• Author / Uploaded
• Raymond Fuentes

• Categories
• Transferencia de calor
• Aislamiento térmico
• Paneles de yeso
• Conductividad térmica
• Calor

Citation preview

3-26.Consider a person standing in a room at 20°C with an exposed surface area of 1.7 m2. The deep body temperature of the human body is 37°C, and the thermal conductivity of the human tissue near the skin is about 0.3 W/m·K. The body is losing heat at a rate of 150 W by natural convection and radiation to the surroundings. Taking the body temperature 0.5 cm beneath the skin to be 37°C, determine the skin temperature of the person.

Answer:



Step 1 of 2

Given data: Room temperature Surface area Thermal conductivity of the tissue near the skin A person is dissipating heat Body temperature 

Step 2 of 2

If steady operating conditions exist, the heat transfer coefficient is constant and uniform over the entire exposed surface of the person and heat generation within the 0.5-cm thick outer layer of the tissue is negligible then The skin temperature can be determined directly from

The skin temperature of the person

3-27.An aluminum pan whose thermal conductivity 237 W/m∙K has a flat bottom with diameter 15 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in the pan through its bottom at a rate of 1400 W. If the inner surface of the bottom of the pan is at I05°C, determine the temperature of the outer surface of the bottom of the pan.

Answer: 

Step 1 of 2

Given data: Thermal conductivity of the aluminum, Diameter of the pan, Thickness, Total heat transfer rate through bottom, Temperature of inner surface of bottom of pan,



Step 2 of 2

Area of the pan,

Under steady state condition, The rate of heat transfer through the bottom of the pan by conduction is given by

Therefore, the temperature of outer surface of bottom of pan is

.

3-28. A wall is constructed of two layers of 0.6-in-thick sheetrock (k= 0.10 Btu/h·ft·°F), which is a plasterboard made of two layers of heavy paper separated by a layer of gypsum, placed 7 in apart. The space between the sheetrocks is filled with fiberglass insulation (k = 0.020 Btu/h·ft·°F), Determine (a) the thermal resistance of the wall and (b) its R-value of insulation in English units.

Answer:



Step 1 of 4

Draw schematic diagram of a plaster board wall.



Step 2 of 4

(a) Draw thermal resistance circuit of the wall.



Step 3 of 4

Calculate the thermal resistance of the wall.

Here, thermal resistance of the sheet rock on left edge is thermal resistance of the sheet rock on right edge is thickness of the fiber glass is

, thermal resistance of the fiber glass is

, thickness of the sheet rock on left edge is

, thickness of the sheet rock on right edge is

conductivity of the sheet rock on left edge is

, the thermal conductivity of the fiber glass is

the thermal conductivity of the sheet rock on the right edge Substitute

for

,

for

,

for

,

for

, and

Thus, the total thermal resistance of the wall is 

, the thermal

Step 4 of 4

(b) Calculate the thermal resistance of the insulation.

.

, for

for .

.

, and

, ,

Substitute

for

and

for

.

3-29.Thus, the thermal resistance of the insulation is The roof of a house consists of a 15-cm-thick concrete slab (k =2 W/m·K) that is 15 m wide and 20 m long. The convection heat transfer coefficient on the inner surface of the roof is 5 W/m2·K. On a clear winter night, the ambient air is reported to be at 10°C, while the night sky temperature is 100 K. The house and the interior surfaces of the wall are maintained at a constant temperature of 20°C. The emissivity of both surfaces of the concrete roof is 0.9. Considering both radiation and convection heat transfer, determine the rate of heat transfer through the roof when wind at 60 km/h is blowing over the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the price of natural gas is $1.20/therm, determine the money lost through the roof that night during a 14-h period.

Answer:



Step 1 of 6

Draw the schematic diagram of the flat plate collector.

Comment 

Step 2 of 6

Consider the following assumptions: 1. Steady operating conditions exist 2. The surfaces are opaque, diffuse, and gray. 3. Air is an ideal gas with constant properties. Obtain the following properties from the table A-15 “Properties of air at 1 atm pressure” and at the film

temperature of Thermal conductivity, Kinematic viscosity, Prandtle number,

.

Volume expansion coefficient, For

, we have horizontal rectangular enclosure. The characteristic length in this case is the

distance between the two glasses Comment 

Step 3 of 6

Calculate the Rayleigh number using the following equation.

Substitute

for

,

and

for

.

for

,

for

,

The Nusselt number correlation for the Natural convention is given by

Comment 

Step 4 of 6

Calculate the surface area as follows:

for

,

for

Here,

is the height of the collector, W is the width of the collector.

Substitute

and

.

Calculate the convection heat transfer coefficient as follows:

Here, k is the thermal conductivity. Comment 

Step 5 of 6

Calculate the rate of heat transfer due to natural convection using the following equation.

Substitute and

for k, for

,

for

,

for

,

for

.

Therefore, the rate of heat loss from the absorber plate by natural convection is

.

Comment 

Step 6 of 6

Calculate the rate of heat transfer due to radiation using the following equation. Here, is the emissivity of the absorber plate, Stephen’s Boltzmann’s constant.

is the emissivity of the glass surface,

is the

Substitute and

for for

,

for

,

for

,

for

,

for

.

Therefore, the rate of heat loss from the absorber plate by radiation is

.

3-30. A 2-m × 1.5-m section of wall of an industrial furnace burning natural gas is not insulated, and the temperature at the outer surface of this section is measured to be 110°C. The temperature of the furnace room is 32°C, and the combined convection and radiation heat transfer coefficient at the surface of the outer furnace is 10 W/m2·K. It is proposed to insulate this section of the furnace wall with glass wool insulation (k = 0.038 W/m·K) in order to reduce the heat loss by 90 percent, Assuming the outer surface temperature of the metal section still remains at about 110°C, determine the thickness of the insulation that needs to be used. The furnace operates continuously and has an efficiency of 78 percent. The price of the natural gas is $1.10/therm (1 therm = 105,500 kJ of energy content). If the installation of the insulation will cost $250 for materials and labor, determine how long it will take for the insulation to pay for itself from the energy it saves.

Answer: 

Step 1 of 7

Consider the Diagram for thermal network.

Comment 

Step 2 of 7

Without having any insulation the rate of heat transfer can be determined as

Here, Combined Convection and radiation heat transfer coefficient is h. Comment 

Step 3 of 7

Substitute

for h.

Comment 

Step 4 of 7

In order to reduce heat loss by 90%, the new heat transfer rate is as follows:

Calculate the thermal resistance.

Here, the difference in temperature is Substitute

.

for

Comment 

Step 5 of 7

Calculate the thickness of the thermal resistance as follows:

Here, k is the thermal conductivity and Area is A. Substitute

for k,

for A,

Therefore, thickness of the insulation is Comment

for

.

.



Step 6 of 7

The amount of heat transfer rate saved by the insulation is follows:

of the total heat transfer rate is as

Comments (1) 

Step 7 of 7

Since the furnace operates continuously, the number of hours per year is the efficiency if the furnace is

, the amount of energy saved per year is as follows:

Amount of energy saved in terms of therms is as follows:

Calculate the money saved as follows:

Calculate the time for which insulation stays.

Therefore, the time taken for payoff is

.

and