Chapter 4 Known Cylinder Extending Between Two

PROBLEM 4.14

KNOWN: Dimensions and temperature of water droplet.

FIND: Time for droplet to freeze completely.

ASSUMPTIONS: (1) Constant properties, (2) Negligible convection and radiation, (3) Isothermal

water particle, (4) Semi-infinite medium.

PROPERTIES: Table A.4, Air (265 K): ka = 0.0235 W/m⋅K. Table A.6, Liquid water (273 K):

ρ

w =

1000 kg/m3.

ANALYSIS: An energy balance on the droplet yields

Combining Equations (1) and (2) with the expression for the droplet volume V = πD3/6 yields

COMMENTS: (1) Solidification might initiate in the lower region of the droplet. The ice that forms

would pose an additional conduction resistance between the cold metal surface and the liquid water.

PROBLEM 4.15

KNOWN: Dimensions and boundary temperatures of a steam pipe embedded in a concrete

casing.

FIND: Heat loss per unit length.

SCHEMATIC:

ASSUMPTIONS: (1) Steady–state conditions, (2) Negligible steam side convection

resistance, pipe wall resistance and contact resistance (T1 = 400K), (3) Constant properties.

PROPERTIES: Table A-3, Concrete (300K): k = 1.4 W/m⋅K.

ANALYSIS: The heat rate can be expressed as

Hence,

COMMENTS: Having neglected the steam side convection resistance, the pipe wall

resistance, and the contact resistance, the foregoing result overestimates the actual heat loss.

PROBLEM 4.16

KNOWN: Power, size and shape of laser beam. Material properties.

FIND: Maximum surface temperature for a Gaussian beam, maximum temperature for a flat

beam, and average temperature for a flat beam.

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Semi-infinite medium,

(4) Negligible heat loss from the top surface.

ANALYSIS: The shape factor is defined in Eq. 4.20 and is q = Sk∆T1-2 (1)

From the problem statement and Section 4.3, the shape factors for the three cases are:

Beam Shape

Shape Factor

T1,avg or T1,max

b

For the Gaussian beam,

-3 -6

S = 2 π × 0.1 × 10 m = 354 × 10 m

12 2

T = T + q/Sk = T + P /Skα

COMMENTS: (1) The maximum temperature occurs at r = 0 for all cases. For the flat beam, the

maximum temperature exceeds the average temperature by 78.1 – 70.0 = 8.1 degrees Celsius.

Flat

Gaussian

Flat

Gaussian

PROBLEM 4.17

KNOWN: Thin-walled copper tube enclosed by an eccentric cylindrical shell; intervening space

filled with insulation.

FIND: Heat loss per unit length of tube; compare result with that of a concentric tube-shell

arrangement.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Thermal resistances of

copper tube wall and outer shell wall are negligible, (4) Two-dimensional conduction in insulation.

ANALYSIS: The heat loss per unit length written in terms of the shape factor S is

Substituting numerical values, all dimensions in mm,

Hence, the heat loss is

using Eq. 3.32. Substituting numerical

values,

COMMENTS: As expected, the heat loss with the eccentric arrangement is larger than that for the

concentric arrangement. The effect of the eccentricity is to increase the heat loss by (10.7–10.2)/10.2

≈ 5.3%.

PROBLEM 4.18

KNOWN: Cubical furnace, 350 mm external dimensions, with 50 mm thick walls.

FIND: The heat loss, q(W).

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Two-dimensional conduction, (3)

Constant properties.

PROPERTIES: Table A-3, Fireclay brick

( )

12

T T T / 2 610K : k 1.1 W/m K.=+= ≈ ⋅

ANALYSIS: Using relations for the shape factor from Table 4.1,

C

The heat rate in terms of the shape factors is

COMMENTS: Note that the restrictions for SE and SC have been met.

PROBLEM 4.19

KNOWN: Dimensions of stainless steel pillar and nominal glass temperatures. Contact resistance

between pillar and glass.

FIND: Conduction rate through the pillar.

ASSUMPTIONS: (1) Steady–state conditions, (2) Constant properties, (3) Negligible radiation, (4)

Two–dimensional conduction, (5) Glass behaves as a semi–infinite medium.

PROPERTIES: Table A.1, AISI 302 stainless steel (300 K): kp = 15.1 W/m⋅K. Table A.3, plate glass

(300 K): kg = 1.4 W/m⋅K.

ANALYSIS: Conduction through the pillar results in a depression of the glass temperature adjacent

to the pillar. This is associated with a constriction resistance within each glass sheet. Therefore, the

Using the shape factor for Case 10 of Table 4.1(a) the resistances are:



Therefore, the total resistance is

COMMENTS: (1) Constriction of the heat flow within the glass poses the largest resistance to heat

transfer. (2) Radiation between the two glass sheets exists, and may be important in determining the

overall heat transfer through the window. (3) Extremely high vacuum between the two glass sheets is

PROBLEM 4.20

KNOWN: Temperature, diameter and burial depth of an insulated pipe.

FIND: Heat loss per unit length of pipe.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through

insulation, two-dimensional through soil, (3) Constant properties, (4) Negligible oil

convection and pipe wall conduction resistances.

PROPERTIES: Table A-3, Soil (300K): k = 0.52 W/m⋅K; Table A-3, Cellular glass (365K):

k = 0.069 W/m⋅K.

ANALYSIS: The heat rate can be expressed as

From Equation 4.21 and Table 4.1,

Hence,

COMMENTS: (1) Contributions of the soil and insulation to the total resistance are

approximately the same. The heat loss may be reduced by burying the pipe deeper or adding

more insulation.

(2) The convection resistance associated with the oil flow through the pipe may be significant,

in which case the foregoing result would overestimate the heat loss. A calculation of this

PROBLEM 4.21

KNOWN: Operating conditions of a buried superconducting cable.

FIND: Required cooling load.

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Two-dimensional

conduction in soil, (4) One-dimensional conduction in insulation, (5) Pipe inner surface is at

liquid nitrogen temperature. (6) Negligible contact resistance.

ANALYSIS: The heat rate per unit length is

where Tables 3.3 and 4.1 have been used to evaluate the insulation and ground resistances,

respectively. Hence,

COMMENTS: The heat gain is small and the dominant contribution to the thermal

resistance is made by the insulation. Inclusion of a contact resistance would further reduce the

heat gain.

PROBLEM 4.22

KNOWN: Dimensions and temperature of thermocouple bead and wires. Manipulator temperature,

distance between bead and surface.

FIND: Surface temperature.

SCHEMATIC:

ASSUMPTIONS: (1) Constant properties, (2) Negligible radiation and convection, (3) Isothermal

thermocouple bead, (4) Air behaves as a semi–infinite medium, (5) Steady state conditions.

PROPERTIES: Table A.4, Air (310 K): ka = 0.027 W/m⋅K.

ANALYSIS: An energy balance on the thermocouple bead yields

44

LL

which may be rearranged to yield

PROBLEM 4.22 (Cont.)

COMMENTS: The required surface temperature to induce the specified thermocouple temperature

and its dependence on the separation distance, z, is shown below. As expected, the required surface

temperature becomes greater as the separation distance increases.

50

PROBLEM 4.23

KNOWN: Dimensions, thermal conductivity and inner surface temperature of furnace wall. Ambient

conditions.

FIND: Heat loss.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Uniform convection coefficient over entire outer surface of

container, (3) Negligible radiation losses.

ANALYSIS: From the thermal circuit, the heat loss is

where the shape factor S must include the effects of conduction through the 8 corners, 12 edges and 6

plane walls. Hence, using the relations for Cases 8 and 9 of Table 4.1,

COMMENTS: The heat loss is extremely large and measures should be taken to insulate the furnace.

Radiation losses may be significant, leading to larger heat losses.

PROBLEM 4.24

KNOWN: Cylinder extending between two walls. Highly polished surfaces. Walls are at specified

temperatures far from the cylinder.

FIND: (a) Thermal resistance network. (b) Expression for shape factor corresponding to conduction

between Tb,1 and Tb,2. (3) Value of shape factor for specified cylinder diameter, length, and thermal

conductivity.

SCHEMATIC:

ASSUMPTIONS: (1) Steady–state conditions. (2) Constant properties. (3) Walls are semi–infinite. (4)

Walls and cylinder have same thermal conductivity.

ANALYSIS: (a) Heat conducting along the cylinder will alter the wall temperature in the vicinity of the

(b) According to the resistance network, when kw = kc = k, the heat transfer rate between the two walls is

given by

Then from Equation 4.20, the shape factor is

(c) The value of the shape factor is

COMMENTS: (1) As long as the materials all have the same thermal conductivity, the value of the shape

factor is independent of the thermal conductivity. (2) The effect of the cylinder on the wall temperature

becomes negligible as the thermal resistance of the cylinder increases, for example by increasing L.

PROBLEM 4.25

KNOWN: Dimensions and surface temperatures of a square channel. Number of chips mounted on

outer surface and chip thermal contact resistance.

FIND: Heat dissipation per chip and chip temperature.

ASSUMPTIONS: (1) Steady state, (2) Approximately uniform channel inner and outer surface

temperatures, (3) Two–dimensional conduction through channel wall (negligible end-wall effects), (4)

Constant thermal conductivity.

ANALYSIS: The total heat rate is determined by the two-dimensional conduction resistance of the

The heat rate per chip is then

COMMENTS: (1) By acting to spread heat flow lines away from a chip, the channel wall provides

an excellent heat sink for dissipating heat generated by the chip. However, recognize that, in practice,

there will be temperature variations on the inner and outer surfaces of the channel, and if the

prescribed values of T1 and T2 represent minimum and maximum inner and outer surface

PROBLEM 4.26

KNOWN: Dimensions and thermal conductivity of concrete duct. Convection conditions of ambient

air. Inlet temperature of water flow through the duct.

FIND: (a) Heat loss per duct length near inlet, (b) Minimum allowable flow rate corresponding to

maximum allowable temperature rise of water.

SCHEMATIC:

ASSUMPTIONS: (1) Steady state, (2) Negligible water–side convection resistance, pipe wall

conduction resistance, and pipe/concrete contact resistance (temperature at inner surface of concrete

corresponds to that of water), (3) Constant properties, (4) Negligible flow work and kinetic and

potential energy changes.

ANALYSIS: (a) From the thermal circuit, the heat loss per unit length near the entrance is

Hence,

(b) From Eq. (1.12d), with

q qL

′

=

and

( )

io

T T 5 C,−=°

io io

COMMENTS: The small reduction in the temperature of the water as it flows from inlet to outlet

induces a slight departure from two-dimensional conditions and a small reduction in the heat rate per

unit length. A slightly conservative value (upper estimate) of

m



is therefore obtained in part (b).

PROBLEM 4.27

KNOWN: Long constantan wire butt-welded to a large copper block forming a thermocouple junction

on the surface of the block.

FIND: (a) The measurement error (Tj – To) for the thermocouple for prescribed conditions, and (b)

Compute and plot (Tj – To) for h = 5, 10 and 25 W/m2⋅K for block thermal conductivity 15 ≤ k ≤ 400

W/m⋅K. When is it advantageous to use smaller diameter wire?

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Thermocouple wire behaves as a fin with constant

heat transfer coefficient, (3) Copper block has uniform temperature, except in the vicinity of the junction.

PROPERTIES: Table A-1, Copper (pure, 400 K), kb = 393 W/m⋅K; Constantan (350 K), kt ≈ 25

W/m⋅K.

ANALYSIS: The thermocouple wire behaves as a long fin permitting heat to flow from the surface

thereby depressing the sensing junction temperature below that of the block To. In the block, heat flows

into the circular region of the wire-block interface; the thermal resistance to heat flow within the block is

block

(b) We keyed the above equations into the IHT workspace, performed a sweep on kb for selected values

of h and created the plot shown. When the block thermal conductivity is low, the error (To – Tj) is larger,

increasing with increasing convection coefficient. A smaller diameter wire will be advantageous for low

values of kb and higher values of h.

5

PROBLEM 4.28

KNOWN: Dimensions, shape factor, and thermal conductivity of square rod with drilled interior hole.

Interior and exterior convection conditions.

FIND: Heat rate and surface temperatures.

ASSUMPTIONS: (1) Steady-state, two-dimensional conduction, (2) Constant properties, (3) Uniform

convection coefficients at inner and outer surfaces.

ANALYSIS: The heat loss can be expressed as

where

Hence,

COMMENTS: The largest resistance is associated with convection at the outer surface, and the

conduction resistance is much smaller than both convection resistances. Hence, (T2 – T∞,2) > (T∞,1 – T1)

>> (T1 – T2).

PROBLEM 4.29

KNOWN: Cylinder of specified diameter within square bakelite coating of dimension w on a side.

Cylinder temperature, environment temperature, and heat transfer coefficient.

FIND: Heat transfer rate per unit length of cylinder. Plot heat transfer rate for 10 mm < w < 100 mm.

Explain dependence of heat transfer rate on bakelite thickness. Find critical value of w that maximizes

heat transfer from cylinder.

SCHEMATIC:

ASSUMPTIONS: (1) Steady–state conditions, (2) Uniform properties.

PROPERTIES: Table A–3, Bakelite, (T = 300 K): k = 1.4 W/m⋅K.

ANALYSIS: The resistance network between the cylinder and the environment is shown below, where

Rt,cond(2D) = 1/Sk (Equation 4.21), and the shape factor S for heat transfer through the bakelite is given in

Table 4.1, Case 6.

Thus the heat transfer rate per unit cylinder length is

The existence of a maximum heat transfer rate is because of the competing effects that as w

increases, the thermal resistance of the bakelite increases, but the thermal resistance for

convection decreases because of the increasing surface area.

Continued…

PROBLEM 4.29 (Cont.)

Heat Transfer Rate per Unit Length

400

375

<

Heat transfer is maximized when thermal resistance is minimized. Therefore the critical bakelite

dimension can be determined by differentiating the thermal resistance with respect to w and

setting the result to zero:

COMMENTS: The critical bakelite thickness is very similar to the concept of a critical thickness of

insulation on a cylinder.

PROBLEM 4.30

KNOWN: Long fin of aluminum alloy with prescribed convection coefficient attached to different base

materials (aluminum alloy or stainless steel) with and without thermal contact resistance

t,j

R′′

at the

junction.

FIND: (a) Heat rate qf and junction temperature Tj for base materials of aluminum and stainless steel,

(b) Repeat calculations considering thermal contact resistance,

t,j

R′′

, and (c) Plot as a function of h for

the range 10 ≤ h ≤ 1000 W/m2⋅K for each base material.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Infinite fin.

PROPERTIES: (Given) Aluminum alloy, k = 240 W/m⋅K, Stainless steel, k = 15 W/m⋅K.

ANALYSIS: (a,b) From the thermal circuits, the heat rate and junction temperature are

and, with P = πD and Ac = πD2/4, from Tables 4.1 and 3.4 find

Without

t,j

R′′

With

t,j

R′′

(c) We used the IHT Model for Extended Surfaces, Performance Calculations, Rectangular Pin Fin to

PROBLEM 4.30 (Cont.)

5

6

COMMENTS: (1) From part (a), the aluminum alloy base material has negligible effect on the fin heat

rate and depresses the base temperature by only 2°C. The effect of the stainless steel base material is

substantial, reducing the heat rate by 27% and depressing the junction temperature by 25°C.

(3) From the plot of qf vs. h, note that at low values of h, the heat rates are nearly the same for both