Question 19.9: Comparison of Energy Use of Different Secondary Systems unde......
Comparison of Energy Use of Different Secondary Systems under Part Load
(a) Single-Duct VAV System (Figure 19.15b)
Consider the same building with two zones as in Example 19.8 for which a VAV system was sized. Analyze this system under part-load operation under the following specications:
Given:
1. Outdoor air conditions: T_{db,o} = 77°F and W_{o} = 0.0126 lb_{w}/lb_{a}.
2. Ventilation airflow rate \dot{m}_{a,o} = 167.2 lb_{a}/min (same as the peak summer design flow rate).
3. The design flow rates to each zone:
\dot{m}_{a,A} = 780.7 lb_{a}/min and \dot{m}_{a,B} = 358.7 lb_{a}/min.
4. Cold deck temperature T_{3} = 54°F and 85% RH (resulting in W_{3} = 0.0077 lb_{w}/lb_{a}).
5. The supply airflow to the zones cannot be reduced to less than 60% of the fullload design value by mass.
6. Fan efficiency is 60%.
Assumptions:
1. The latent load in the interior zone is unchanged with outdoor temperature.
2. Ignore factors not included in the list given earlier, such as duct heat losses and gains.
3. The location is assumed to be at sea level.
4. Peak loads are coincident; no diversity adjustment is used.
5. Supply fan air temperature rise is 1°F, and assume the same 3 inWG pressure drop in the duct.
6. The preheat coil is inactive.
Find: Coil loads and fan power
Lookup values: Specic volume v_{0} = 13.95 ft^{3}/lb_{a} and v_{3}= 13.13 ft_{3}/lb_{a}
(i) Outdoor airow rate is specied: \dot{m}_{a,0} = 167.2 lb_{a}/min
(ii) Calculate zone supply airflows assuming supply temperatures to be equal to cold deck temperature T_{3} = 55°F plus fan reheat of 1°F:
= max[468.4,469.9] = 496.9 lb_{a}/min
Note that minimum flow condition has not been reached.
Similarly,
= max[215.2,316.9] = 316.9 lb_{a}/min
i.e., minimum flow condition has not been reached.
Total supply airflow is
\dot{m}_{a,2} = 469.9 + 316.9 = 813.8 lb_{a}/min
or 10,685 ft³ /min (assuming v_{3} = 13.13 ft^{3}/min)
(iii) Determine zone supply air dry-bulb temperatures:
Since the minimum flow conditions have not been reached, we have
T_{supply,A} = 55°F and T_{supply,B} = 55°F
(iv) Determine fan brake horse power from Equation 16.34 (actually the static pressure will drop a little, but this is neglected):
\dot{W}_{shaft} = \frac{\dot{V} ft^{3}/min \times H inWG}{6356 \times \eta_{fan}} (hp) (16.34 IP)
\dot{W}= \frac{10,685 ft^{3}/min \times 3.00 inWG}{6356 \times 0.6} = 8.41 hp
Note that this does not include fan motor efficiency
(v) Check whether average space humidity levels are acceptable:
\bar{W}_{6} = 0.0077 lb_{w}/lb_{a} + \frac{(36,000 + 20,000) Btu/h}{813.8 lb_{a}/min \times 1075 Btu/lb_{w} \times 60 min/h}
= 0.00877 lb_{w}/lb_{a}
which for a space at 75°F dry-bulb temperature corresponds to RH_{5} = 48% (acceptable).
(vi) Determine mixed-air condition:
Dry-bulb temperature:
Humidity ratio:
W_{1} = \left\lgroup \frac{167.2}{813.8} \right\rgroup \times 0.0126 lb_{w}/lb_{a} + \left\lgroup 1 – \frac{167.2}{813.8} \right\rgroup \times 0.00877 lb_{w}/lb_{a}= 0.0096 lb_{w}/lb_{a}
(vii) Calculate cooling coil loads:
\dot{Q}_{cc,sen} = 813.8 lb_{a}/min \times 60 min/h \times 0.24 Btu/(lb_{a} \cdot °F) \times (75.4 – 54)°F = 250,893 Btu/h = 20.9 tons\dot{Q}_{cc,lat} = 813.8 lb_{a}/min \times 60 min/h \times 1075 Btu/(lb_{w}) \times (0.0096 – 0.0077)lb_{w}/lb_{a} = 97,337 Btu/h = 8.11 tons
Total cooling coil load Q_{cc,tot} = 348,231 Btu/h or 29.0 tons
(viii) Calculate reheat at terminals for Zones A and B:
\dot{Q}_{hc,A} = 0 and \dot{Q}_{hc,B} = 0
(b) Single-Duct CAV System (Figure 19.15b)
Consider the same building with two zones as for the single-duct VAV system analyzed earlier.
Compute the cooling and heating loads if the secondary system used to condition the two zones is a single-duct CAV system.
0. Same outdoor airflow rate \dot{m}_{a,0} = 167.2 lb_{a}/min. Also, total supply air mass flow rate \dot{m}_{a} = 1139 lb_{a}/min while the design flow rates to each zone:
\dot{m}_{a,A} = 780.7 lb_{a}/min and \dot{m}_{a,B} = 358.7 lb_{a,B} = 358.7 lb_{a}/min
| Zone A (Exterior) | Zone B (Interior) | |
| Sensible cooling load | \dot{Q}_{sen,A} = 143,100 Btu/h | \dot{Q}_{sen,B} = 91,260 Btu/h |
| Latent cooling load | \dot{Q}_{lat,A} = 36,000 Btu/h | \dot{Q}_{lat,B} = 20,000 Btu/h |
| Zone temperature Tzone | T_{6} = 75°F | T_{6} = 75°F |
1. Determine zone supply air dry-bulb temperatures assuming the airflows to individual zones to be equal to those determined under peak conditions (Example 19.8):
The supply temperature of zone A is
= 62.3°F
and that of zone B is
T_{db,B} = 75°F – \frac{91,260 Btu/h}{358.7 lb_{a}/h \times 0.24 Btu/(lb_{a} \cdot °F) \times 60 min/h}= 57.3°F
2. Fan power is 12 hp (same as under peak load since flow is unchanged).
3. Check whether average space humidity levels are acceptable:
= 0.0085 lb_{w}/lb_{a}
which corresponds to RH_{5} = 45% (acceptable)
4. Determine mixed-air condition:
Dry-bulb temperature:
Humidity ratio:
W_{1} = \left\lgroup \frac{167.2}{1139.4} \right\rgroup \times 0.0126 lb_{w}/lb_{a} + \left\lgroup 1 – \frac{167.2}{1139.4} \right\rgroup \times 0.0085 lb_{w}/lb_{a} = 0.0091 lb_{w}/lb_{a}5. Calculate cooling coil loads:
\dot{Q}_{cc,sen} = 1139.4 lb_{a}/min \times 60 min/h \times 0.24 Btu/(lb_{a} \cdot °F) \times (75.3 – 54)°F = 349,477 Btu/h = 29.12 tons\dot{Q}_{cc,lat} = 1,139.4 lb_{a}/min \times 60 min/h \times 1,075 Btu/lb_{w} \times (0.0091 – 0.0077)lb_{w}/lb_{a} = 102,888 Btu/h
= 8.57 tons
Total cooling coil load Q_{cc,tot} = 37.7 tons.
6. Calculate reheat at terminals for zones A and B:
\dot{Q}_{hc,B} = 358.7 lb_{a}/min \times 60 min/h \times 0.24 Btu/(lb_{a} \cdot °F) \times (57.3 – 55)°F = 11,880 Btu/h
Total load on reheat coils:
\dot{Q}_{hc,tot} = \dot{Q}_{hc,A} + \dot{Q}_{hc,B} = 11,880 Btu/h(c) Dual-Duct CAV System (Figure 19.16b)
Consider the same building with two zones as for the single-duct CAV system analyzed earlier.
Compute the cooling and heating loads if the secondary system used to condition the two zones is a dual-duct CAV system with the cold deck set at the same cooling coil conditions as previously.
Because of the location of the fan, the fan heats up the mixed airstream (T_{2} = 75.3°F + 1°F = 76.3°F), while, in order to be consistent, the cold deck is set at T_{db,3} = 55°F. The hot deck temperature is taken to be T_{db,4} = 105°F to be consistent with the peak design procedure described in Example 19.8.
Steps 1–5 are identical to the single-duct CAV system case.
6. Compute the airflow rates using Equation 19.18:
\dot{m}_{a,cc,A} = \dot{m}_{a,A} \frac{(T_{db,4} – T_{db,5A})}{(T_{db,4} – T_{db,3})} and \dot{m}_{a,hc,A} = \dot{m}_{a,A} – \dot{m}_{a,cc,A}
and
\dot{m}_{a,cc,B} = \dot{m}_{a,B} \frac{(T_{db,4} – T_{db,5B})}{(T_{db,4} – T_{db,3})} and \dot{m}_{a,hc,B} = \dot{m}_{a,B} – \dot{m}_{a,cc,B} (19.18)
\dot{m}_{a,cc,A} = 780.7 lb_{a}/min \times \frac{(105 – 62.3)°F}{(105 – 55)°F} = 667.7 lb_{a}/minand \dot{m}_{a,hc,A} = 780.7 – 6666.7 = 1140.0 lb_{a}/min
\dot{m}_{a,cc,B} = 358.7 lb_{a}/min \times \frac{(105 – 57.3)°F}{(105 – 55)°F} = 342.2 lb_{a}/min
and \dot{m}_{a,hc,B} = 358.7 – 342.2 = 16.5 lb_{a}/min
7. Determine the airflow over the cooling and heating coils:
\dot{m}_{a,cc} = 666.7 + 342.2 = 1008.9 lb_{a}/minand \dot{m}_{a,hc} = 130.5 lb_{a}/min
8. Calculate cooling coil load:
\dot{Q}_{cc,sen} = 1,008.9 lb_{a}/min \times 60 min/h \times 0.24 Btu/(lb_{a} \cdot °F) \times (76.3 – 55)°F = 322,343 Btu/h = 26.86 tons
\dot{Q}_{cc,lat} = 1,008.9 lb_{a}/min \times 60 min/h \times 1,075 Btu/lb_{w} \times (0.0091 – 0.0077) lb_{w}/lb_{a} = 91,104 Btu/h= 7.6 tons
Total cooling coil load Q_{cc,tot} = 34.46 tons.
9. Calculate heating coil load:
Comments
The secondary system is meant to meet the zone loads plus the energy needed to condition the ventilation outdoor air to the space condition.
We will assume that the ideal indoor space condition is 75°F and 50% RH (at which condition, the humidity ratio is 0.0093 lb_{w}/lb_{a}). Hence, the “ideal” load is given by
\dot{Q}_{cc,ideal} = (Sensible + latent loads) for Zone A and B + (Sensible + latent loads) for ventilation air
= (143,100 + 91,260 + 36,000 + 20,000) Btu/h + 167.2 kg/min × 60 min/h
× [0.24 Btu/(lb_{a} \cdot °F) \times (77 – 75) °F + 1,075 Btu/lb_{w} × (0.0126 – 0.0093) lb_{w}/lb_{a}]
= 330,763 Btu/h = 27.6 tons
The following table assembles the summary results of all three all-air systems under the partload operation specified in Example 19.9.
We note that the VAV system outperforms the other two systems and is very close to the ideal loads. It is the best choice since it requires the least cooling energy, no heating energy, and lower fan electricity use. If we dene efficiency in terms of the ideal loads, then for the VAV system, it is (27.6/29) = 95%, while that for the single-duct CAV = (27.6 tons × 12,000 Btu/h · ton)/(37.7 tons × 12,000 Btu/h · ton + 93,947 Btu/h) = 0.61. The two CAV systems are similar thermodynamically, and the differences in energy use determined are partly due to the manner in which we selected the hot deck temperature, due to the way we adjusted for the supply fan being upstream, and also due to slightly different resulting humidity levels in the space that affect the latent loads. The single-duct CAV dehumidies the entire airstream, while, in the dual-duct system, it is only the portion of the airstream flowing over the cooling coil that is dehumidied. This has an impact on humidity levels inside the two zones.
| Ideal Loads |
SingleDuct CAV | SingleDuct VAV | Dual-Duct CAV | |
| Cooling (tons) |
27.6 | 37.7 | 29.0 | 34.5 |
| Heating (Btu/h) |
– | 93,947 | 0 | 53,726 |
| Fan power (hp) |
– | 12.0 | 7.41 | 12.0 |
| Thermal efficiency |
– | 0.61 | 0.95 | 0.71 |
