| ToC, Chapter 17 |Course Calendar|

**D17.1** A goose-down sleeping bag has a surface area of 2.25 m^{2}
and is filled with a layer of down 5 cm thick. What is the heat-transfer rate
(H_{cd}) through it from a person with skin temperature of 35°C
to the outside air at -5°C. How does this rate compare with the body’s
minimal metabolism rate, about 100 W?

The thermal conductivity of goose down isK = 0.023 W/m C°

(Hmmm, . . . go ahead and use e = 1.00, even in this case).

The first part of this questions requires exactly the same reasoning, using exactly the same kind of diagrams, as the previous question..

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CHANGE NEEDED

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D17.5Calculate the R value of an insulated wall like that in Figure 13.30 if 2 x 6’s are used for wall studs instead of 2 x 4’s. The larger board allow 5.5 in of fiberglass insulation instead of 3.5 in.By what percentage does this increase in thickness change the R value?

How does the increased thickness change the heat loss calculated in Example 13.14?

The R-values for the components of the wall in Figure 13.30 are given in Table 13.4, on page 481. By increasing the fiberglass insulation from 3.5 inches to 5.5 inches, the R-value for this insulation increases from 10.90 to 18.80, according to Table 13.3, on page 480. This increases the total R-value from 14.33 to22.23.

This is an increase of^{7.9}/_{14.33}= 0.55 =55%for the R-value of the wall.

In terms of R-values, the conduction heat transfer rate is given by equation 13.9,H _{cd}= A (T_{2}- T_{1}) / R = (^{1}/_{R}) [A (T_{2}- T_{1})]In Example 13.14, we found the conduction heat transfer rate to be

H _{cd}(un) = 3.8 kWfor the

uninsulated wall,H _{cd}(3.5) = 1.2 kWfor the insulated wall with 2x4’s and 3.5 in of insulation

Now, with 2x6’s and 5.5 in of insulation,H _{cd}(5.5) = (^{1}/_{22.23})[(960)(60)] = 2591 Btu/h = 0.76 kWThis is a reduction of heat flow by

H = 1.18 kW - 0.76 kW = 0.42 kW

H / H =^{0.42 kW}/_{1.18 kW}= 0.36 =36%

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D17.7If concrete roadway sections are poured butted against each other when the temperature is 15°C, what will the thermal stress be when the temperature reaches 40°C? Young’s modulus for concrete is about 20 x 10^{9}Pa.

D17.8If 0.60 kg of boiling hot coffee is poured into a 0.250-kg steel camping cup initially at 20°C, what is the final temperature of the cup-coffee system?

D17.9A 0.050-kg ice cube, initially at - 5.0°C, is placed in 0.30 kg of water at 25°C. What is the final temperature of the water? Or, if melting is not complete, how much ice remains unmelted at thermal equilibrium? Assume no heat is lost to the environment.

D17.10A 0.15-kg ice cube initially at -15°C is placed in 0.30 kg of water at 25°C. What is the final temperature of the water? Or, if melting is not complete, how much ice remains unmelted at thermal equilibrium? Assume no heat is lost to the environment.

This should look very similar to the previous question or problem. Now we have a larger and colder ice cube.

D17.11The inside diameter of a steel lid and the outside diameter of a glass peanut butter jar are both exactly 11.50 cm at room temperature, 21.0°C. If the lid is stuck and you run 80°C hot water over it until the lid and the jar top both come to 80°C, what will the new diameters be?

D17.12If concrete roadway sections are poured butted against each other when the temperature is 15°C, what will the thermal stress be when the temperature reaches 40°C? Young’s modulus for concrete is about 20 x 10^{9}Pa.

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