cooling load. Problems of this type may be solved using the program PSYCH given on the website.