Transferencia de calor -7th-Kreith-Sol

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(Parte 2 de 5)

Area = 6 cm × 6 cm = 36 cm2 = 0.0036 m2

Power dissipation rate of the heater (qh) = 10 W Surface temperatures Thot = 322 K Tcold = 300 K

The thermal conductivity of the sample at the mean temperature in W/(m K)

One dimensional, steady state conduction No heat loss from the edges of the apparatus

Guard Ring and InsulationSE Heater

Wattmeter

Similar Specimen

Thot Tcold http://librosysolucionarios.net

By conservation of energy, the heat loss through the two specimens must equal the power dissipation of the heater. Therefore the heat transfer through one of the specimens is qh/2. For one dimensional, steady state conduction (from Equation (1.3) qk = kAL ΔT = 2

Solving for the thermal conductivity k = 2 hq L k = 0.63 W/(mK)

In the construction of the apparatus care must be taken to avoid edge losses so all the heat generated will be conducted through the two specimens.

PROBLEM 1.5

To determine the thermal conductivity of a structural material, a large 15 cm-thick slab of the material was subjected to a uniform heat flux of 2500 W/m2, while thermocouples embedded in the wall 2.5 cm apart were read over a period of time. After the system had reached equilibrium, an operator recorded the readings of the thermocouples as shown below for two different environmental conditions.

Distance from the surface (cm) Temperature (°C)

Test 1: 0 5 10 15

Test 2: 0 5 10 15

From these data, determine an approximate expression for the thermal conductivity as a function of temperature between 40 and 208°C.

Thermal conductivity test on a large, 15 cm slab Thermocouples are embedded in the wall, 2.5 cm apart Heat flux (q/A) = 2500 W/m2 Two equilibrium conditions were recorded (shown in Table above) http://librosysolucionarios.net

An approximate expression for thermal conductivity as a function of temperature between 40 and 208°C.

ASSUMPTIONS One-dimensional conduction

5 10 15Distance (cm) Thermocouples

0 SOLUTION

The thermal conductivity can be calculated for each pair of adjacent thermocouples using the equation for one-dimensional conduction q = k A TL Δ

Solving for k k = qL ATΔ

This will give a thermal conductivity for each pair of adjacent thermocouples which are assigned to the average temperature of the pair of thermocouples. As an example, for the first pair of thermocouples in Test 1, the thermal conductivity (ko) is

The average temperature for this pair of thermocouples is

The average temperature and the thermal conductivity for all other pairs of thermocouples are given in the table below.

n (°C) Thermal Conductivity W/(m K)

1 52.5 5 2 81 3.9 3 114.5 3.57 4 112.5 3.38 5 149 3.29 6 188 3.125 http://librosysolucionarios.net

These points are displayed graphically.

Temperature (°C) k (W/(mK))

MeanV ariation of with Temperature k

We will use the best fit quadratic function to represent the relationship between thermal conductivity and temperature k (T) = a + b T + c T 2

The constants a, b, and c can be found using a least squares fit.

Let the experimental thermal conductivity at data point n be designated as kn. A least squares fit of the data can be obtained as follows

The sum of the squares of the errors is

By setting the derivatives of S (with respect to a, b, and c) equal to zero, the following equations result

N a + Σ Tnb + Σ Tn2 c = Σ kn Σ Tn a + Σ Tn2 b + Σ Tn3 c = Σ kn Tn

Σ Tn2 a + Σ Tn3 b + Σ Tn4 c = Σ kn Tn2 For this problem

Σ Tn = 697.5 Σ Tn2 = 9.263 × 104 Σ Tn3 = 1.3554 × 107 Σ Tn4 = 2.125 × 109 Σ kn = 2.41

Σ kn Tn = 2445.12

Σ kn Tn2 = 3.124 × 104 Solving for a, b, and c a = 6.9722 b = – 4.7213 × 10–2 c = 1.443 × 10–4 http://librosysolucionarios.net

Therefore the expression for thermal conductivity as a function of temperature between 40 and 208°C is k (T) = 6.9722 – 4.7213 × 10–2 T + 1.443 × 10–4 T 2 This is plotted in the following graph

Temperature (degC )

T h er m al

Condu ctivit

[W/mK] k

Note that the derived empirical expression is only valid within the temperature range of the experimental data.

PROBLEM 1.6

A square silicone chip 7 m by 7 m in size and 0.5 m thick is mounted on a plastic substrate with its front surface cooled by a synthetic liquid flowing over it. Electronic circuits in the back of the chip generate heat at a rate of 5 watts that have to be transferred through the chip. Estimate the steady state temperature difference between the front and back surfaces of the chip. The thermal conductivity of silicone is 150 W/(m K).

A 0.007 m by 0.007 m silicone chip Thickness of the chip (L) = 0.5 m = 0.0005 m

Heat generated at the back of the chip (Gq ) = 5 W The thermal conductivity of silicon (k) = 150 W/(m K)

The steady state temperature difference (ΔT)

One dimensional conduction (edge effects are negligible) The thermal conductivity is constant The heat lost through the plastic substrate is negligible

Substrate

0.5 CNIP

7mm http://librosysolucionarios.net

For steady state the rate of heat loss through the chip, given by Equation (1.3), must equal the rate of heat generation qk = AkL (ΔT) = Gq

Solving this for the temperature difference

ΔT = GL qkA

ΔT = 0.34°C PROBLEM 1.7

A warehouse is to be designed for keeping perishable foods cool prior to transportation to grocery stores. The warehouse has an effective surface area of 1860 m2 exposed to an ambient air temperature of 32°C. The warehouse wall insulation (k = 0.17 W/(m K) is 7.5 cm thick. Determine the rate at which heat must be removed from the warehouse to maintain the food at 4°C.

Cooled warehouse Effective area (A) = 1860 m2 Temperatures Outside air = 32°C

Food inside = 4°C

Thickness of wall insulation (L) = 7.5 cm Thermal conductivity of insulation (k) = 0.17 W/(m K)

FIND Rate at which heat must be removed (q)

One dimensional, steady state heat flow The food and the air inside the warehouse are at the same temperature The thermal resistance of the wall is approximately equal to the thermal resistance of the wall insulation alone

Warehouse T• = 32°C Ti =4 °Cq http://librosysolucionarios.net

The rate at which heat must be removed is equal to the rate at which heat flows into the warehouse. There will be convective resistance to heat flow on the inside and outside of the wall. To estimate the upper limit of the rate at which heat must be removed these convective resistances will be neglected. Therefore the inside and outside wall surfaces are assumed to be at the same temperature as the air inside and outside of the wall. Then the heat flow, from Equation (1.2), is q = kAL ΔT q = 118 kW PROBLEM 1.8

With increasing emphasis on energy conservation, the heat loss from buildings has become a major concern. For a small tract house the typical exterior surface areas and

R-factors (area × thermal resistance) are listed below

Element Area (m2) R-Factors = Area × Thermal Resistance [(m2 K/W)]

Walls 150 2.0 Ceiling 120 2.8 Floor 120 2.0 Windows 20 0.1 Doors 5 0.5

(a) Calculate the rate of heat loss from the house when the interior temperature is 22°C and the exterior is –5°C.

(b) Suggest ways and means to reduce the heat loss and show quantitatively the effect of doubling the wall insulation and the substitution of double glazed windows (thermal resistance = 0.2 m2 K/W) for the single glazed type in the table above.

Small house Areas and thermal resistances shown in the table above Interior temperature = 22°C Exterior temperature = –5°C

(a) Heat loss from the house (qa)

(b) Heat loss from the house with doubled wall insulation and double glazed windows (qb). Suggest improvements.

All heat transfer can be treated as one dimensional Steady state has been reached The temperatures given are wall surface temperatures Infiltration is negligible The exterior temperature of the floor is the same as the rest of the house http://librosysolucionarios.net

(a) The rate of heat transfer through each element of the house is given by Equations (1.3) and (1.34) q = thTR Δ

The total rate of heat loss from the house is simply the sum of the loss through each element:

q = ΔT wall ceiling floor windows doors

1 1 1 1 AR AR AR AR AR q = (22°C – –5°C) (75 + 42.8 + 60 + 200 + 10) W/K q = 10,500 W (b) Doubling the resistance of the walls and windows and recalculating the total heat loss: q = (22°C – –5°C) q = (22°C – –5°C) (37.5 + 42.8 + 60 + 100 + 10) W/K q = 6800 W

Doubling the wall and window insulation led to a 35% reduction in the total rate of heat loss.

Notice that the single glazed windows account for slightly over half of the total heat lost in case (a) and that the majority of the heat loss reduction in case (b) is due to the double glazed windows. Therefore double glazed windows are strongly suggested.

PROBLEM 1.9

Heat is transferred at a rate of 0.1 kW through glass wool insulation (density = 100 kg/m3) of 5 cm thickness and 2 m2 area. If the hot surface is at 70°C, determine the temperature of the cooler surface.

http://librosysolucionarios.net

Glass wool insulation with a density (ρ) = 100 kg/m3 Thickness (L) = 5 cm = 0.05 m Area (A) = 2 m2

Temperature of the hot surface (Th) = 70°C Rate of heat transfer (qk) = 0.1 kW = 100 W

The temperature of the cooler surface (Tc) ASSUMPTIONS

One dimensional, steady state conduction Constant thermal conductivity

L = 0.05 m Glass Wool

From Appendix 2, Table 1 The thermal conductivity of glass wool at 20°C (k) = 0.036 W/(m K)

SOLUTION For one dimensional, steady state conduction, the rate of heat transfer, from Equation (1.2), is qk = Ak

L (Th – Tc)

Solving this for Tc

Tc = Th – kqL Ak

Tc = 0.6°C PROBLEM 1.10

A heat flux meter at the outer (cold) wall of a concrete building indicates that the heat loss through a wall of 10 cm thickness is 20 W/m2. If a thermocouple at the inner surface of the wall indicates a temperature of 22°C while another at the outer surface shows 6°C, calculate the thermal conductivity of the concrete and compare your result with the value in Appendix 2, Table 1.

Concrete wall Thickness (L) = 100 cm = 0.1 m Heat loss (q/A) = 20 W/m2

Surface temperature Inner (Ti) = 22°C Outer (To) = 6°C http://librosysolucionarios.net

FIND The thermal conductivity (k) and compare it to the tabulated value

One dimensional heat flow through the wall Steady state conditions exist

SKETCH L = 0.1 m

To=6°CTi=22°C SOLUTION

The rate of heat transfer for steady state, one dimensional conduction, from Equation (1.2), is qk = kAL (Thot – Tcold)

Solving for the thermal conductivity k = ()kio

= 0.125 W/(m K)

This result is very close to the tabulated value in Appendix 2, Table 1 where the thermal conductivity of concrete is given as 0.128 W/(m K).

PROBLEM 1.1

Calculate the heat loss through a 1-m by 3-m glass window 7 m thick if the inner surface temperature is 20°C and the outer surface temperature is 17°C. Comment on the possible effect of radiation on your answer.

Window: 1 m by 3 m Thickness (L) = 7 m = 0.007 m

Surface temperature Inner (Ti) = 20°C Outer (To) = 17°C

The rate of heat loss through the window (q)

One dimensional, steady state conduction through the glass Constant thermal conductivity http://librosysolucionarios.net

Glass

From Appendix 2, Table 1 Thermal conductivity of glass (k) = 0.81 W/(m K)

SOLUTION The heat loss by conduction through the window is given by Equation (1.2) qk = kAL (Thot – Tcold)

(20°C – 17°C) qk = 1040 W COMMENTS

Window glass is transparent to certain wavelengths of radiation, therefore some heat may be lost by radiation through the glass.

During the day sunlight may pass through the glass creating a net heat gain through the window.

PROBLEM 1.12

If in Problem 1.1 the outer air temperature is –2°C, calculate the convective heat transfer coefficient between the outer surface of the window and the air assuming radiation is negligible.

Problem 1.1: Calculate the heat loss through a 1 m by 3 m glass window 7 m thick if the inner surface temperature is 20°C and the outer surface temperature is 17°C. Comment on the possible effect of radiation on your answer.

Window: 1 m by 3 m Thickness (L) = 7 m = 0.007 m

Surface temperatures Inner (Ti) = 20°C

Outer (To) = 17°C The rate of heat loss = 1040 W (from the solution to Problem 1.1)

The outside air temperature = –2°C

The convective heat transfer coefficient at the outer surface of the window (ch) ASSUMPTIONS The system is in steady state and radiative loss through the window is negligible http://librosysolucionarios.net

For steady state the rate of heat transfer by convection (Equation (1.10)) from the outer surface must be the same as the rate of heat transfer by conduction through the glass qc = ch A ΔT = qk Solving for ch ch = ()koq AT T∞ ch = o 1040W ch = 18.2 W/(m2 K) COMMENTS

This value for the convective heat transfer coefficient falls within the range given for the free convection of air in Table 1.4.

PROBLEM 1.13

Using Table 1.4 as a guide, prepare a similar table showing the order of magnitudes of the thermal resistances of a unit area for convection between a surface and various fluids.

Table 1.4— The order of magnitude of convective heat transfer coefficient (ch) FIND

The order of magnitudes of the thermal resistance of a unit area (A Rc) SOLUTION

The thermal resistance for convection is defined by Equation (1.14) as

Rc = 1 chA

Therefore the thermal resistances of a unit area are simply the reciprocal of the convective heat transfer coefficient

A Rc = 1 ch http://librosysolucionarios.net

(Parte 2 de 5)

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Transferencia de calor -7th-Kreith-Sol

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