Thermal Resistance

This problem set deals with thermal resistance.

1. Thermal resistance is the opposition to flow of heat energy.

2. A defining equation for thermal resistance is RT = DT/QH.

3. What is the thermal resistance if you know the thermal conductivity, k, the thickness, L, and the surface area, A? (Note: the capital "L" should be lowercase and cursive)

RT = L/(k·A)

4. The thermal conductivity for a given material is a constant value. It depends on (b.) type of material.

5. Thermal resistance in English units may be expressed as F°/(BTU/hr).

6. A building has a thermal resistance of 0.05 C°/(kcal/hr). What is the building's heat flow rate when the temperature difference between inside and outside is 30 C°?
 Knowns Unknowns Equations/Remarks RT = 0.05 C°/(kcal/hr) DT = 30 C° QH = ? RT = DT/QH RT = DT/QH QH = DT/RT = (30 C°)/(0.05 C°/(kcal/hr)) QH = 600 kcal/hr Rearrange the equation to solve for the heat flowrate, then substitute in the known values and calculate the answer.

7. In the equation for heat flow rate, QH = kADT/L, increasing the thermal conductivity, area, or temperature difference will increase heat flow rate. (Note: the capital "L" should be lowercase and cursive)

8. In the equation for heat flow rate, QH = kADT/L, increasing the thickness will decrease the heat flow rate. (Note: the capital "L" should be lowercase and cursive)

9. The higher the R-value rating for a thermal material is, the stronger its insulating effect is.

10. Find the thermal resistance, RT, of the plate, given that it is 0.25 inches thick, has an area of 2 square feet, and is made of copper. Use the thermal conductivity given in Table II-1.

 Knowns Unknowns Equations/Remarks L = 0.25 in A = 2 ft2 k = 2784 (BTU·in)/(hr·ft2·F°) RT = ? RT = L/(k·A) RT = L/(k·A) = (0.25 in)/[(2784 BTU·in/hr·ft2·F°)(2 ft2)] RT = 4.49 ×10-5 F°/(BTU/hr) Substitute in the known values and calculate the answer.

11. Find the heat flow rate, QH, through the plate if DT across the plate is 40 F°. (Refer to problem 10).

 Knowns Unknowns Equations/Remarks DT = 40 F° RT = 4.49 ×10-5 F°/(BTU/hr) QH = ? RT = DT/QH RT = DT/QH QH = DT/RT = (40 F°)/(4.49 ×10-5 F°/(BTU/hr)) QH = 891,000 BTU/hr Rearrange the equation to solve for the heat flowrate, then substitute in the known values and calculate the answer.

12. The temperature difference between the inside and outside of an electric oven is 300 F°. The heating element must supply 1200 BTU/hr of heat energy to the oven to overcome the heat flow out of the oven. Find the thermal resistance of the oven walls.

 Knowns Unknowns Equations/Remarks DT = 300 F° QH = 1200 BTU/hr RT = ? RT = DT/QH RT = DT/QH = (300 F°)/(1200 BTU/hr)RT = 0.25 F°/(BTU/hr) Substitute in the known values and calculate the answer.

13. The thermal resistance of a plate of glass is 0.006 F°/(BTU/hr). 9600 BTU of heat energy flow through the plate glass every hour. Find the temperature difference across the glass that is causing the flow of heat energy.

 Knowns Unknowns Equations/Remarks RT = 0.006 F°/(BTU/hr) QH = 9600 BTU/hr DT = ? RT = DT/QH RT = DT/QH DT = RT · QH = (0.006 F°/BTU/hr)(9600 BTU/hr) DT = 57.6 F° Rearrange the equation to solve for the temperature difference, substitute in the known values and calculate the answer.

14. The thermal resistance of a 16 ft2 piece of 0.5 inch thick aluminum is 19.1 ×10-6 F°/(BTU/hr). Find the thermal conductivity for the aluminum plate.

 Knowns Unknowns Equations/Remarks RT = 19.1 ×10-6 F°/(BTU/hr) L = 0.5 in A = 16 ft2 k = ? RT = L/(k·A) RT = L/(k·A) k  = L/(RT·A) = (0.5 in)/[(19.1 ×10-6 F°/BTU/hr)(16 ft2)] k  = 1636 (BTU·in)/(hr·ft2·F°) Rearrange the equation to solve for the thermal conductivity, substitute in the known values and calculate the answer.

15. A firefighter's suit has a thermal resistance of 50 C°/(kcal/hr). When exposed to a fire, the temperature difference from outside to inside the suit may be as high as 1000 C°. Find the heat flow rate through the suit at 1000 C°.

 Knowns Unknowns Equations/Remarks DT = 1000 C° RT = 50 C°/(kcal/hr) QH = RT = DT/QH RT = DT/QH QH = DT/RT = (1000 C°)/(50 C°/kcal/hr) QH = 20 kcal/hr Rearrange the equation to solve for the heat flowrate, then substitute in the known values and calculate the answer.

16. Given the conditions of problem 14 and a constant temperature difference of 50 F° across the material,
1. find the heat flow rate using the equation QH = kADT/L, (Note: the capital "L" should be lowercase and cursive)
2. Is there an easier way to do this? If so solve for QH the easiest way.

 Knowns Unknowns Equations/Remarks RT = 19.1 ×10-6 F°/(BTU/hr) L = 0.5 in A = 16 ft2 DT = 50 F° k  = 1636 (BTU·in)/(hr·ft2·F°) a. QH = ? b. QH = ? by the easiest way QH = kADT/L  RT = DT/QH a. QH = kADT/L      QH = (1636 BTU·in/hr·ft2·F°)(16 ft2)(50 F°)/(0.5 in)     QH = 2,620,000 BTU/hr b. RT = DT/QH      QH = DT/RT = (50 F°)/(19.1 ×10-6 F°/BTU/hr)     QH = 2,620,000 BTU/hr Substitute in the known values and calculate the answer.The easier way is to use the thermal resistance equation.

17.
1. Using the table of thermal conductivity (Table II-1), find the thermal resistance of a 36" x 40" windowpane which is 0.125 in thick.
2. If the inside of the window is at 70 °F and the outside window surface is at 65 °F, find the heat flow rate through the window.
3. How many BTU's will escape through the window in 8 hours?

 Knowns Unknowns Equations/Remarks k  = 5.8 (BTU·in)/(hr·ft2·F°) W = 36 in H = 40 in L = 0.125 in T1 = 70 °F T2 = 65 °F t = 8 hrs a. RT = ? b. QH = ? c. H = ? RT = L/(k·A)  A = W·H DT = T1 – T2  RT = DT/QH  QH = H/t a. A = W·H = (36 in)(40 in)(1 ft2/144 in2) = 10 ft2     RT = L/(k·A) = (0.125 in)/[(5.8 BTU·in/hr·ft2·F°)(10 ft2)]     RT = 2.16 ×10-3 F°/(BTU/hr) b. DT = T1 – T2 = 70 °F – 65 °F = 5 F°     RT = DT/QH      QH = DT/RT = (5 F°)/(2.155 ×10-3 F°/BTU/hr)     QH = 2320 BTU/hr c. QH = H/t      H = QH·t = (2320 BTU/hr)(8 hr)     H = 18,600 BTU Find the area of the window and use it along with the thickness and thermal conductivity to find the thermal resistance. Notice that the results do not match those in the book. The possible cause could be a different k.Find the temperature difference across the window, then use it with the thermal resistance to find the heat flowrate. Solve the next equation for H, substitute in the numbers, and calculate the answer. Note that all three answers differ from those in the book.

18. A container is filled with 800 milliliters of water at 95 °C. After sitting for 40 minutes in a room whose temperature is 25 °C, the water has cooled to 80 °C.
1. Find the temperature change of the water, DT = Tf - Ti. What is the meaning of a negative temperature change?
2. How many calories did the water lose?
3. Find the heat flow rate in cal/sec.
4. What was the average temperature of the water as it cooled?
5. What was the temperature difference between the room and the average water temperature? (DT = Tav - Troom)
6. Find the thermal resistance of the container.
7. Explain the difference in meanings for the DT in part a and part e above.

 Knowns Unknowns Equations/Remarks V = 800 mL t = 40 min Ti = 95 °C Tf = 80 °C Troom = 25 °C c = 1.0 cal/g·C° r = 1.0 g/cm3 1 cm3 = 1 mL 1 min = 60 sec a. DT = ? b. H = ? c. QH = ? d. Tav = ? e. DT = ? f. RT = ? g. Explain the different DT's DT = Tf – Ti  r = m/V  H = mcDT  QH = H/t  Tav = (Tf + Ti)/2 DT = Tav - Troom  RT = DT/QH a. DT = Tf – Ti = 80 °C – 95 °C = –15 C°. The negative sign means that the water cooled down.b. r = m/V      m = rV     H = mcDT = rVcDT      H = (1.0 g/cm3)(800 cm3)(1.0 cal/g·C°)(|–15 C°|)     H = 12,000 cal c. QH = H/t = (12000 cal)/[(40 min)(60 sec/1 min)]     QH = 5 cal/s d. Tav = (Tf + Ti)/2 = (80 °C + 95 °C)/2     Tav = 87.5 °C e. DT = Tav - Troom = 87.5 °C – 25 °C      DT = 62.5 C° f. RT = DT/QH = (62.5 C°)/(5 cal/s)     RT = 12.5 C°/(cal/s) g. The DT in part (a) is the change in temperature of the water (one substance). The DT in part (e) is the difference between the water and the air in the room (two substances). It is this DT that drives the flow of heat. Find the change in temperature of the water.Use the density equation to find the mass of the water from the given volume and subsitute that in the heat equation. Use the absolute value of the temperature difference since we know that the water lost the heat. For the next few parts, substitute in the known values and calculate the answers. Convert the minutes to seconds.

19. Using the same process as above, find the thermal resistance of a container which holds 5 gallons of water at 200 °F if the water cools to 170 °F in 15 minutes. The room temperature is 75 °F.

 Knowns Unknowns Equations/Remarks V = 5 gal t = 15 min Ti = 200 °F Tf = 170 °F Troom = 75 °F c = 1.0 BTU/lb·F° rw = 62.4 lb/ft3 1 ft3 = 7.48 gal 1 hr = 60 min RT = ? DT = Tf – Ti  rw = w/V  H = wcDT  QH = H/t  Tav = (Tf + Ti)/2 DT = Tav - Troom  RT = DT/QH DT = Tf – Ti = 170 °F – 200 °F = –30 F°rw = w/V  w = rw·V  H = wcDT = rwVcDT  H = (62.4 lb/ft3)(5 gal)(1 ft3/7.48 gal)(1.0 BTU/lb·F°)(|–30 F°|) H = 1251 BTU QH = H/t = (1251 BTU)/[(15 min)(1 hr/60 min)] QH = 5004 BTU/hr Tav = (Tf + Ti)/2 = (170 °F + 200 °F)/2 Tav = 185 °F DT = Tav - Troom = 185 °F – 75 °F  DT = 110 F° RT = DT/QH = (110 F°)/(5004 BTU/hr) RT = 0.0220 F°/(BTU/hr) Find the change in water temperature (negative means the water cooled down).Find the amount of heat transferred out of the water. Use the absolute value of the temperature change since we know that the heat is going out of the water. Go through the next few steps, substituting in the known values and calculating the results. Convert minutes to hours since BTU/hr is a common unit. Finally we get to the end result.