Question 24.2: Use of the Solar Load Ratio Method Find the auxiliary heatin......
Use of the Solar Load Ratio Method
Find the auxiliary heating energy for the building of Figure 24.7 in Denver, Colorado, in January.
Given: T_{i} = 68°F (20°C)
\dot{Q}_{gain} = 4200 Btu/h (1231 W)Sunspace design type D1 (12 in. [0.3 m] masonry wall between sunspace and conditioned space)
Sunspace being unconditioned and Semienclosed by conditioned space
Aperture of sunspace double glazed with an area of 550 ft² (51.1 m²), its normal being tilted at 50° from vertical
South facing, without night insulation
The envelope has the following characteristics:
The heat loss coefficient without the sunspace is indicated in italic as 525.0 Btu/(h· °F) (277 W/K).
Hence, the NLC is
NLC = 24 h/day \times 525.0 Btu/(h \cdot °F)
= 12,600 Btu/(°F . day)
Find: Q_{aux}
Lookup values: We need to look up solar radiation S and degree-days D_{h} for January in Denver.
(For simplicity, we take these values as known; they could be calculated by using the methods of Sections 4.9 and 10.3.)
S = 43,600 Btu/ft²
= Monthly solar radiation on aperture
For the degree-days, we first determine the balance-point temperature from Equation 24.3:
T_{bal} = T_{i} – \frac{\dot{Q}_{gain,day}}{NLC} (24.3)
T_{bal} = 68° – \frac{24 h \times 4,200 Btu/h}{12,600 Btu/(°F \cdot day)} = 60°F
The corresponding degree-days are
= Monthly degree-days D_{h}(T_{bal}) for base T_{bal}
| A or \dot{V},ft^{2} or ft³/min | U, Btu/ (h· °F·ft²) | UA or \rho c_{p} \dot{V}, Btu/(h· °F) | |
| Opaque walls | 2,000 | 0.040 | 80.0 |
| Roof | 3,000 | 0.030 | 90.0 |
| Floor (over crawl space) |
3,000 | 0.040 | 120.0 |
| Windows (east, west, north) |
100 | 0.550 | 55.0 |
| Subtotal | 345.0 | ||
| Infiltration | 200 ft³/min (at 0.5 ach with V = 24,000 ft³) | \rho c_{p} = 0.015 Btu/(ft³ · °F) in Denver | 180.0 |
| Subtotal | 525.0 | ||
| Sunspace | 550 | 0.55 | 302.2 |
| Total | 827.2 |
The projected area is
A_{proj} = 550 ft^{2} \times \sin 50° = 421 ft^{2}From Equation 24.4,
LCR = \frac{NLC}{A_{proj}} (23.4 IP)
LCR = \frac{12,600 Btu/(°F \cdot day \cdot ft^{2})}{421 ft^{2}} = 29.9 Btu/(°F \cdot day)For the ratio S/D_{h}, we have
\frac{S}{D_{h}} = \frac{43.600 Btu/ft^{2}}{930°F \cdot day} = 46.9 Btu/(°F \cdot day \cdot ft^{2})Reading Figure 24.8 at these values, we find
SSF = 0.51
This means that 51% of the heating requirement of the building is provided by the sunspace in January. Equation 24.5 yields the auxiliary heating requirement for the month:
Q_{aux} = NLC \times D_{h} \times (1 – SSF) (24.5)
\dot{Q}_{aux} = NLC \times D_{h} \times (1 – SSF)= 12,600 Btu/(°F . day) × 930 °F . days × (1 – 0.51)
= 5.74 MBtu/month
Comments
In accordance with the assumptions of the method, this result presupposes that the thermostat dead band is 65°F–75°F for the conditioned part of the building and 45°F–95°F for the sunspace. The designer should check whether this is agreeable with the future occupants of the building
