Question 3.10: A space is to be maintained at 75 F and 50 percent relative ......

Figure 3-19 is a schematic for the problem; it contains the given information and reference points. First consider the conditioned space:

SHF  =  \frac{225,000}{225,000  +  56,250}  =  0.80

The state of the supply air lies on a line drawn through state point 3 parallel to the SHF = 0.8 line on the protractor of Chart 1a. Figure 3-20 shows this construction. State 2 is located at 120 F dry bulb and the intersection of this line. An energy balance on the space gives

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

or

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

From Chart 1a, i_{2}  =  42  Btu/lbma,  i_{3} =  28.2 Btu/lbma, and

\dot{m}_{a2}  =  \frac{\dot{q}}{i_{2}  –  i_{3}}  =  \frac{281,250}{42  –  28.2}  =  20,400  lbma/ hr

From Chart 1a, \text{v}_{2} = 14.89 ft³/lbma, and

\dot{Q}_{2}  =  \frac{20,400}{60}  ×  14.89  =  5060  cfm

To find the conditions at state 1, the mixing process must be considered. A mass balance on the mixing section yields

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

or

\dot{m}_{a4}  =  \dot{m}_{a2}  –  \dot{m}_{a0}

\dot{m}_{a0}  =  \frac{\dot{Q}_{0}}{v_{0}}   and  v_{0}  =  12.54  ft³ /lbma

\dot{m}_{a0}  =  \frac{1000(60)}{12.54}  =  4800  lbma/hr

\dot{m}_{a4}  =  20,400  –  4800  =  15,600  lbma/hr

Using the graphical technique and referring to Fig. 3-20, we obtain

\bar{31}  =  \frac{\dot{m}_{a0}}{\dot{m}_{a1}}  \bar{30}  =  \frac{4800}{20,400}  \bar{30}  =  0.235  (\bar{30})

State 1 is then located at 65.5 F db and 57 F wb. A line \bar{12} could be constructed on Chart 1a, Fig. 3-20, representing the combination heating and humidifying process that must take place in the heating and humidifying unit. However, in practice the processes must be carried out separately. Assume that saturated vapor at 200 F is used in the humidifier. Then i_{w} = 1145.8 Btu/lbm from Table A-1a. The required sensible heating is

\dot{q}_{1x}  =  \dot{q}_{s}  =  \dot{m}_{a}  c_{pa}  (t_{x}  –  t_{1})

\dot{q}_{s}  =  20,400(0.244)(119  –  65.5)  =  266,000  Btu/hr

The amount of water vapor supplied to the humidifier is given by

\dot{m}_{\text{v}}  =  \dot{m}_{a}  (W_{2}  –  W_{1})

where W_{2} =  0.012  lbv/lba  and  W_{1}  =  0.0078  lbv/lba from Chart 1a, so that

\dot{m}_{\text{v}}  =  20,400(0.012  –  0.0078)  =  86  lbv/hr

It is usually necessary to use a preheat coil to heat the outdoor air to a temperature above the dew point of the air in the equipment room so that condensation will not form on the air ducts upstream of the regular heating coil. Figure 3-21 shows this arrangement. The outdoor air is heated to state 0′, where it is mixed with return air, resulting in state 1. The mixed air is then heated to state x, where it is humidified to state 2 on the condition line for supply to the space. Figure 3-22 shows the states on Chart 1a.

Example CD3-5 illustrates a system with preheat of outdoor air. Examples of other single-path systems such as VAV or multizone could be presented here; however, under the full-flow design condition, these systems operate the same as the simple system of Figs. 3-12 and 3-13. They will be discussed further in the following section on part-load operation.