Heating BTU worksheet

1. Heat needed (heater output): 32,572 BTU/h

1. Change to feet and °F, the units the insulation values use (1 ft² = 0.09290304 m², 1 ft = 0.3048 m, 1 °C of difference = 1.8 °F)

   $\frac{150\text{\hspace{0.17em}m²}}{0.09290304}$$=\mathrm{1,614.59}\text{\hspace{0.17em}ft²}$

2. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $17-\left(\mathrm{-12}\right)\times 1.8$$=52.2\text{\hspace{0.17em}°F}$

3. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height, times the 50% that faces outdoors

   $4\times \sqrt{\mathrm{1,614.59}}\times 6.5617\times 0.5$$=527.32\text{\hspace{0.17em}ft²}$

4. Heat through the solid walls (85% of the wall area at R-4): area × temperature rise ÷ R-value

   $\frac{448.22\times 52.2}{4}$$=\mathrm{5,849.32}\text{\hspace{0.17em}BTU/h}$

5. Heat through windows and doors (15% of the wall area at U-1): area × U-factor × temperature rise

   $79.1\times 1\times 52.2$$=\mathrm{4,128.93}\text{\hspace{0.17em}BTU/h}$

6. Heat through the ceiling into the attic (R-11)

   $\frac{\mathrm{1,614.59}\times 52.2}{11}$$=\mathrm{7,661.95}\text{\hspace{0.17em}BTU/h}$

7. Heat carried off by air leaking in and out: 0.018 × volume × 1.5 air changes an hour × temperature rise

   $0.018\times \mathrm{10,594.4}\times 1.5\times 52.2$$=\mathrm{14,931.75}\text{\hspace{0.17em}BTU/h}$

8. Add the four parts. A heated space below loses no heat through the floor

   $\mathrm{5,849.32}+\mathrm{4,128.93}+\mathrm{7,661.95}+\mathrm{14,931.75}$$=\mathrm{32,571.94}\text{\hspace{0.17em}BTU/h}$

9. With no ducts, the heat goes straight into the room

   $\text{heat needed}$$=\mathrm{32,571.94}\text{\hspace{0.17em}BTU/h}$

10. Change to kilowatts (1 kW = 3,412.14 BTU/h)

    $\frac{\mathrm{32,571.94}}{\mathrm{3,412.14}}$$=9.546\text{\hspace{0.17em}kW}$

11. A furnace that is 80% efficient must burn more fuel than the heat it delivers, so divide by its efficiency

    $\frac{\mathrm{32,571.94}}{0.8}$$=\mathrm{40,714.93}\text{\hspace{0.17em}BTU/h}$

12. Round the input up to the next 10,000 BTU/h to pick a size

    $\text{size}$$=\text{50,000 BTU/h furnace (input)}$

2. Heat needed (heater output): 54,950 BTU/h

1. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $55-8$$=47\text{\hspace{0.17em}°F}$

2. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height, times the 50% that faces outdoors

   $4\times \sqrt{\mathrm{1,800}}\times 7\times 0.5$$=593.97\text{\hspace{0.17em}ft²}$

3. Heat through the solid walls (85% of the wall area at R-4): area × temperature rise ÷ R-value

   $\frac{504.87\times 47}{4}$$=\mathrm{5,932.27}\text{\hspace{0.17em}BTU/h}$

4. Heat through windows and doors (15% of the wall area at U-1): area × U-factor × temperature rise

   $89.1\times 1\times 47$$=\mathrm{4,187.49}\text{\hspace{0.17em}BTU/h}$

5. Heat through the ceiling into the attic (R-11)

   $\frac{\mathrm{1,800}\times 47}{11}$$=\mathrm{7,690.91}\text{\hspace{0.17em}BTU/h}$

6. Heat carried off by air leaking in and out: 0.018 × volume × 1.5 air changes an hour × temperature rise

   $0.018\times \mathrm{12,600}\times 1.5\times 47$$=\mathrm{15,989.4}\text{\hspace{0.17em}BTU/h}$

7. Heat lost down through the floor (R-4). The full temperature rise is used, because a vented crawlspace can get nearly as cold as outdoors

   $\frac{\mathrm{1,800}\times 47}{4}$$=\mathrm{21,150}\text{\hspace{0.17em}BTU/h}$

8. Add the parts to get the heat the house loses on a design-cold day

   $\mathrm{5,932.27}+\mathrm{4,187.49}+\mathrm{7,690.91}+\mathrm{15,989.4}+\mathrm{21,150}$$=\mathrm{54,950.07}\text{\hspace{0.17em}BTU/h}$

9. With no ducts, the heat goes straight into the room

   $\text{heat needed}$$=\mathrm{54,950.07}\text{\hspace{0.17em}BTU/h}$

10. Change to kilowatts (1 kW = 3,412.14 BTU/h)

    $\frac{\mathrm{54,950.07}}{\mathrm{3,412.14}}$$=16.104\text{\hspace{0.17em}kW}$

11. A furnace that is 100% efficient must burn more fuel than the heat it delivers, so divide by its efficiency

    $\frac{\mathrm{54,950.07}}{1}$$=\mathrm{54,950.07}\text{\hspace{0.17em}BTU/h}$

12. Round the input up to the next 10,000 BTU/h to pick a size

    $\text{size}$$=\text{60,000 BTU/h furnace (input)}$

3. Heat needed (heater output): 31,986 BTU/h

1. Change to feet and °F, the units the insulation values use (1 ft² = 0.09290304 m², 1 ft = 0.3048 m, 1 °C of difference = 1.8 °F)

   $\frac{200\text{\hspace{0.17em}m²}}{0.09290304}$$=\mathrm{2,152.78}\text{\hspace{0.17em}ft²}$

2. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $28-\left(\mathrm{-13}\right)\times 1.8$$=73.8\text{\hspace{0.17em}°F}$

3. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height

   $4\times \sqrt{\mathrm{2,152.78}}\times 6.5617$$=\mathrm{1,217.8}\text{\hspace{0.17em}ft²}$

4. Heat through the solid walls (85% of the wall area at R-13): area × temperature rise ÷ R-value

   $\frac{\mathrm{1,035.13}\times 73.8}{13}$$=\mathrm{5,876.34}\text{\hspace{0.17em}BTU/h}$

5. Heat through windows and doors (15% of the wall area at U-0.5): area × U-factor × temperature rise

   $182.67\times 0.5\times 73.8$$=\mathrm{6,740.51}\text{\hspace{0.17em}BTU/h}$

6. Heat through the ceiling into the attic (R-30)

   $\frac{\mathrm{2,152.78}\times 73.8}{30}$$=\mathrm{5,295.84}\text{\hspace{0.17em}BTU/h}$

7. Heat carried off by air leaking in and out: 0.018 × volume × 0.75 air changes an hour × temperature rise

   $0.018\times \mathrm{14,125.87}\times 0.75\times 73.8$$=\mathrm{14,073.6}\text{\hspace{0.17em}BTU/h}$

8. Add the four parts. A heated space below loses no heat through the floor

   $\mathrm{5,876.34}+\mathrm{6,740.51}+\mathrm{5,295.84}+\mathrm{14,073.6}$$=\mathrm{31,986.29}\text{\hspace{0.17em}BTU/h}$

9. With no ducts, the heat goes straight into the room

   $\text{heat needed}$$=\mathrm{31,986.29}\text{\hspace{0.17em}BTU/h}$

10. Change to kilowatts (1 kW = 3,412.14 BTU/h)

    $\frac{\mathrm{31,986.29}}{\mathrm{3,412.14}}$$=9.374\text{\hspace{0.17em}kW}$

11. Electric heat turns all its power into heat. Round the watts up to the next 500 W to pick a size

    $\text{9,374 W → size}$$=\text{9,500 W electric heat}$

4. Heat needed (heater output): 52,450 BTU/h

1. Change to feet and °F, the units the insulation values use (1 ft² = 0.09290304 m², 1 ft = 0.3048 m, 1 °C of difference = 1.8 °F)

   $\frac{152\text{\hspace{0.17em}m²}}{0.09290304}$$=\mathrm{1,636.11}\text{\hspace{0.17em}ft²}$

2. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $21-\left(\mathrm{-12}\right)\times 1.8$$=59.4\text{\hspace{0.17em}°F}$

3. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height, times the 50% that faces outdoors

   $4\times \sqrt{\mathrm{1,636.11}}\times 9.1864\times 0.5$$=743.16\text{\hspace{0.17em}ft²}$

4. Heat through the solid walls (85% of the wall area at R-4): area × temperature rise ÷ R-value

   $\frac{631.68\times 59.4}{4}$$=\mathrm{9,380.48}\text{\hspace{0.17em}BTU/h}$

5. Heat through windows and doors (15% of the wall area at U-1): area × U-factor × temperature rise

   $111.47\times 1\times 59.4$$=\mathrm{6,621.52}\text{\hspace{0.17em}BTU/h}$

6. Heat through the ceiling into the attic (R-11)

   $\frac{\mathrm{1,636.11}\times 59.4}{11}$$=\mathrm{8,835.02}\text{\hspace{0.17em}BTU/h}$

7. Heat carried off by air leaking in and out: 0.018 × volume × 1.5 air changes an hour × temperature rise

   $0.018\times \mathrm{15,029.92}\times 1.5\times 59.4$$=\mathrm{24,104.99}\text{\hspace{0.17em}BTU/h}$

8. Heat lost through the edge of the slab: 0.73 BTU/h for each foot of outside edge and each degree (no edge insulation)

   $0.73\times 80.9\times 59.4$$=\mathrm{3,507.89}\text{\hspace{0.17em}BTU/h}$

9. Add the parts to get the heat the house loses on a design-cold day

   $\mathrm{9,380.48}+\mathrm{6,621.52}+\mathrm{8,835.02}+\mathrm{24,104.99}+\mathrm{3,507.89}$$=\mathrm{52,449.9}\text{\hspace{0.17em}BTU/h}$

10. The ducts are inside the heated space, so any heat they lose stays in the house

    $\text{heat needed}$$=\mathrm{52,449.9}\text{\hspace{0.17em}BTU/h}$

11. Change to kilowatts (1 kW = 3,412.14 BTU/h)

    $\frac{\mathrm{52,449.9}}{\mathrm{3,412.14}}$$=15.372\text{\hspace{0.17em}kW}$

12. A furnace that is 65% efficient must burn more fuel than the heat it delivers, so divide by its efficiency

    $\frac{\mathrm{52,449.9}}{0.65}$$=\mathrm{80,692.16}\text{\hspace{0.17em}BTU/h}$

13. Round the input up to the next 10,000 BTU/h to pick a size

    $\text{size}$$=\text{90,000 BTU/h furnace (input)}$

5. Heat needed (heater output): 99,134 BTU/h

1. Change to feet and °F, the units the insulation values use (1 ft² = 0.09290304 m², 1 ft = 0.3048 m, 1 °C of difference = 1.8 °F)

   $\frac{180\text{\hspace{0.17em}m²}}{0.09290304}$$=\mathrm{1,937.5}\text{\hspace{0.17em}ft²}$

2. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $24-\left(\mathrm{-8}\right)\times 1.8$$=57.6\text{\hspace{0.17em}°F}$

3. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height, times the 50% that faces outdoors

   $4\times \sqrt{\mathrm{1,937.5}}\times 9.1864\times 0.5$$=808.71\text{\hspace{0.17em}ft²}$

4. Heat through the solid walls (85% of the wall area at R-4): area × temperature rise ÷ R-value

   $\frac{687.41\times 57.6}{4}$$=\mathrm{9,898.65}\text{\hspace{0.17em}BTU/h}$

5. Heat through windows and doors (15% of the wall area at U-1): area × U-factor × temperature rise

   $121.31\times 1\times 57.6$$=\mathrm{6,987.28}\text{\hspace{0.17em}BTU/h}$

6. Heat through the ceiling into the attic (R-11)

   $\frac{\mathrm{1,937.5}\times 57.6}{11}$$=\mathrm{10,145.47}\text{\hspace{0.17em}BTU/h}$

7. Heat carried off by air leaking in and out: 0.018 × volume × 1.5 air changes an hour × temperature rise

   $0.018\times \mathrm{17,798.59}\times 1.5\times 57.6$$=\mathrm{27,680.37}\text{\hspace{0.17em}BTU/h}$

8. Heat lost down through the floor (R-4). The full temperature rise is used, because a vented crawlspace can get nearly as cold as outdoors

   $\frac{\mathrm{1,937.5}\times 57.6}{4}$$=\mathrm{27,900.06}\text{\hspace{0.17em}BTU/h}$

9. Add the parts to get the heat the house loses on a design-cold day

   $\mathrm{9,898.65}+\mathrm{6,987.28}+\mathrm{10,145.47}+\mathrm{27,680.37}+\mathrm{27,900.06}$$=\mathrm{82,611.83}\text{\hspace{0.17em}BTU/h}$

10. Ducts in an unheated attic, crawlspace or garage leak and lose heat on the way. Add 20%, a rule of thumb, to cover that

    $\mathrm{82,611.83}\times 1.2$$=\mathrm{99,134.19}\text{\hspace{0.17em}BTU/h}$

11. Change to kilowatts (1 kW = 3,412.14 BTU/h)

    $\frac{\mathrm{99,134.19}}{\mathrm{3,412.14}}$$=29.053\text{\hspace{0.17em}kW}$

12. Electric heat turns all its power into heat. Round the watts up to the next 500 W to pick a size

    $\text{29,053 W → size}$$=\text{29,500 W electric heat}$

6. Heat needed (heater output): 28,150 BTU/h

1. Change to feet and °F, the units the insulation values use (1 ft² = 0.09290304 m², 1 ft = 0.3048 m, 1 °C of difference = 1.8 °F)

   $\frac{140\text{\hspace{0.17em}m²}}{0.09290304}$$=\mathrm{1,506.95}\text{\hspace{0.17em}ft²}$

2. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $21-\left(\mathrm{-11}\right)\times 1.8$$=57.6\text{\hspace{0.17em}°F}$

3. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height, times the 25% that faces outdoors

   $4\times \sqrt{\mathrm{1,506.95}}\times 5.9055\times 0.25$$=229.25\text{\hspace{0.17em}ft²}$

4. Heat through the solid walls (85% of the wall area at R-4): area × temperature rise ÷ R-value

   $\frac{194.86\times 57.6}{4}$$=\mathrm{2,806}\text{\hspace{0.17em}BTU/h}$

5. Heat through windows and doors (15% of the wall area at U-1): area × U-factor × temperature rise

   $34.39\times 1\times 57.6$$=\mathrm{1,980.71}\text{\hspace{0.17em}BTU/h}$

6. Heat through the ceiling into the attic (R-11)

   $\frac{\mathrm{1,506.95}\times 57.6}{11}$$=\mathrm{7,890.92}\text{\hspace{0.17em}BTU/h}$

7. Heat carried off by air leaking in and out: 0.018 × volume × 1.5 air changes an hour × temperature rise

   $0.018\times \mathrm{8,899.3}\times 1.5\times 57.6$$=\mathrm{13,840.19}\text{\hspace{0.17em}BTU/h}$

8. Heat lost through the edge of the slab: 0.73 BTU/h for each foot of outside edge and each degree (no edge insulation)

   $0.73\times 38.82\times 57.6$$=\mathrm{1,632.28}\text{\hspace{0.17em}BTU/h}$

9. Add the parts to get the heat the house loses on a design-cold day

   $\mathrm{2,806}+\mathrm{1,980.71}+\mathrm{7,890.92}+\mathrm{13,840.19}+\mathrm{1,632.28}$$=\mathrm{28,150.1}\text{\hspace{0.17em}BTU/h}$

10. With no ducts, the heat goes straight into the room

    $\text{heat needed}$$=\mathrm{28,150.1}\text{\hspace{0.17em}BTU/h}$

11. Change to kilowatts (1 kW = 3,412.14 BTU/h)

    $\frac{\mathrm{28,150.1}}{\mathrm{3,412.14}}$$=8.25\text{\hspace{0.17em}kW}$

12. Electric heat turns all its power into heat. Round the watts up to the next 500 W to pick a size

    $\text{8,250 W → size}$$=\text{8,500 W electric heat}$

7. Heat needed (heater output): 10,880 BTU/h

1. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $66-12$$=54\text{\hspace{0.17em}°F}$

2. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height

   $4\times \sqrt{\mathrm{1,400}}\times 8$$=\mathrm{1,197.33}\text{\hspace{0.17em}ft²}$

3. Heat through the solid walls (85% of the wall area at R-21): area × temperature rise ÷ R-value

   $\frac{\mathrm{1,017.73}\times 54}{21}$$=\mathrm{2,617.02}\text{\hspace{0.17em}BTU/h}$

4. Heat through windows and doors (15% of the wall area at U-0.3): area × U-factor × temperature rise

   $179.6\times 0.3\times 54$$=\mathrm{2,909.51}\text{\hspace{0.17em}BTU/h}$

5. Heat through the ceiling into the attic (R-49)

   $\frac{\mathrm{1,400}\times 54}{49}$$=\mathrm{1,542.86}\text{\hspace{0.17em}BTU/h}$

6. Heat carried off by air leaking in and out: 0.018 × volume × 0.35 air changes an hour × temperature rise

   $0.018\times \mathrm{11,200}\times 0.35\times 54$$=\mathrm{3,810.24}\text{\hspace{0.17em}BTU/h}$

7. Add the four parts. A heated space below loses no heat through the floor

   $\mathrm{2,617.02}+\mathrm{2,909.51}+\mathrm{1,542.86}+\mathrm{3,810.24}$$=\mathrm{10,879.63}\text{\hspace{0.17em}BTU/h}$

8. The ducts are inside the heated space, so any heat they lose stays in the house

   $\text{heat needed}$$=\mathrm{10,879.63}\text{\hspace{0.17em}BTU/h}$

9. Change to kilowatts (1 kW = 3,412.14 BTU/h)

   $\frac{\mathrm{10,879.63}}{\mathrm{3,412.14}}$$=3.189\text{\hspace{0.17em}kW}$

10. A furnace that is 80% efficient must burn more fuel than the heat it delivers, so divide by its efficiency

    $\frac{\mathrm{10,879.63}}{0.8}$$=\mathrm{13,599.54}\text{\hspace{0.17em}BTU/h}$

11. Round the input up to the next 10,000 BTU/h to pick a size

    $\text{size}$$=\text{20,000 BTU/h furnace (input)}$

8. Heat needed (heater output): 21,992 BTU/h

1. Change to feet and °F, the units the insulation values use (1 ft² = 0.09290304 m², 1 ft = 0.3048 m, 1 °C of difference = 1.8 °F)

   $\frac{120\text{\hspace{0.17em}m²}}{0.09290304}$$=\mathrm{1,291.67}\text{\hspace{0.17em}ft²}$

2. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $17-\left(\mathrm{-13}\right)\times 1.8$$=54\text{\hspace{0.17em}°F}$

3. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height, times the 50% that faces outdoors

   $4\times \sqrt{\mathrm{1,291.67}}\times 8.8583\times 0.5$$=636.73\text{\hspace{0.17em}ft²}$

4. Heat through the solid walls (85% of the wall area at R-13): area × temperature rise ÷ R-value

   $\frac{541.22\times 54}{13}$$=\mathrm{2,248.14}\text{\hspace{0.17em}BTU/h}$

5. Heat through windows and doors (15% of the wall area at U-0.5): area × U-factor × temperature rise

   $95.51\times 0.5\times 54$$=\mathrm{2,578.75}\text{\hspace{0.17em}BTU/h}$

6. Heat through the ceiling into the attic (R-30)

   $\frac{\mathrm{1,291.67}\times 54}{30}$$=\mathrm{2,325}\text{\hspace{0.17em}BTU/h}$

7. Heat carried off by air leaking in and out: 0.018 × volume × 0.75 air changes an hour × temperature rise

   $0.018\times \mathrm{11,441.95}\times 0.75\times 54$$=\mathrm{8,341.18}\text{\hspace{0.17em}BTU/h}$

8. Heat lost through the edge of the slab: 0.73 BTU/h for each foot of outside edge and each degree (no edge insulation)

   $0.73\times 71.88\times 54$$=\mathrm{2,833.49}\text{\hspace{0.17em}BTU/h}$

9. Add the parts to get the heat the house loses on a design-cold day

   $\mathrm{2,248.14}+\mathrm{2,578.75}+\mathrm{2,325}+\mathrm{8,341.18}+\mathrm{2,833.49}$$=\mathrm{18,326.58}\text{\hspace{0.17em}BTU/h}$

10. Ducts in an unheated attic, crawlspace or garage leak and lose heat on the way. Add 20%, a rule of thumb, to cover that

    $\mathrm{18,326.58}\times 1.2$$=\mathrm{21,991.89}\text{\hspace{0.17em}BTU/h}$

11. Change to kilowatts (1 kW = 3,412.14 BTU/h)

    $\frac{\mathrm{21,991.89}}{\mathrm{3,412.14}}$$=6.445\text{\hspace{0.17em}kW}$

12. Electric heat turns all its power into heat. Round the watts up to the next 500 W to pick a size

    $\text{6,445 W → size}$$=\text{6,500 W electric heat}$

9. Heat needed (heater output): 7,888 BTU/h

1. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $64-8$$=56\text{\hspace{0.17em}°F}$

2. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height, times the 25% that faces outdoors

   $4\times \sqrt{\mathrm{1,500}}\times 7\times 0.25$$=271.11\text{\hspace{0.17em}ft²}$

3. Heat through the solid walls (85% of the wall area at R-21): area × temperature rise ÷ R-value

   $\frac{230.44\times 56}{21}$$=614.51\text{\hspace{0.17em}BTU/h}$

4. Heat through windows and doors (15% of the wall area at U-0.3): area × U-factor × temperature rise

   $40.67\times 0.3\times 56$$=683.19\text{\hspace{0.17em}BTU/h}$

5. Heat through the ceiling into the attic (R-49)

   $\frac{\mathrm{1,500}\times 56}{49}$$=\mathrm{1,714.29}\text{\hspace{0.17em}BTU/h}$

6. Heat carried off by air leaking in and out: 0.018 × volume × 0.35 air changes an hour × temperature rise

   $0.018\times \mathrm{10,500}\times 0.35\times 56$$=\mathrm{3,704.4}\text{\hspace{0.17em}BTU/h}$

7. Heat lost through the edge of the slab: 0.54 BTU/h for each foot of outside edge and each degree (insulated edge)

   $0.54\times 38.73\times 56$$=\mathrm{1,171.19}\text{\hspace{0.17em}BTU/h}$

8. Add the parts to get the heat the house loses on a design-cold day

   $614.51+683.19+\mathrm{1,714.29}+\mathrm{3,704.4}+\mathrm{1,171.19}$$=\mathrm{7,887.58}\text{\hspace{0.17em}BTU/h}$

9. With no ducts, the heat goes straight into the room

   $\text{heat needed}$$=\mathrm{7,887.58}\text{\hspace{0.17em}BTU/h}$

10. Change to kilowatts (1 kW = 3,412.14 BTU/h)

    $\frac{\mathrm{7,887.58}}{\mathrm{3,412.14}}$$=2.312\text{\hspace{0.17em}kW}$

11. A furnace that is 70% efficient must burn more fuel than the heat it delivers, so divide by its efficiency

    $\frac{\mathrm{7,887.58}}{0.7}$$=\mathrm{11,267.98}\text{\hspace{0.17em}BTU/h}$

12. Round the input up to the next 10,000 BTU/h to pick a size

    $\text{size}$$=\text{20,000 BTU/h furnace (input)}$

10. Heat needed (heater output): 7,790 BTU/h

1. Find how much warmer you want it inside than outside. This temperature rise drives every heat loss below

   $50-8$$=42\text{\hspace{0.17em}°F}$

2. Picture the space as a square box. Its outside wall area is the distance around (4 × √area) times the height, times the 50% that faces outdoors

   $4\times \sqrt{\mathrm{1,800}}\times 8\times 0.5$$=678.82\text{\hspace{0.17em}ft²}$

3. Heat through the solid walls (85% of the wall area at R-21): area × temperature rise ÷ R-value

   $\frac{577\times 42}{21}$$=\mathrm{1,154}\text{\hspace{0.17em}BTU/h}$

4. Heat through windows and doors (15% of the wall area at U-0.3): area × U-factor × temperature rise

   $101.82\times 0.3\times 42$$=\mathrm{1,282.97}\text{\hspace{0.17em}BTU/h}$

5. Heat through the ceiling into the attic (R-49)

   $\frac{\mathrm{1,800}\times 42}{49}$$=\mathrm{1,542.86}\text{\hspace{0.17em}BTU/h}$

6. Heat carried off by air leaking in and out: 0.018 × volume × 0.35 air changes an hour × temperature rise

   $0.018\times \mathrm{14,400}\times 0.35\times 42$$=\mathrm{3,810.24}\text{\hspace{0.17em}BTU/h}$

7. Add the four parts. A heated space below loses no heat through the floor

   $\mathrm{1,154}+\mathrm{1,282.97}+\mathrm{1,542.86}+\mathrm{3,810.24}$$=\mathrm{7,790.07}\text{\hspace{0.17em}BTU/h}$

8. The ducts are inside the heated space, so any heat they lose stays in the house

   $\text{heat needed}$$=\mathrm{7,790.07}\text{\hspace{0.17em}BTU/h}$

9. Change to kilowatts (1 kW = 3,412.14 BTU/h)

   $\frac{\mathrm{7,790.07}}{\mathrm{3,412.14}}$$=2.283\text{\hspace{0.17em}kW}$

10. Electric heat turns all its power into heat. Round the watts up to the next 500 W to pick a size

    $\text{2,283 W → size}$$=\text{2,500 W electric heat}$