Question 19.8: Single-Duct VAV Secondary System Sizing (Design or Peak Cond......
Single-Duct VAV Secondary System Sizing (Design or Peak Conditions)
A two-zone building is to be equipped with a VAV system with preheat and reheat under the load conditions noted as follows. Size the supply fan, cooling coil, preheat coil, and reheat coil (or baseboard heating) for this building. Note that these will also apply to a CAV system under peak design conditions.
Given: The following data are known from the load calculation and by the designer’s specification (these loads exclude the contribution from outdoor air):
Other:
(a) Summer design dry- and wet-bulb temperatures: 97°F and 76°F (resulting in 0.0144 lb_{w}/lb_{a} humidity ratio at peak conditions).
(b) Winter design temperature: 7°F and air assumed to be totally dry.
(c) Supply system pressure drop at full airflow: 3.0 inWG.
(d) Fan efficiency: 60%.
(e) Supply fan air temperature rise: 1°F.
(f) Return air fan air temperature* rise: none.
(g) Ventilation airow rate: 2400 ft³/min.
(h) Cooling coil outlet humidity ratio (at typical 85% RH): 0.0077 lb_{w}/lb_{a}.
(i) Zone primary supply air temperature difference ∆T_{supply}: 20°F.
(j) The maximum temperature to which space supply air can be heated is 105°F.
Figure: See Figure 19.15b.
Assumptions:
(i) Ignore factors not included in the list given earlier, such as duct heat losses and gains.
(ii) The location is assumed to be at sea level.
(iii) Rather than use enthalpy balances on the zones, the flows will be designed based
on sensible heat loads (this is meant to simplify the problem).
(iv) Peak loads are coincident; no diversity adjustment is used.
(v) During peak heating, latent loads on the space are negligible.
(vi) During winter, room humidifiers are used to maintain the necessary humidity levels.
(vii) The supply airflow to the zones cannot be reduced to less than 60% by mass of the full-load design value by mass.
Lookup value: Outdoor-specific volume = v_{0} = 14.35 ft^{3}/lb_{a}, c_{a} = 0.24 Btu/(lb_{a}·°F), and h_{v} = 1075 Btu/lb_{w}.
Find: Coil loads (preheat, cooling, and terminal reheat) and fan size.
* The return fan heat is often less than the supply fan heat because airflow is less (local exhausts remove air locally), and return ducting via ceiling plenums has smaller pressure drop than found in supply ducts.
| Zone A (Exterior) | Zone B (Interior) | |
| Sensible peak summer design cooling load |
\dot{Q}_{sen,A} = 224,844 Btu/h | \dot{Q}_{sen,B} = 103,308 Btu/h |
| Latent peak summer design cooling load | \dot{Q}_{lat,A} = 56,000 Btu/h | \dot{Q}_{lat,B} = 20,000 Btu/h |
| Sensible peak winter design load | \dot{Q}_{heat,A} = 143,000 Btu/h | \dot{Q}_{cool,B} = 49,092 Btu/h |
| Zone temperature Tzone | 75°F | 75°F |
The solution will follow the numbering system used in the preceding discussion.
1. Calculate ventilation airflow rate.
Cooling Design Day (T_{db,0} = 97°F)
2. Determine air mass flow rates (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,A} = \frac{224,844 Btu/h}{0.24 Btu/(lb \cdot °F) \times 60 min/h \times 20°F} = 780.7 lb_{a}/min\dot{m}_{a,B} = \frac{103,308 Btu/h}{0.24 Btu/(lb \cdot °F) \times 60 min/h \times 20°F} = 358.7 lb_{a}/min
Thus, the total airflow is \dot{m}_{a} = 1,139.4 lb_{a}/min or 14,983 ft³/min (based on cooling coil leaving condition, specific volume = 13.15 ft^{3}/lb_{a})
3. The system pressure drop is 3.0 inWG.
Therefore, the fan power is given by Equation 16.34 IP:
Note that this is the fan brake hp and does not include the electric motor efficiency
4. Determine cooling air supply temperature.
5. Determine cooling coil leaving temperature.
T_{coil,out} = T_{3} = T_{supply} – \Delta T_{db,fan} – \Delta T_{db,duct}= 55°F – 1°F – 0°F = 54°F
6. Determine zone return air condition.
According to the problem statement, there are no return duct heat gains; therefore, the return dry-bulb temperature for each zone is 75°F.
7. Compute average return air humidity and temperature.
The return air average moisture content under steady state is the given zone supply air humidity ratio (0.0077 lb_{w}/lb_{a}) increased by the total latent load:
+ \frac{(56,000 + 20,000) Btu/h}{1,139.4 lb/min \times 60 min/h \times 1,075 Btu/lb}
= 0.00873 lb_{w}/lb_{a}
The average return air temperature is T_{7} = T_{9} = 75°F. This condition corresponds to an average space RH_{6} = 48% that meets human comfort criteria.
8. Compute mixed-air condition. To find the mixed-air temperature entering the cooling coil, an energy balance is performed at point 1 where the return and outside air ducts connect. Since the ventilation airow rate is 167.2 lb_{a}/min, the mixed-air temperature under design conditions is* as follows:
= 78.2°F
Likewise, the mixed-air humidity ratio is found by a water vapor mass balance:
W_{m} = 0.00873 lb_{w}/lb_{a} \times \left\lgroup 1 – \frac{167.2}{1139.4} \right\rgroup + 0.0144 lb_{w}/lb_{a} \times \left\lgroup \frac{167.2}{1139.4} \right\rgroup = 0.0096 lb_{w}/lb_{a}9. Compute cooling coil loads (preheat coil is inactive in summer). The coil sensible load is
\dot{Q}_{cc,sen} = \dot{m}_{a} c \Delta T_{coil} = 1139.4 lb_{a}/min \times 60 min/h \times 0.24 Btu/(lb \cdot °F) \times (78.2 – 54)°F= 397,529 Btu/h or 33.13 tons
The latent load is found from:
\dot{Q}_{cc,lat} = m_{a} \cdot h_{vap} \cdot \Delta W_{cc} = 1,139.4 lb_{a}/min \times 60 min/h \times 1075 Btu/lb \times (0.00957 – 0.0077 ) lb_{w}/lb_{a}= 137,103 Btu/h = 11.45 tons
The total coil load or coil cooling rate is
\dot{Q}_{cc,tot} = \dot{Q}_{cc,sen} + \dot{Q}_{cc,lat} = 534,632 Btu/h = 44.55 tonsHeating Design Day (T_{db,0} = 7°F)
10. Determine preheat coil load. The peak load on the preheat coil is determined based on the load needed to heat the cold outdoor ventilation air on the winter design day to the nominal coil outlet temperature (54°F in this example).
Therefore, the preheat (subscript “ph”) coil size is given by (assuming that outdoor air intake is the same as under peak summer condition)
It must, however, be noted that the preheat coil will be activated if T_{1} < 54°F and will raise the supply airstream temperature to 54°F.
11. Determine airflow rates from Equation 19.11
a. Zone A requires heating: assume maximum allowable supply air temperature of 105°F:
= 331.0 lb_{a}/min
Recall the stipulation that minimum airflow rate should not be less than 60% of the peak, or \dot{m}_{a,min} = 0.6 \times 780.7 = 468.4 lb_{a}/min . So, we set m_{a,A} = 468.4 lb_{a}/min.
b. Zone B requires cooling: assume minimum coil set point temperature
of 54°F:
= max(215.2,170.4) = 215.2 lb_{a}/min
Note that even for zone B, the stipulated minimum is reached.
Total airflow rate is 468.4 + 215.2 = 683.7 lb_{a}/min (rounded).
12. Determine supply air temperatures to the zones based on sensible heat balances.
Zone A: heating
T_{db,5A} = 75°F + \frac{143,000 Btu/h}{0.24 Btu/(lb_{a} \cdot °F) \times 60 min/h \times 468.4 lb_{a}/min}= 96.2°F
Zone B: cooling
T_{db,5B} = 75°F – \frac{49,092 Btu/h}{0.24 Btu/(lb_{a} \cdot °F) \times 60 min/h \times 215.2 lb_{a}/min}= 59.2°F
13. Determine mixed-air temperature.
T_{2} = 75.0°F \times \left\lgroup 1 – \frac{167.2}{683.7} \right\rgroup + 7°F \times \left\lgroup \frac{167.2}{683.7} \right\rgroup = 58.4°FLikewise, the mixed-air humidity ratio can be found assuming room air to be at the cooling coil leaving humidity ratio W_{1} = W_{3} = 0.0077 lb_{w}/lb_{a}. This is not needed, however, for our calculations, since there will not be any latent load on the coil during the winter design condition.
14. Compute cooling coil load. The needed supply air temperature to Zone B is 59.2°F (from step 12) and 1°F temperature rise by the fan, which should be taken into account. Hence, the coil leaving temperature should be 58.2°F.
\dot{Q}_{cc,sen} = 683.7 lb_{a}/min \times 60 min/h \times 0.24 Btu/(lb_{a} \cdot °F) \times (58 – 58.2)°F = 2060 Btu/h or 0.172 tons
(in practice such a small cooling load would not justify switching on the chiller)
15. Compute reheating (assuming the chiller to be activated).
Zone A:
Zone B:
\dot{Q}_{hc,B} = 215.2 lb_{a}/min \times 60 min/h \times 0.24 Btu/(lb_{a} \cdot °F) \times (59.2 – 59.2)°F = 0 Btu/hThe total peak heating requirement is the sum of the preheat and reheat amounts:
\dot{Q}_{heat,tot} = 113,161 + 249,846 + 0 = 363,007 Btu/hThis heat rate must be increased by pickup loads from night setback, safety factors, and piping losses to size the coils and the boiler.
Comments
Since this is a simple two-zone building, no diversity was assumed in the cooling or heating loads.
Although diversity is rarely taken into account in heating loads unless internal or solar gains coincide with peak heating conditions, some diversity can be assumed in cooling loads when the building consists of a larger number of zones. With two zones, it is safer not to assume any diversity.
This example shows that the secondary systems have a considerable effect on the sizing of the primary heating and cooling plants. In any building more complex than a single-family residence, the secondary system loads are usually greater (and in many instances, much greater) than space zone
loads. This effect must be considered in sizing the central plant. In this example at design conditions, the zone cooling load is about 330 kBtu/h, while the coil cooling load is 535 kBtu/h. The difference is due to the outdoor air that is conditioned at the central air handler coil.
In this example, we have used a given value of fan temperature rise. The actual rise can be computed as discussed in Section 19.1.3. Although the return fan has not been sized in this example, it can be done exactly as it was done with the supply fan. The duct pressure drop and return fan efciency must be known to complete the calculation.
* Although mass flow rates should theoretically be used, engineers often use volumetric flow rates instead during preliminary analysis, since the density of all airstreams is about the same and the inaccuracy introduced is generally small.