The Degree-Day Formula

The degree-day formula was devised some 40-odd years ago by the American Gas Association and other groups and has since been revised (see pages 627-28 of the 1970 ASHRAE Guide) to reflect

Table 4-7 Reduction in Standard Degree Days for All Areas Except the Pacific Coast

Degree Days at 65°F Base

Reduce by This Percentage

New Total

10,000

3.15

9685

9500

3.20

9196

9000

3.22

8710

8500

3.23

8226

8000

3.25

7740

7500

3.80

7215

7000

3.85

6720

6500

4.35

6218

6000

4.55

5727

5500

4.60

5247

5000

4.65

4768

4500

4.70

4289

4000

5.30

3788

3500

5.35

3313

3000

6.15

2816

2500

7.15

2321

2000

8.00

1840

1500

9.50

1358

1000

11.90

881

500

15.10

425

Courtesy National Oil Fuel Institute

Table 4-8 Reduction in Degree-Day (DD) Base When Calculated Heat Loss (Btu per Degree Temperature Difference) is Less Than 1000

Calculated Heat Loss

Revised DD Base, °F

200

55

300

60

400

61

500

62

600

63

700

64

800

64

900

64

1000

65

Table 4-9 Using Corrected Heat Loss Formula for Determining Comparative Fuel Requirements

Assume 500 Btu/h per degree temperature difference (35,000 Btu/h at 70°F difference); degree days (dd) with 65°F base, 5542; average indoor temperature 73°F.

Correction factor from Table 4-7 is 4.60%.

From Table 4-5, revised degree-day base is 62°F.

Reduction in base is (65 — 62) 3°F.

Multiply 0.0460 X 3 = 0.1380 and deduct from 1.0000 = 0.8620. Multiply 5542 (dd at 65° base) by 0.8620 = 4557.

Example:

No. 2 Oil:

35 X 0.00304 X 4557 = 485 gallons per year Natural Gas:

35 X 0.00429 X 4557 = 684 therms per year

Courtesy National Oil Fuel Institute

Internal heat gains and the levels of insulation. The correction factors involved in the revision are given in Tables 4-10 and 4-11. These correction factors should also be applied to the corrected heat loss formula (see the previous section).

The degree-day formula is based on the assumption that heat for the interior of a house or building will be obtained from sources other than the heating system (e. g., sunlight and body heat of the occupants) until the outside temperature declines to 65°F. At this point the heating system begins to operate. The con­sumption of fuel will be directly proportionate to the difference

Table 4-10 Unit Fuel Consumption per Degree Day per 1000 Btu Design Heat Loss

Utilization Efficiency

Fuel

60% 70%

80%

Gas in therms

Oil in gallons (141,000 Btu)

0.00572 0.00490 0.00405 0.00347

0.00429

0.00304

Table 4-11 Correction Factors for Outdoor Design Temperatures*

Outside Design Temp., °F

-20 -10 0 +10 +20

Correction factor 0.778 0.875 1.000 1.167 1.400

*To be applied to Table 4-1 when design temperature is higher or lower than 0‘F. Courtesy National Oil Fuel Institute

Between the 65°F base temperature and the mean outdoor tem­perature. In other words, three times as much fuel will be used when the mean outdoor temperature is 35°F than when it is 55°F. The mean outdoor temperature can be determined by taking the sum of the highest and lowest outside temperatures during a 24- hour period, beginning at midnight, and dividing it by 2. Each degree in temperature below 54°F is regarded as 1 degree day.

The degree-day formula is applied by dividing the heat loss fig­ure by 1000 and multiplying the result by the figure for the unit fuel consumption per degree day per 1000 Btu design heat loss, which, in turn, is multiplied by the total number of degree days in the heat­ing season (calculated on a 65°F base) and then by the correction factors given in Tables 4-10 and 4-11. The application of this for­mula is illustrated in Table 4-12.

Table 4-12 Application of Degree-Day Formula

35,000 Btu/h loss; outside design temperature 0; no correction factor needed; degree days, 5542.

No. 2 oil, 140,000 Btu per therm, 80% efficiency:

35 X 0.00304 X 5542 = 590 gallons per year Natural gas, 100,000 Btu per therm, 80% efficiency:

35 X 0.00429 X 5542 = 832 therms per year Electricity, required Btu/year

832 (therms) X 100,000 (Bru/therm) X 0.80 (efficiency)

66,560,000 = 66,560,000 = 19,502 kWh

KWh = heat loss 3413

X annual degree days X constant (usually 18.5) divided by difference between indoor and outdoor design temperature. Example:

35,0 Btuh

3413

1030 X 5542 X 18.5

подпись: 3413
1030 x 5542 x 18.5
= 1030

= 15,086 kWh

70 — 0

Courtesy National Oil Fuel Institute

Posted in Audel HVAC Fundamentals Volume 1 Heating Systems, Furnaces, and Boilers