Question 19.5: Analysis of CAV Systems under Part-Load Operation Consider t......

Analysis of CAV Systems under Part-Load Operation

Consider the CAV system that was analyzed under peak-load condition in Example 19.1. This example will serve to illustrate how the CAV system needs to be operated under part-load condition so as to maintain satisfactory indoor space conditions. We conditions. We will determine the cooling coil loads as well as the reheat energy required in this case. We assume the same zone and cooling coil outlet temperatures and the amount of ventilation air as stipulated in Example 19.1. The outdoor air conditions are 80°F and 60% RH, while the space cooling load is 70,000 Btu/h with SHR_{space} = 0.7.
Figure: See Figures 19.1 and 19.4.*
Assumptions: The location is at sea level. The supply air mass flow rate = 17,140  lb_{a}/h. The duct heat transfer and the fan air temperature rise are ignored for simplicity. The preheat and steam humidifier are inactive.
Given: SHR_{space} = 0.70, \dot{Q}_{space,tot} = 7 0,000  Btu/h, \dot{Q}_{space,sen} = 70,000 \times 0.7 = 49,000  Btu/h
Outdoor conditions: T_{db,o} = 80°F, \phi_{o} = 0.6, \dot{V}{0} = \dot{V}_{0} = 1000  ft^{3}/min
Cooling coil conditions: T_{db,3} = 58°F and \phi_{3} = 0.8.
Space condition: T_{db,6} = 78°F and supply air mass flow rate = 17,140  lb_{a}/h
Find: \dot{Q}_{cc,tot}, \dot{Q}_{cc,sen}, and \dot{Q}_{hc,tot}

Lookup values: Specific volume v_{0} = 13.9  ft^{3}/lb_{m}, humidity ratio W_{0} = 0.01325  lb_{w}/lb_{a}, W_{3} = 0.0082  lb_{w}/lb_{a}, specific heat of air c_{a} = 0.24  Btu/(lb_{a} · °F), and latent heat of vaporization h_{v} = 1075  Btu/lb_{w}

* Numerical values shown in the figure do not correspond to this example.

19.1
19.4
Step-by-Step
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1. Calculate outdoor air mass flow rate

\dot{m}_{a,0} = \frac{\dot{V}_{0}}{v_{0}} = \frac{1000  ft^{3}/min  \times 60  min/h}{13.9  ft^{3}/lb_{a}} = 4320  lb_{a}/h

2. Calculate supply air temperature point 5 using sensible heat balance since the room thermostatic control is based on dry-bulb temperature.
Using sensible heat balance equation:

\dot{Q}_{space,sen} = \dot{m}_{a} \cdot c_{a} \cdot (T_{db,6}  –  T_{db,5})

 

T_{db,5} = T_{db,6}  –  \frac{\dot{Q}_{space,sen}}{\dot{m}_{a}  \times  c_{a}}

 

= 78°F  –  \frac{49,000  Btu/h}{17,140  lb_{a}/h  \times  0.24  Btu/(lb_{a} \cdot °F)}

= 66.1°F

3. Verify indoor comfort.
Calculate humidity ratio of air leaving room using a latent heat balance

\dot{Q}_{space,lat} = \dot{m}_{a} \cdot h_{v } \cdot (W_{6}  –  W_{5})

or

W_{6} = W_{5} + \frac{\dot{Q}_{space,lat}}{\dot{m}_{a}  \times  h_{vap}} = 0.0082  lb_{w}/lb_{a}

 

+ \frac{70,000  Btu/h  \times  (1  –  0.7)}{17,140  lb_{a}/h  \times  1,075  Btu/lb_{w}} = 0.00934  lb_{W}/lb_{a}

This corresponds to an indoor relative humidity of \phi_{o} = 45%, which along with a dry-bulb temperature of 78°F is satisfactory for indoor human comfort.
4. Calculate mixed-air condition point 1.
A common assumption that simplifies the analysis is to assume a constant-specific heat of moist air c_{a}. The energy balance equation can then be expressed as

\dot{m}_{a,1} T_{db,1} = (\dot{m}_{a,1}  –  \dot{m}_{o} ) T_{db,7} + \dot{m}_{o} T_{db,o}

or

17,140  lb_{a}/h \times T_{db,1} = (17,140  –  7,320)  lb_{a}/h \times 78°F + 4,320  lb_{a}/h \times 80°F

resulting in T_{db,1} = 78.5°F.
Similarly, the humidity is determined as W_{1} = 0.0103  lb_{w}/lb_{a}.
5. Determine cooling coil loads(process 1–3).
Using the simplified expressions for sensible and latent loads, we have

a. Sensible load (Equation 13.45)
\dot{Q}_{cc,sen} = 17,140  lb_{a}/h \times 0.24  Btu/(lb_{a} \cdot °F) \times (78.5  –  58)°F
= 84,329 Btu/h = 7.03 tons
b. Latent load (Equation 13.46)

\dot{Q}_{lat} = \dot{m}_{a} h_{vap} (W_{1}  –  W_{2} )            (13.46)
\dot{Q}_{cc,lat} = 17,140  lb_{a}/h  \times 1075  Btu/lb_{w} \times (0.0103  –  0.0082) lb_{w}/lb_{a}
= 38,931 Btu/h = 3.24 tons
c. Total load
\dot{Q}_{cc,tot} = 84,329 + 38,931 = 123,260  Btu/h =10.27  tons
6. Determine reheat coil load.
Using the sensible heat equation, reheat load
\dot{Q}_{rh,tot} = 17,140  lb/h \times 0.24  Btu/(lb_{a} \cdot °F) \times (66.1  –  58) °F
= 73,320  Btu/h
Comments
The reheat is an energy penalty in that the single airstream supply to the space has to be overcooled in order to provide the dehumidication necessary for occupant comfort and then reheated in order to negate this overcooling. This is a good illustration of the concept of coil bucking stated earlier.

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