Heat Transfer Solutions to first Chater

PROBLEM 1.1

A teaspoon is placed in a cup of hot coffee.

[a] Discuss the flow of heat through the spoon and show directions on a sketch.[b] What modes of heat transfer are taking place in this example?[c] What is the approximate temperature of the spoon at the coffee-air interface ?

[a] Heat flows by convection from the coffee to the surface of the submerged part of the spoon. This energy is transmitted by conduction towards the exposed part. Heat is convected from the exposed surface to the surrounding air. Heat is also transferred by radiation from the exposed surface to the surroundings.

[b] Three modes of heat transfer take place in this example: conduction, convection and radiation.

[c] There is a small temperature drop between the coffee and the submerged part of the spoon. At the coffee-air interface the temperature of the spoon is slightly lower than the coffee temperature.

PROBLEM 1.2

A hand drill is used to drill a hole in a very hard material. Because the bit is very dull it spins without advancing. After a few minutes it is observed that the bit is hot. Discuss the direction and modes of heat flow in this situation. At what location is the temperature highest?

(1) Observations. (i) Heat generated by friction at the tip of the bit is transferred by conduction through bit and the hard material. (ii) Heat leaves the surface of the bit by convection to the surrounding air. (iii) Heat loss from the surface of the bit also takes place by radiation to the surroundings.

(2) Problem Definition. Determine, qualitatively, the temperature distribution in the bit and the mode of heat transfer.

(3) Solution Plan. Identify the source of heat and direction of heat flow.

(4) Plan Execution.

(i) Assumptions. The drill and hard material are initially at the ambient air temperature.

(ii) Analysis. Since heat is generated by friction at the tip of the bit, it follows that this is the location of the highest temperature. From this location heat is transferred by conduction through the bit and the hard material. Heat is lost by convection and radiation from the bit surface to the surroundings. As a result of this heat loss the temperature of the bit drops as the distance from the tip is increased towards the chuck.

(5) Learning and Generalizing. The source of energy is the electric power supplied to the drill. A portion of this energy is stored in the hard material and the drill itself. The remainder is transferred to the surroundings by convection and radiation. At steady state no energy is stored.

hard material

hand drill

drill bit

PROBLEM 1.3

Consider the door handle of a wood burning stove. Discuss the direction and modes of heat transfer in this configuration.

Solution

Heat flows by conduction and radiation from the stove door to rod A. Heat is lost from rod A to the surroundings by convection and radiation and is transmitted to handle B by conduction. Heat is conducted through handle A and convected and radiated to the surroundings. Handle A also receive heat by radiation from the stove and rod A. Thus three modes of heat transfer take place in this example: conduction, convection and radiation.

PROBLEM 1.4

A very thin metallic sheet is placed between two wood plates of different thicknesses. The plates are firmly pressed together and electricity is passed through the sheet. The exposed surfaces of the two plates lose heat to the ambient fluid by convection. Assume uniform heating at the interface. Neglect end effects and assume steady state.

[a] Will the heat transfer through the two plates be the same? Explain.[b] Will the exposed surfaces be at the same temperature? Explain.

(1) Observations. (i) The electric energy generated in the metallic sheet is transferred outwardly through the two wood plates. (ii) Since the plates are of identical material but of different thicknesses, it follows that this is an asymmetrical problem. (iii) According to Fourier’s law of conduction, material thickness plays a role in the rate of heat conducted. (iv) Heat is removed from the exposed surfaces by convection to the surroundings.

(2) Problem Definition. Determine if the flow of heat is symmetrical with respect to the interface metallic plane.

(3) Solution Plan. Examine the effect of plate thickness on the rate of heat transfer. Use Fourier’s law of conduction and Newton’s law of cooling.

(4) Plan Execution.(i) Assumptions. (1) Steady state, (2) one-dimensional conduction, (3) the two plates are of

identical material, (4) constant thermal conductivity, (5) the heat transfer coefficients on the two exposed sides are equal, (6) the ambient temperature is the same on both sides and (7) negligible radiation.

(ii) Analysis. [a] According to Fourier’s law for one-dimensional conduction in a plate, the rate of heat transfer is inversely proportional to the plate thickness. Since the two plates do not have the same thickness, it follows that the rate of heat transfer through the plates is not the same.

Ts1

current

+

-

Ts2

q1q2

[b] Application of Newton’s law of cooling gives

q = h (Ts -T ) (a)where

h = heat transfer coefficient, W/m2-oC q = surface heat flux, W/m2

Ts = exposed surface temperature, oCT = ambient temperature, oC

Solving (a) for the surface temperature yields

Ts = T + q

h(b)

Equation (b) shows that if h and T are the same for both plates but q is different, then Ts will be different.

PROBLEM 1.4 (continued)

(iii) Checking. Dimensional check: Each term in (b) must have the units of temperature.

The units of the term q

h are

CC)(W/m

)(W/m oo2

2

h

q

(5) Learning and Generalizing. (i) More heat is transferred through the thinner of the two plates. (ii) The outside surface temperature of the thinner plate is higher than that of the thicker plate.

PROBLEM 1.5

A rod is perfectly insulated along half its length and is exposed to a fluid at T along the other half. The end of the rod at the insulated half is heated such that the temperature is To. If To > T , what is the temperature of the rod at the midpoint relative to To and T ?

Solution

According to the second law of thermodynamics, heat transfers from high to low temperature. The temperature of the rod at the end of the insulated half is To. Heat transfers from this end by conduction towards the uninsulated half. The temperature drops through this half such that at the mid-point of the rod it is lower than To but higher than T. In the uninsulated half heat is conducted axially and radially and then transferred by convection and radiation from the surface to the surroundings.

PROBLEM 1.7

At a given section of an insulated lead bar the temperature is 300oC and the temperature gradient is 448oC/m. Taking into consideration variation of thermal conductivity with temperature and assuming steady state,

[a] Calculate the heat flux at this section.[b] Determine the gradient at another section where the temperature is 20oC.

(1) Observations. (i) Heat is conducted axially along the bar. (ii) The temperature varies along the bar. (iii) Since the temperature is not constant it follows that the thermal conductivity varies along the bar. (iv) Heat flux is determined from Fourier’s law.

(2) Problem Definition. Establish a relationship between heat flux, temperature gradient and thermal conductivity.

(3) Solution Plan. Apply Fourier's law of conduction.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) one-dimensional conduction, (3) constant cross section area and (4) perfectly insulated surface.

(ii) Analysis. [a] Fourier’s law gives

dx

Tdkq

(a)where

x0

1 2

T1=300oC

dT1/dx = 448oC/m

T2 = 0oC

insulation

k = thermal conductivity, W/m- o C q" = heat flux, W/m2

T = temperature, oC x = distance along bar, m

Applying (a) to the first section

1

11

dx

dTkq

(b)

where subscript 1 refers to the first section. Since the temperature at section 1 is known the corresponding thermal conductivity of lead can be determined from Appendix A. With the gradient at this section given, equation (b) gives the flux.

[b] Equation (a) is applied to determine the temperature gradient at section 2

222

k/qdx

dT

(c)


However, conservation of energy between sections 1 and 2 requires that the heat transfer rate at section 1 be equal to that at section 2. Since the cross section area is constant, the flux at the two sections is the same. Thus

21qq (d)

Substituting (d) into (c)

212

k/qdx

dT

(e)

where 1q is given in (b).

(iii) Computations. [a] Flux at section 1. The conductivity of lead at 300oC is obtained from Appendix A

k1 = 31.75 W/m-oC

Substituting into (b) and noting that 1

dx

dT= 448 (oC/m)

1q = -31.75 (W/m-oC) 448 (oC/m) = - 14,224 W/m2

[b] The thermal conductivity at section 2 where the temperature is at 0oC is obtained from Appendix A:

k2 = 35.7 W/m-oC

The temperature gradient at section 2 is given by (e)

2

dx

dT= - (-14,224) (W/m2) / 35.7 (W/m -oC) = 398.4 oC/m

(iv) Checking. Dimensional check: Computations showed that units of equations (b) and (e) are correct.

(5) Learning and Generalizing. If the thermal conductivity is treated as constant, the error in determining the temperature gradient at section 2 will be 12.5%. This error is associated with a temperature change of 300oC. A smaller change in temperature results in a smaller error in assuming constant conductivity.

PROBLEM 1.10

Heat is removed from a rectangular surface by convection to an ambient fluid at . The heat transfer coefficient is h. Surface temperature is given by

=

where A is constant. Determine the steady state heat transfer rate from the plate.

(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling may be helpful. (ii) Surface temperature is not uniform. It varies the plate in a known manner. (iii) The heat transfer coefficient and the ambient temperature are known.

(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface temperature.

(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem the surface temperature is not constant. This means that the rate of heat by convection varies along the surface. Therefore, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer.

(4) Plan Execution.

L

x

dx

Wdqs0

(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform heat transfer coefficient and (4) uniform ambient fluid temperature.

(ii) Analysis. Newton's law of cooling states that

qs = h As (Ts - T) (a)where

As = surface area, m2

h = heat transfer coefficient, W/m2-oCqs = rate of surface heat transfer by convection, WTs = surface temperature, oCT = ambient temperature, oC

Applying (a) to an infinitesimal area dAs = W dx

d qs = h (Ts - T) W dx (b)where

W = width of plate, mx = distance along plate’s length, m

To determine the total heat transfer rate, equation (b) is integrated from x = 0 to x = L


qs = sdq =

L

s dx WTTh0

) ( (c)

In this equation, h, Ts and W are constants. However, surface temperature Ts varies with x according to

Ts = A

x1 2/ (d)

where

A = constant, m1/2-oC


qs = h W dxTx

AL

01/2

(e)

Performing the integration givesqs = h W (2 A L1/2 - T L) (f)

(iii) Checking. Dimensional check: According to (d) the units of A are m1/2-oC. Therefore units of qs in (f) are

qs = h(W/m2-oC)W(m) [A(oC-m1/2)L1/2(m1/2) T (oC)L(m)] = W.

Limiting checks: (1) If h = 0 or W = 0, then qs = 0. Equation (f) satisfies these limiting cases.

(2) If the surface is at a uniform temperature Ts = 0, equation (a) can be applied directly to give

qs = h As ( 0 T) = h As T (g)

According to (d), setting A = 0 gives Ts = 0. Substituting Ts = 0 into (a) gives

qs = h As T (h)which agrees with (g).

Learning and Generalizing. Integration was necessary because surface temperature is not uniform. The same procedure can be followed if the ambient temperature or the heat transfer coefficient or the width varies along the plate.

PROBLEM 1.11

A semi-transparent plate of thickness L and conductivity k is heated by a laser. The temperature distribution in the plate is

T(x) = A

k a exp(-ax) + B x + C

where A, a, B, C and k are known constants. Determine the conduction heat flux at the two surfaces x = 0 and x = L.

(1) Observations. (i) Temperature distribution in the plate is known. (ii) Heat flux is given by Fourier’s law.

(2) Problem Definition. Determine the heat flux at specified locations in the plate.

(3) Solution Plan. Apply Fourier's law of conduction.

(4) Plan Execution.

(i) Assumptions. (1) One-dimensional conduction and (2) constant k.

(ii) Analysis. Fourier's law gives

q" = kd T

d x (a)

where

k = thermal conductivity, W/m-oC q" = heat flux in the x direction, W/m2

T = temperature, oC x = coordinate, m

Temperature distribution in the plate is given by

T(x) = A

k a exp( ax ) + B x + C

(b)

where a, A, B and C are constants. Temperature gradient is obtained from (b):

d T

d x = ak

A exp( ax ) + B (c)

Evaluating (c) at x = 0 and substituting into (a) gives the flux at x = 0

q"(0) = a

A k B (d)

Similarly, at x = L the flux is

q"(L) = kBaLa

A )exp( (e)

(iii) Checking. Dimensional check: (1) Units of (a) should be W/m2

= W/m2


(2) To check units of (d), units of a, A and B must be determined. Since the exponent of the exponential in (b) must be dimensionless, a must have units of (1/m). Since each term in (b) must have units of oC, it follows that units of A are (W/m3) and units of B are (oC/m). Substituting into (d)

)C/m()CW/m-()m/1(

)W/m()0( oo

3

Bka

Aq = W/m2

Limiting check: For the special case where A = 0, temperature distribution in equation (b) becomes linear. This case corresponds to one-dimensional conduction in which energy added at x = 0 must be equal to energy removed at x = L. That is

q"(0) = q"(L) (f)

Setting A = 0 in (d) and (e) gives

q"(0) = kB

and

q"(L) = kB

This result satisfies (f).

(5) Learning and Generalizing. Since temperature distribution is independent of time, it follows that this is a steady state problem. However, the flux at x = 0 is not equal to that at x = L. This does not mean that conservation of energy is violated. The difference is due to the fact that laser energy is added volumetrically and not just at the surface x = 0.

PROBLEM 1.12

A right angle triangle is at a uniform surface temperature Ts. Heat is removed by convection to an ambient fluid at T . The heat transfer coefficient h varies along the surface according to

h = C

x1 2/

where C is constant and x is distance along the base measured from the apex. Determine the total rate of heat transfer from the triangle.

(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling may be helpful. (ii) The ambient temperature and surface temperature are uniform. (iii) The surface area and heat transfer coefficient vary along the triangle.

(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable surface area and heat transfer coefficient.

(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem surface area and heat transfer coefficient are not uniform. This means that the rate of heat transfer varies along the surface. Thus, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer.

W

x

L

dx

dqs

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) negligible radiation and (3) uniform ambient fluid temperature.


qs = h As (Ts - T) (a)where



Applying (a) to an infinitesimal area dAs

dqs = h (Ts - T) dAs (b)

The next step is to express h and dAs in terms of distance x along the triangle. The heat transfer coefficient h is given by

h = C

x1 2/ (c)

The infinitesimal area dAs is given by


dAs = y(x) dx (d)where

x = distance along base of triangle, my(x) = height of the element dAs, m

Similarity of triangles give

y(x) = W

Lx (e)

where

L = base of triangle, mW = height of triangle, m

Substituting (c), (d) and (e) into (b)

dqs = C

x1 2/ (Ts - T) W

Lx dx (f)

Integration of (f) gives qs. Keeping in mind that C, L, W, Ts and T are constants (f) gives

qs = sdq = )( TTL

WCs

L

x

x

02/1 dx (g)

Evaluating the integral in (g)

qs = 2

3C W L1/2 (Ts - T) (h)

(iii) Checking. Dimensional check: According to (c) the units of C are W/m3/2-oC. Therefore units of qs in (h) are

qs = C(W/m3/2-oC) W(m) L1/2(m1/2) (Ts - T)(oC) = W

Limiting checks: If h = 0 (that is C = 0) then qs = 0. Similarly, if W = 0 or L = 0 or Ts = T

then qs = 0. Equation (h) satisfies these limiting cases.

Learning and Generalizing. Integration was necessary because both the heat transfer coefficient and area vary with distance along the triangle. The same procedure can be followed if the ambient temperature or surface temperature is variable.

PROBLEM 1.13

Design consideration requires that the surface of a small electronic package be maintained at a temperature not to exceed 82 oC. Noise constraints rule out the use of fans. The power dissipated in the package is 35 watts and the surface area is 520 cm2. The ambient temperature and surrounding walls are assumed to be at 24 oC. The heat transfer coefficient is estimated to

be 9.2 W/m2-oC and surface emissivity is 0.7. Will the package dissipate the required power without violating design constraints?

(1) Observations. (i) The electronic package loses heat by convection and radiation. Both Newton's law of cooling and Stefan-Boltzmann radiation law may be needed. (ii) In a steady state process the power dissipated is equal to the heat loss from the surface. (iii) Increasing the power, increases surface temperature. (iv) The maximum allowable surface temperature is 82oC. However, the operating surface temperature is unknown.

(2) Problem Definition. Determine the relationship between total heat transfer rate from the package and its surface temperature.

(3) Solution Plan. Apply Newton's law of cooling and Stefan-Boltzmann law to the surface of the package to obtain a relationship between total heat transfer rate and surface temperature.

surroundings

+ -

electronic package

Tsur

P

sTTh,

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) uniform surface temperature, ambient temperature and heat transfer coefficient, (3) the package is assumed to be a small surface surrounded by a much larger surface, (4) the ambient air does not interfere with radiation and (5) the package radiates energy to a surrounding which is at the same temperature as the ambient air.

(ii) Analysis. Applying Newton's law of cooling and Stefan-Boltzmann law to the surface of the package gives

P = q = h A (Ts T ) + A T Ts4 4 (a)

where:

A = surface area = 0.052 m2

h = heat transfer coefficient = 9.2 W/m2-KP = Power dissipated in package = 35 W

q = total surface heat transfer rate = 35 W Ts = surface temperature = 82oC + 273.15 = 355.15 K T = ambient air temperature = surroundings temperature = 24oC + 273.15 = 297.15 K

= emissivity = 0.7 = Stefan-Boltzmann constant = 5.6710-8 W/m2-K4


Of interest is the determination of the surface temperature Ts. However, because of the Ts4 term

in (a), this equation can not be solved explicitly for Ts. The solution can be obtained by a trial and error procedure. Another approach is to calculate the power dissipated corresponding to the maximum allowable surface temperature. If the calculated power is less than design level, it follows that the package cannot be operated safely.

(iii) Computations. Assuming that surface temperature is at the maximum allowable level of 355.15 K, equation (a) is used to compute the corresponding power

P = q = 9.2 (W/m2-K) 0.052(m2) (355.15 15.297 )(K) +

0.7 x 5.67x10-8 (W/m2-K4) 0.052 (m2) [(355.15)4 4)15.297( (297.15)4 ] (K4)

P = q = 27.75 W + 16.74 W = 44.49 W

Since the resulting power level is more than the operating value of 35 W, it follows that the package can dissipate the required power without violating design constraints.

(iv) Checking. Dimensional check: Computations showed that equation (a) is dimensionally correct. Qualitative checks: Increasing h or Ts or A or , increases the allowable power level. Equation (a) exhibits this behavior. (5) Learning and Generalizing. (i) Including radiation is important since it represents 37.6% of the total heat loss. Neglecting radiation leads to the wrong conclusion the package can not be operated. (ii) To solve for the surface temperature corresponding to P = 35 W by trial and error, a

value for Ts is assumed and substituted into (a). The calculated P is then compared with 35 W. The process is repeated until a satisfactory agreement between calculated and given values is obtained. Using this approach gives Ts = 343.7 K , or 70.6 oC. This is less than design value.

PROBLEM 1.14 The temperature of a rod varies along its length according to

T(x) = + (To )

where To is the temperature at x = 0, is the ambient temperature, m is a constant and x is measured along the rod. The length of the rod is L and its radius is ro. The heat transfer coefficient is h.

[a] Determine the heat conducted through the rod at x=0 and x=L.[b] Determine the total heat transfer from the cylindrical surface to the ambient fluid .

(1) Observations. (i) Temperature distribution in the rod is known. (ii) This is a steady state problem since time does not appear in the temperature distribution equation. (iii) Heat transfer can take place at the ends as well as along the cylindrical surface of the rod. (iv) Mathematically this is a one-dimensional problem since temperature varies with x only. (v) Fourier’s law gives heat transfer rate by conduction.

(2) Problem Definition. Determine: [a] the heat conducted through the rod at the two ends. [b] The total heat transfer rate from the cylindrical surface.

h ,

L

Tor

0 x

oT

(3) Solution Plan. Apply Fourier’s law of conduction to determine the heat conducted at the two ends. Apply conservation of energy to the entire rod to determine the heat transfer rate from the cylindrical surface.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) no energy generation and (3) no radiation.

(ii) Analysis. [a] Knowing the temperature distribution, Fourier’s law gives the rate of heat flow by conduction

q k AdT

dxx k rdT

dxo 2 (a)

where

A = cross section area = r o2 , m2

k = thermal conductivity, W/m-oCqx = heat transfer rate, Wro = radius of rod, mT = temperature, oCx = distance along rod, m

Temperature gradient in (a) is obtained from the temperature solution which is given by

cosh

cosh

m L x

mL

(b)

PROBLEM 1.14 (continued)where

L = length of rod, m m = constant, 1/m To = temperature at x = 0, oC T = ambient temperature, oC

Differentiating (b) with respect to x gives the temperature gradient

d T

d x m ( To - T)

sinh

cosh

m L x

mL

(c)

To determine the heat transfer rate at x = 0 equation (c) is evaluated at x = 0 and the result is substituted in (a)

qx ( )0 = = ro2 k m (To - T) tanh mL (d)

Similarly, at x = L

qx (L) = - k = ro2 k m( To - T)

sinh

cosh

0

mL = 0 (e)

[b] The heat transfer from the cylindrical surface is obtained by applying conservation of energy to the rod Rate of energy in = Rate of energy out

or qx (0) = qx (L) + sq (f)

where

= rate of heat transfer from the cylindrical surface, W

Solving (f) for and substituting (d) and (e) into the resulting equation

qs = k m ( To - T) tanh mL (g)

(iii) Checking. Dimensional check: Units of in (g) should be W

= (m2) m(1/m) k(W/m-oC) m(1/m)( To - T)(oC) tanh m(1/m)L(m) = W

Limiting check: For the limiting cases of ro = 0 or To = T, heat transfer from the cylindrical surface must vanish, i.e. = 0. Setting ro = 0 or To = T in (g) gives = 0.

Learning and Generalizing. (i) q x (L) = 0 means that the end of the rod at x = L is perfectly insulated. (ii) Although the given temperature distribution in (a) is one-dimensional, the problem is in fact two-dimensional since the flow of heat is in both the axial and radial directions. Thus (b) is an approximation. (iii) An alternate method for determining qs is by applying Newton’s law of cooling to the cylindrical surface. Since T(x) varies along the surface, Newton’s law must be applied to an infinitesimal surface element. Integration gives qs.

PROBLEM 1.17

A small electronic package with a surface area of 820 cm2 is placed in a room where the air temperature is 28oC. The heat transfer coefficient is 7.3 W/m2-oC. You are asked to determine if it is justified to neglect heat loss from the package by radiation. Assume a uniform surface temperature of 78oC and surface emissivity of 0.65 Assume further that room’s walls and ceiling are at a uniform temperature of 16 oC.

(1) Observations. (i) The electronic package loses heat by convection and radiation. Thus both Newton's law of cooling and Stefan-Boltzmann radiation law apply. (ii) Radiation heat loss may be neglected if it is small compared to heat loss by convection.

(2) Problem Definition. Determine the rate of heat loss by convection and radiation and compare the two.

surroundings

+ -

electronic package

Tsur

P

qrqc

Ts h

(3) Solution Plan. Apply Newton’s law of cooling and Stefan-Boltzmann radiation law to the electronic package.

(4) Plan Execution.

(i) Assumptions. (1) Uniform surface temperature, (2) uniform ambient air and surroundings (walls and ceiling) temperature, (3) package surface is small compared to that surroundings and (4) atmospheric air does not interfere with radiation.

(ii) Analysis. Application of Newton’s law of cooling gives

qc = h A (Ts -T) (a)where

A = surface area of package = 0.082 m2

h = heat transfer coefficient = 7.3 W/m2-oCqc = convection heat transfer rate, WTs = surface temperature = 78oC = 78oC + 273.15 = 351.15 K T = ambient air temperature = 28oC + 273.15 = 301.15 K

Application of Stefan-Boltzmann law gives

qr = A T Ts sur4 4 (b)

where

qr = radiation heat transfer rate, WTsur = surroundings temperature (walls and ceiling) = 16oC = 16oC + 273.15 = 289.15 K = emissivity = 0.65

= Stefan-Boltzmann constant = 5.67 x 10-8 W/m2-K4

(iii) Computations. Substituting numerical values into (a)


qc = 7.3(W/m2-K) 0.082(m2) (351.15 15.301 )(K) = 29.93 W

Similarly, equation (b) gives

qr = 0.65 5.6710-8 (W/m2-K4) 0.082(m2) [(351.15)4 - (289.15)4] (K4) = 24.82 W

(iv) Checking. Dimensional check: Equations (a) and (b) are dimensionally correct since they give the correct units for heat transfer rate.

(5) Learning and Generalizing. (i) Heat loss by radiation can not be neglected since it is of the same order of magnitude as heat loss by convection. (ii) Examination of equation (b) shows that radiation heat transfer is directly proportional to emissivity. Radiation heat loss can be increased by coating the surface with a high emissivity paint. (iii) When carrying out computations of radiation heat loss, all temperatures must be expressed in absolute degrees.

PROBLEM 1.18

Consider radiation from a small surface at 100 oC which is enclosed by a much larger surface at 24 oC. Determine the percent increase in the radiation heat transfer if the temperature of the small surface is doubled.

(1) Observations. (i) This is a radiation heat transfer problem. (ii) Radiation is from a small surface which is enclosed by a much larger surface. (iii) Of interest is the effect of surface temperature on the rate of heat transfer. (iv) Increasing surface temperature increases heat transfer by radiation.

(2) Problem Definition. Determine the radiation heat transfer rate corresponding to two surface temperatures.

(3) Solution Plan. Apply Stefan-Boltzmann law.

(4) Plan Execution.

(i) Assumptions. (1) The heat transfer surface is a small body which is completely enclosed

by a much larger surface and (2) constant emissivity.

(ii) Analysis. Stefan-Boltzmann law gives

44sursr TTAq (a)

where

A = surface area, m2

qr = radiation heat transfer rate, WTs = surface temperature = 100oC + 273.15 = 373.15 K

surT = surroundings temperature = 24oC + 273.15 = 297.15 K = emissivity = Stefan-Boltzmann constant = 5.6710-8 W/m2-K4

The percent change in radiation heat due to doubling of the surface temperature is given by

% change = PC = 100

1100

1

2

1

12

r

r

r

rr

q

q

q

qq(b)

where the subscripts 1 and 2 correspond to Ts1 and Ts 2 , respectively. Substituting (a) into (b) and

noting that A cancels out, gives

PC = 100 T T

T Ts

s

24 4

14 4 1

(c)

(iii) Computations. For T = 297.15 K, Ts1 = 373.15 K and Ts 2 = 2100oC + 273.15 =

473.15 K, equation (c) gives

PC = 100 47315 297 15

37315 297 151

4 4

4 4

. .

. .

= 265


(iv) Checking. Dimensional check: Percent change should be dimensionless. Equations (b) and (c) are dimensionless.

Limiting check: For the special case of Ts 2 = Ts1 , there is no change inqr . Letting Ts 2 = Ts1 in

(c) gives PC = 0, as expected.

(5) Learning and Generalizing. (i) A 100 % increase in surface temperature results in 265% increase in radiation heat flux. This is a reflection of the non-linear nature of radiation as indicated by the fourth power of temperature in the Stefan-Boltzmann law. (ii) Doubling temperature on the Celsius scale is not the same as doubling temperature on the kelvin scale.

Tsur

Ts2

qr1

qr2

surroundings

Ts1

(iii) It is possible to solve this problem without knowing the emissivity of surface area because both cancel out in equation (c).

PROBLEM 1.19

A sphere of radius 10 cm is maintained at uniform surface temperature of 220 . The sphere is suspended in a large room whose surfaces are at 18 . Determine the steady state radiation heat transfer rate from the sphere if its emissivity is 0.85.(1) Observations. (i) The sphere loses heat by radiation. Thus, Stefan-Boltzmann radiation law applies.(ii) Sphere surface area is small compared to the room walls, floor and ceiling area.

(2) Problem Definition. Determine the net rate of heat exchanged by radiation between a small surface and a much larger surface enclosing it.

(3) Solution Plan. Apply Stefan-Boltzmann radiation law.

(4) Plan Execution.

(i) Assumptions. (1) Uniform sphere surface temperature, (2) uniform surroundings (walls, floor and ceiling) temperature, (3) sphere surface area is small compared to surroundings area and (4) room atmosphere does not interfere with radiation.

(ii) Analysis. Application of Stefan-Boltzmann law gives

qr = A T Ts sur4 4 (a)

where

A = surface area of sphere, m2

qr = radiation heat transfer rate, WTs = sphere surface temperature = 220 + 273.15 = 493.15 K Tsur = surroundings temperature = 18 + 273.15 = 291.15 K = emissivity = 0.85

= Stefan-Boltzmann constant = 5.67 x 10-8

Sphere surface area is (b)

where

= sphere radius = 0.1 m

(iii) Computations. Substituting (b) into (a) and using numerical values

qr = 0.85 5.6710-8 ( ) 4 ( 0.1)2(m2) [(493.15)4 - (291.15)4] (K4) = 314.7 W


Limiting check: If the sphere is at the same temperature as the surroundings no heat transfer can take place. Setting in (a) gives


(5) Learning and Generalizing. (i) Heat loss by radiation can not be neglected unless it is small compared to heat loss by convection. (ii) Examination of equation (a) shows that radiation heat transfer is directly proportional to emissivity. It can be significantly changed using paint to alter surface emissivity. (iii) When carrying out computations of radiation heat loss, all temperatures must be expressed in absolute degrees.

PROBLEM A-1.3

The steady state temperature distribution along an insulated rod of radius 1.5 cm is found to be linear. The hot end of the rod is at 120 and the cold end is at 40 . Determine the heat transfer rate through the rod. The rod length is 10 cm and its conductivity is 230 .

(1) Observations. (i) Rod surface is perfectly insulated and thus heat flows axially by conduction from the hot end to the cold end . (ii) Temperature distribution in the rod is one-dimensional. (iii) Fourier’s law gives the rate of heat conducted through the rod.

(2) Problem Definition. Determine the rate of heat conducted through the rod.

(3) Solution Plan. Apply Fourier’s law of conduction.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) the cylindrical surface is perfectly insulated, (3) constant thermal conductivity k and (4) uniform conditions at any cross section (one-dimensional).

(ii) Analysis. Fourier's law gives

(a)

where

A = cross section area = m2

k = thermal conductivity = 230 W/m-oC L = rod length = 10 cm = 0.1 m qx = rate of heat transfer, W ro = radius of rod = 1.5 cm = 0.015 m Tsi = hot end temperature = 120oC Tsi = cold end temperature = 40oC

(iii) Computations. Substituting into (a)

(iv) Checking. Dimensional check: Units of (a) are consistent and indicated in the computation of Limiting check: (1) If the conductivity k is zero, the heat transfer rate should also be zero. Setting k = 0 in (a) gives qx = 0. (2) If the two ends are at the same temperature, no heat flow will take place. Setting in (a) gives qx = 0.

(5) Learning and Generalizing. Equation (a) is valid for the assumptions made above. PROBLEM A-1.4

A rectangular plate is cooled by convection. The heat transfer coefficient varies along the plate according to

where is constant. The plate surface is maintained at uniform temperature and the ambient fluid is at The width of the plate is W and its length is L. Determine the steady state heat transfer rate from the plate.

(1) Observations. (i) Heat is removed from the surface by convection. Therefore, Newton's law of cooling should be helpful. (ii) The heat transfer coefficient is not

uniform. It varies along the plate in a known manner. (iii) Total heat transfer can be obtained by integration of Newton’s law of cooling.

(2) Problem Definition. Find the total heat transfer rate by convection from the surface of a plate with a variable heat transfer coefficient.

(3) Solution Plan. Newton's law of cooling gives the rate of heat transfer by convection. However, in this problem the heat transfer coefficient is not uniform. This means that the rate of heat by convection varies along the surface. Therefore, Newton’s law should be applied to an infinitesimal area dAs and integrated over the entire surface to obtain the total heat transfer rate.

(4) Plan Execution.

(i) Assumptions. (1) Steady state, (2) negligible radiation, (3) uniform surface temperature and (4) uniform ambient fluid temperature.


qs = h (Ts - T) As (a)where



Applying (a) to an infinitesimal area dAs = W dx

d qs = h (Ts - T) W dx (b)where

W = width of plate, mPROBLEM A-1.4 (continued)

x = distance along plate’s length, m

To determine the total heat transfer rate, equation (b) is integrated from x = 0 to x = L

qs = sdq = (c)

In this equation, Ts, and W are constant. However, the heat transfer coefficient h varies with x according to

(d)where

= constant,


(e)

Performing the integration gives

(f)

(iii) Checking. Dimensional check: Units of qs in (f) are

= W

Limiting checks: If h = 0 or W = 0, then qs = 0. Setting or in (f) gives

(5) Learning and Generalizing. Integration is necessary in this problem because the heat transfer coefficient is not uniform. The same procedure can be followed if the ambient temperature or surface temperature or the width varies along the plate.

PROBLEM A-1.5

A sphere of radius 10 cm is maintained at uniform surface temperature of 220 . The sphere is suspended in a large room whose surfaces are at 18 . Determine the steady state radiation heat transfer rate from the sphere if its emissivity is 0.85.

(1) Observations. (i) The sphere loses heat by radiation. Thus, Stefan-Boltzmann radiation law applies.(ii) Sphere surface area is small compared to the room walls, floor and ceiling area.

(2) Problem Definition. Determine the net rate of heat exchanged by radiation between a small surface and a much larger surface enclosing it.

(3) Solution Plan. Apply Stefan-Boltzmann radiation law.

(4) Plan Execution.

(i) Assumptions. (1) Uniform sphere surface temperature, (2) uniform surroundings (walls, floor and ceiling) temperature, (3) sphere surface area is small compared to surroundings area and (4) room atmospheric does not interfere with radiation.

(ii) Analysis. Application of Stefan-Boltzmann law gives

qr = A T Ts sur4 4 (a)

where

A = surface area of sphere, m2

qr = radiation heat transfer rate, WTs = sphere surface temperature = 220 + 273.15 = 493.15 K Tsur = surroundings temperature = 18 + 273.15 = 291.15 K = emissivity = 0.85

= Stefan-Boltzmann constant = 5.67 x 10-8 Sphere surface area is

(b)

where

= sphere radius = 0.1 m

(iii) Computations. Substituting (b) into (a) and using numerical values

PROBLEM A-1.5 (continued)

qr = 0.85 5.6710-8 ( ) 4 ( 0.1)2(m2) [(493.15)4 - (291.15)4] (K4) = 314.7 W


Limiting check: If the sphere is at the same temperature as the surroundings no heat transfer can take place. Setting in (a) gives

(5) Learning and Generalizing. (i) Heat loss by radiation can not be neglected unless it is small compared to heat loss by convection. (ii) Examination of equation (a) shows that radiation heat transfer is directly proportional to emissivity. It can be significantly changed using paint to alter surface emissivity. (iii) When carrying out computations of radiation heat loss, all temperatures must be expressed in absolute degrees.

Heat Transfer Solutions to first Chater

Documents

heat transfer

total heat

setting sur

larger surface

cooling states

infinitesimal

wherea surface

sphere loses