Question 19.6: VAV Analysis under Part-Load Operation Consider Example 19.5......
VAV Analysis under Part-Load Operation
Consider Example 19.5 where a CAV system was assumed. We will solve this problem assuming VAV system operation.
Figure: Figures 19.1 and 19.3.*
Assumptions: The location is at sea level. The duct heat transfer and the fan air temperature rise are ignored for simplicity. The preheat coil is inactive.
Given: \dot{Q}_{space tot} = 70,000 Btu/h, \dot{Q}_{space,sen} = 49,000 Btu/h
Outdoor conditions: T_{o} = 80°F, \phi_{o} = 0.6, \dot{V}_{0} = \dot{V}_{0} = 1000 ft^{3}/min, or \dot{m}_{0} = 4320 lb_{a}/h
Cooling coil conditions:T_{db,3} = 58°F and \phi_{3} = 0.8.
Space condition: T_{db,6} = 78°F.
Find: \dot{Q}_{cc,tot}, \dot{Q}_{cc,sen}, and \dot{Q}_{hc,tot}
Lookup values: 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_{vap} = 1075 Btu/lb_{w}.
* Numerical values shown in the figure do not correspond to this example.
1. Calculate supply air mass flow rate with supply air temperature assumed to be equal to that of the cooling coil set temperature. From a sensible heat balance equation (Equation 19.11),
\dot{Q}_{space,sen} = \dot{m}_{a} \cdot c_{a} \cdot (T_{db,space} – T_{db,supply}) (19.11)
\dot{m}_{a} = \frac{\dot{Q}_{space,sen}}{(T_{db,6} – T_{db,6}) \times c_{a}} = \frac{49,000 Btu/h}{(78 – 58)°F \times 0.24 Btu/(lb_{a} \cdot °F)}= 10,205.3 lb_{a}/h
2. Verify indoor comfort. Calculate humidity ratio of air leaving room from a latent heat balance:
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)}{10,208.3 lb_{a}/h \times 1075 Btu/lb_{w}} = 0.0101 lb_{w}/lb_{a}
This condition corresponds to an indoor relative humidity of \phi_{o} = 49%, which along with a dry-bulb temperature of 75°F is within the indoor human comfort range.
3. Calculate mixed-air condition.
or
10,208.3 lb_{a}/h \times T_{db,1} = (10,208.3 – 4,320) lb_{a}/h \times 78°F + 4,320 lb_{a}/h \times 80°F
resulting in T_{db,1} = 78.85°F.
Similarly, the humidity ratio is determined as W_{1} = 0.0114 lb_{w}/lb_{a}
4. Determine cooling coil loads.
Using the simplified expressions from Section 13.6.5, we have
a. Sensible load (Equation 13.45)
\dot{Q}_{sen} = \dot{m}_{a} c_{p,m} (T_{db1} – T_{db2}) (13.45)
\dot{Q}_{cc,sen} = 10,208.3 lb_{a}/h \times 0.24 Btu/(lb_{a} \cdot °F) \times (78.85 – 58)°F
= 51,082 Btu/h = 4.26 tons (down from 7.03 tons in Example 19.5)
b. Latent load (Equation 13.46)
\dot{Q}_{lat} = \dot{m}_{a} h_{vap} (W_{1} – W_{2}) (13.46)
\dot{Q}_{cc,lat} = 10,208.3 lb_{a}/h \times 1,075 Btu/lb_{w} \times (0.0114 – 0.0082) lb_{w}/lb_{a}
= 35,116 Btu/h
= 2.93 tons (down from 3.24 tons)
c. Total load
\dot{Q}_{cc,tot} = 51,082 + 35,116 = 86,198 Btu/h
= 7.18 tons (down from 10.27 tons)
5. Calculate reheat coil load.
In this case, reheat coil load is zero.
Comments
In the aforementioned example, the reduced airflow has eliminated the need for reheat but has somewhat compromised occupant comfort due to the slightly elevated humidity in the space (RH increased from 45% to 49%). The flow reduction of (10,208.3/17,140) = 0.595 is right at the lower end of the norm allowed.
Had it been lower, the proper operating strategy would be to adopt mode b shown in Figure 19.9. This would have entailed a small amount of reheat and increased cooling energy use as well.
