Chapter 1 Plate is of uniform surface temperature

PROBLEM 1.58

KNOWN: Temperatures at 15 mm and 30 mm from the surface and in the adjoining airflow for a

thick stainless steel casting.

FIND: Surface convection coefficient, h.

SCHEMATIC:

ANALYSIS: From a surface energy balance, it follows that

′′ =′′

q q

cond conv

where the convection rate equation has the form

( )

0

conv

q h T T ,

∞

′′ = −

To = 70°C.

Hence, the convection coefficient is

COMMENTS: The accuracy of this procedure for measuring h depends strongly on the validity of

the assumed conditions.

PROBLEM 1.59

KNOWN: Conditions associated with surface cooling of plate glass which is initially at 600°C.

Maximum allowable temperature gradient in the glass.

FIND: Lowest allowable air temperature, T∞.

SCHEMATIC:

ASSUMPTIONS: (1) Surface of glass exchanges radiation with large surroundings at Tsur = T∞,

(2) One-dimensional conduction in the x-direction.

ANALYSIS: The maximum temperature gradient will exist at the surface of the glass and at the

instant that cooling is initiated. From the surface energy balance, Eq. 1.13, and the rate equations,

Eqs. 1.1, 1.3a and 1.7, it follows that

T∞ may be obtained from a trial-and-error solution, from which it follows that, for T∞ = 618K,

∞

COMMENTS: (1) Initially, cooling is determined primarily by radiation effects.

(2) For fixed T∞, the surface temperature gradient would decrease with increasing time into the

PROBLEM 1.60

KNOWN: Surface temperature, diameter and emissivity of a hot plate. Temperature of surroundings

and ambient air. Expression for convection coefficient.

SCHEMATIC:

ANALYSIS: (a) From an energy balance on the hot plate, Pelec = qconv + qrad = Ap

( )

conv rad

q q.

′′ ′′

+

Substituting for the area of the plate and from Eqs. (1.3a) and (1.7), with h = 0.80 (Ts

– T∞)1/3, it follows that



(b) As shown graphically, both the radiation and convection heat rates, and hence the requisite electric

power, increase with increasing surface temperature.

COMMENTS: Radiation losses could be reduced by applying a low emissivity coating to the

surface, which would have to maintain its integrity over the range of operating temperatures.

Effect of surface temperature on electric power and heat rates

500

T

s

q

rad

q

conv

Air

T

sur

T

s

q

rad

q

conv

Air

T

sur

PROBLEM 1.61

KNOWN: Solar collector designed to heat water operating under prescribed solar irradiation and

loss conditions.

FIND: (a) Useful heat collected per unit area of the collector,

′′

qu,

(b) Temperature rise of the water

flow,

T T

o i

−,

and (c) Collector efficiency.

SCHEMATIC:

PROPERTIES: Table A.6, Water (300K): cp = 4179 J/kg⋅K.

ANALYSIS: (a) Defining the collector as the control volume and writing the conservation of energy

requirement on a per unit area basis, find that

    .E E E E

in out gen st

− + =

Identifying processes as per above right sketch,

(b) The total useful heat collected is

′′ ⋅qA.

u

Defining a control volume about the water tubing, the

useful heat causes an enthalpy change of the flowing water. That is,

COMMENTS: Note how the sky has been treated as large surroundings at a uniform temperature

Tsky.

PROBLEM 1.62(a)

KNOWN: Solar radiation is incident on an asphalt paving.

FIND: Relevant heat transfer processes.

SCHEMATIC:

The relevant processes shown on the schematic include:

COMMENTS: (1)

′′

q

cond

and

′′

q

conv

could be evaluated from Eqs. 1.1 and 1.3, respectively.

(2) It has been assumed that the pavement surface temperature is higher than that of the

underlying pavement and the air, in which case heat transfer by conduction and convection

are from the surface.

′′

conv

′′

cond

PROBLEM 1.62(b)

KNOWN: Physical mechanism for microwave heating.

FIND: Comparison of (i) cooking in a microwave oven with a conventional radiant or

convection oven and (ii) a microwave clothes dryer with a conventional dryer.

(i) Microwave cooking of food that contains water molecules occurs as a result of volumetric

thermal energy generation throughout the food, without heating of the food container or the

(ii) In a microwave dryer, the microwave radiation would heat the water, but not the fabric,

directly (the fabric would be heated indirectly by thermal energy transfer from the water). By

PROBLEM 1.62(c)

KNOWN: Double-pane windows with foamed insulation inside or outside. Cold, dry air outside and

warm, moist air inside.

FIND: Identify heat transfer processes. Which configuration is preferred to avoid condensation?

SCHEMATIC:

Insulation on inside of window.

ASSUMPTIONS: (1) Steady–state conditions, (2) One–dimensional heat transfer through window

and insulation.

ANALYSIS: With the insulation on the inside, heat is transferred from the warm room air to the

Cold, dry

night air

Cold, dry

night air

Exterior

pane

Warm, moist

room air

Interior

pane

Insulation

Air gap Air gap

q

conv,1

q

rad,1

q

rad,4

q

conv,4

q

cond,1

q

cond,2

q

cond,3

q

conv,2

q

conv,3

q

rad,2

q

rad,3

Moist air

Cold, dry

night air

Cold, dry

night air

Warm, moist

room air

Air gapAir gap

PROBLEM 1.62(c) (Cont.)

Condensation may occur on any surface that is exposed to moist air if the surface temperature is below

the dewpoint temperature. Condensation causes an additional heat transfer mechanism because when

water vapor condenses it releases the enthalpy of vaporization (qcondense ), which heats the surface on

which condensation is occurring. For example, if condensation occurs on the inside surface of the

window, this will increase the temperature of that surface and the rate of heat transfer through that

window pane. The condensation heat transfer processes are not shown on the schematics.

COMMENTS: (1) The potential water damage is not caused by window leakage. Any condensation

problem would be exacerbated by adding more insulation to the inside of the window. (2) The

potential for condensation damage would be reduced by lowering the humidity in the room, at the risk

PROBLEM 1.62(d)

KNOWN: Geometry of a composite insulation consisting of a honeycomb core.

FIND: Relevant heat transfer processes.

SCHEMATIC:

Heat may be transferred to the inner surface by convection and radiation, whereupon it is

transferred through the composite by

qcond,i

Conduction through the inner solid slab,

Heat may then be transferred from the outer surface by convection and radiation. Note that

for a single cell under steady state conditions,

COMMENTS: Performance would be enhanced by using materials of low thermal

conductivity, k, and emissivity, ε. Evacuating the airspace would enhance performance by

eliminating heat transfer due to free convection.

PROBLEM 1.62(e)

KNOWN: A thermocouple junction is used, with or without a radiation shield, to measure

the temperature of a gas flowing through a channel. The wall of the channel is at a

temperature much less than that of the gas.

FIND: (a) Relevant heat transfer processes, (b) Temperature of junction relative to that of

gas, (c) Effect of radiation shield.

SCHEMATIC:

ASSUMPTIONS: (1) Junction is small relative to channel walls, (2) Steady–state conditions,

(3) Negligible heat transfer by conduction through the thermocouple leads.

ANALYSIS: (a) The relevant heat transfer processes are:

(b) From a surface energy balance on the junction,

s j g

That is, the junction assumes a temperature between that of the channel wall and the gas,

thereby sensing a temperature which is less than that of the gas.

(c) The measurement error

( )

gj

TT−

is reduced by using a radiation shield as shown in the

PROBLEM 1.62(f)

KNOWN: Fireplace cavity is separated from room air by two glass plates, open at both ends.

FIND: Relevant heat transfer processes.

SCHEMATIC:

The relevant heat transfer processes associated with the double-glazed, glass fire screen are:

qrad,1

Radiation from flames and cavity wall, portions of which are absorbed and

transmitted by the two panes,

qrad,3

qconv,1

Convection between cavity gases and inner pane,

q

conv2

Convection across air space between panes,

qconv,3

qcond,1

qcond,2

COMMENTS: (1) Much of the luminous portion of the flame radiation is transmitted to the

room interior.

PROBLEM 1.62(g)

KNOWN: Thermocouple junction held in small hole in solid material by epoxy. Solid is hotter than

surroundings.

FIND: Identify heat transfer processes. Will thermocouple junction sense temperature less than,

equal to, or greater than solid temperature? How will thermal conductivity of epoxy affect junction

temperature?

SCHEMATIC:

ANALYSIS: Heat is transferred from the solid material through the epoxy to the thermocouple

junction by conduction, qcond,1 . Heat is also transferred from the junction along the thermocouple

wires and their sheathing by conduction (qcond,2 and qcond,3 ), and from there to the surroundings by

convection (qconv) and radiation (qrad). Thus, the junction is heated by the solid and cooled by the

surroundings, and its temperature will be between the solid temperature and the temperature of the

cool gases.

The junction temperature will be less than the solid temperature. <

Under steady-state conditions, the rate at which heat is transferred to the junction from the solid

COMMENTS: (1) High thermal conductivity epoxies are formulated specifically for the purpose of

affixing thermocouples. Their thermal conductivity is increased by adding small particles of high

Thermocouple

Hot solid

PROBLEM 1.63(a)

KNOWN: Room air is separated from ambient air by one or two glass panes.

FIND: Relevant heat transfer processes.

SCHEMATIC:

The relevant processes associated with single (above left schematic) and double (above right

schematic) glass panes include.

qconv,1

Convection from room air to inner surface of first pane,

qrad,1

qcond,1

qrad,s

Net radiation exchange between outer surface of first pane and inner surface of

second pane (across airspace),

qcond,2

qconv,2

COMMENTS: Heat loss from the room is significantly reduced by the double pane

construction.

PROBLEM 1.63(b)

KNOWN: Configuration of a flat plate solar collector.

FIND: Relevant heat transfer processes with and without a cover plate.

SCHEMATIC:

The relevant processes without (above left schematic) and with (above right schematic)

include:

q

S

Incident solar radiation, a large portion of which is absorbed by the absorber

plate. Reduced with use of cover plate (primarily due to reflection off cover

plate).

COMMENTS: The cover plate acts to significantly reduce heat losses by convection and

radiation from the absorber plate to the surroundings.

PROBLEM 1.63(c)

KNOWN: Configuration of a solar collector used to heat air for agricultural applications.

FIND: Relevant heat transfer processes.

SCHEMATIC:

Assume the temperature of the absorber plates exceeds the ambient air temperature. At the

cover plates, the relevant processes are:

q

conv,a–i

Convection from inside air to inner surface,

Additional processes relevant to the absorber plates and airspace are:

q

S,t

Solar radiation transmitted by cover plates,

PROBLEM 1.63(d)

KNOWN: Features of an evacuated tube solar collector.

FIND: Relevant heat transfer processes for one of the tubes.

SCHEMATIC:

The relevant heat transfer processes for one of the evacuated tube solar collectors includes:

qS

Incident solar radiation including contribution due to reflection off panel (most

is transmitted),

There is also conduction heat transfer through the inner and outer tube walls. If the walls are

thin, the temperature drop across the walls will be small.