Heat Transfer Analysis and Applications, Ejercicios de Física

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Chapter 1

INTRODUCTION AND BASIC CONCEPTS

Thermodynamics and Heat Transfer

1-1C Thermodynamics deals with the amount of heat transfer as a system undergoes a process from one equilibrium state to another. Heat transfer, on the other hand, deals with the rate of heat transfer as well as the temperature distribution within the system at a specified time.

1-2C The description of most scientific problems involves equations that relate the changes in some key variables to each other, and the smaller the increment chosen in the changing variables, the more accurate the description. In the limiting case of infinitesimal changes in variables , we obtain differential equations , which provide precise mathematical formulations for the physical principles and laws by representing the rates of changes as derivatives.

As we shall see in later chapters, the differential equations of fluid mechanics are known, but very difficult to solve except for very simple geometries. Computers are extremely helpful in this area.

1-3C (a) The driving force for heat transfer is the temperature difference. (b) The driving force for electric current flow is the electric potential difference (voltage). (a) The driving force for fluid flow is the pressure difference.

1-4C The caloric theory is based on the assumption that heat is a fluid-like substance called the "caloric" which is a massless, colorless, odorless substance. It was abandoned in the middle of the nineteenth century after it was shown that there is no such thing as the caloric.

1-5C The rating problems deal with the determination of the heat transfer rate for an existing system at a specified temperature difference. The sizing problems deal with the determination of the size of a system in order to transfer heat at a specified rate for a specified temperature difference.

1-6C The experimental approach (testing and taking measurements) has the advantage of dealing with the actual physical system, and getting a physical value within the limits of experimental error. However, this approach is expensive, time consuming, and often impractical. The analytical approach (analysis or calculations) has the advantage that it is fast and inexpensive, but the results obtained are subject to the accuracy of the assumptions and idealizations made in the analysis.

1-7C Modeling makes it possible to predict the course of an event before it actually occurs, or to study various aspects of an event mathematically without actually running expensive and time-consuming experiments. When preparing a mathematical model, all the variables that affect the phenomena are identified, reasonable assumptions and approximations are made, and the interdependence of these variables are studied. The relevant physical laws and principles are invoked, and the problem is formulated mathematically. Finally, the problem is solved using an appropriate approach, and the results are interpreted.

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1-15 A cylindrical resistor on a circuit board dissipates 1.2 W of power. The amount of heat dissipated in 24 h, the heat flux, and the fraction of heat dissipated from the top and bottom surfaces are to be determined.

Assumptions Heat is transferred uniformly from all surfaces.

Q &

Resistor 1.2 W

Analysis ( a ) The amount of heat this resistor dissipates during a 24-hour period is

Q = Q &^ ∆ t =( 1. 2 W)(24h)= 28.8Wh= 104 kJ (since 1 Wh = 3600 Ws = 3.6 kJ)

( b ) The heat flux on the surface of the resistor is

2 2 2 ( 0. 4 cm)(2cm) 0. 251 2. 513 2. 764 cm 4

( 0. 4 cm) 2 4

DL

D

As

= = = 0.434W/cm^2

1. 764 cm^2

1. 2 W

s

s (^) A

Q

q

( c ) Assuming the heat transfer coefficient to be uniform, heat transfer is proportional to the surface area. Then the fraction of heat dissipated from the top and bottom surfaces of the resistor becomes

or (9.1%)

1. 764

total

topbase total

top base = = = 0. − − A

A

Q

Q

Discussion Heat transfer from the top and bottom surfaces is small relative to that transferred from the side surface.

1-16E A logic chip in a computer dissipates 3 W of power. The amount heat dissipated in 8 h and the heat flux on the surface of the chip are to be determined.

Assumptions Heat transfer from the surface is uniform.

Analysis ( a ) The amount of heat the chip dissipates during an 8-hour period is

Logic chip Q &= 3 W

Q = Q &∆ t =( 3 W)( 8 h)= 24 Wh= 0.024 kWh

( b ) The heat flux on the surface of the chip is

= = = 37.5W/in^2

1. 08 in^2

3 W

A

Q

q

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1-17 A house is heated from 10°C to 22°C by an electric heater, and some air escapes through the cracks as the heated air in the house expands at constant pressure. The amount of heat transfer to the air and its cost are to be determined.

Assumptions 1 Air as an ideal gas with a constant specific heats at room temperature. 2 The volume occupied by the furniture and other belongings is negligible. 3 The pressure in the house remains constant at all times. 4 Heat loss from the house to the outdoors is negligible during heating. 5 The air leaks out at 22°C.

Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg ⋅ °C.

22 °C

10 °C AIR

Analysis The volume and mass of the air in the house are

V=(floor space)(height)=( 200 m^2 )(3m)= 600 m^3

1. 9 kg ( 0. 287 kPa m/kgK)(10+273.15K)

( 101. 3 kPa)(600m ) 3

3

⋅ ⋅

RT

P

m

V

Noting that the pressure in the house remains constant during heating, the amount of heat that must be transferred to the air in the house as it is heated from 10 to 22°C is determined to be

Q = mcp ( T 2 − T 1 )=( 747. 9 kg)(1.007kJ/kg⋅°C)(22− 10 )°C= 9038 kJ

Noting that 1 kWh = 3600 kJ, the cost of this electrical energy at a unit cost of $0.075/kWh is

EnegyCost=(Energyused)(Unitcostofenergy)= ( 9038 / 3600 kWh)($0.075/kWh)= $0.

Therefore, it will cost the homeowner about 19 cents to raise the temperature in his house from 10 to 22°C.

1-18 A 800 W iron is left on the ironing board with its base exposed to the air. The amount of heat the iron dissipates in 2 h, the heat flux on the surface of the iron base, and the cost of the electricity are to be determined.

Assumptions Heat transfer from the surface is uniform. Iron Analysis ( a ) The amount of heat the iron dissipates during a 2-h period is 800 W

Q = Q &∆ t =( 0. 8 kW)(2h)= 1.6 kWh

( b ) The heat flux on the surface of the iron base is

Q &base =( 0. 85 )( 800 W)= 680 W

= = = 45,300W/m^2 2 base

base

1. 015 m

680 W

A

Q

q

( c ) The cost of electricity consumed during this period is

Costofelectricity=(1.6kWh)× ($0.07/kWh)= $0.

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1-21 An electrically heated house maintained at 22°C experiences infiltration losses at a rate of 0.7 ACH. The amount of energy loss from the house due to infiltration per day and its cost are to be determined.

Assumptions 1 Air as an ideal gas with a constant specific heats at room temperature. 2 The volume occupied by the furniture and other belongings is negligible. 3 The house is maintained at a constant temperature and pressure at all times. 4 The infiltrating air exfiltrates at the indoors temperature of 22°C.

Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg ⋅ °C.

5 °C

0.7 ACH

22 °C

AIR

Analysis The volume of the air in the house is

V =(floor space)(height)=( 150 m^2 )(3m)= 450 m^3

Noting that the infiltration rate is 0.7 ACH (air changes per hour) and thus the air in the house is completely replaced by the outdoor air 0.7×24 = 16.8 times per day, the mass flow rate of air through the house due to infiltration is

8485 kg/day ( 0. 287 kPa m/kg K)(5+273.15K)

( 89. 6 kPa)(16.8 450 m /day)

(ACH )

3

3

air house air

×

×

o

o o

o RT

P

RT

P

m

V& V

Noting that outdoor air enters at 5°C and leaves at 22°C, the energy loss of this house per day is

( 8485 kg/day)(1. 007 kJ/kg.C)(22 5 )C 145 , 260 kJ/day= 40.4 kWh/day

infilt air (^ indoors outdoors) = ° − ° =

Q & = m & cp T − T

At a unit cost of $0.082/kWh, the cost of this electrical energy lost by infiltration is

EnegyCost=(Energyused)(Unitcostofenergy)= ( 40. 4 kWh/day)($0.082/kWh)= $3.31/day

1-22 The filament of a 150 W incandescent lamp is 5 cm long and has a diameter of 0.5 mm. The heat flux on the surface of the filament, the heat flux on the surface of the glass bulb, and the annual electricity cost of the bulb are to be determined.

Assumptions Heat transfer from the surface of the filament and the bulb of the lamp is uniform.

Analysis ( a ) The heat transfer surface area and the heat flux on the surface of the filament are

A =π DL =π( 0. 05 cm)( 5 cm)= 0. 785 cm^2

s

= = =^2 = 1.91 × 106 W/m^2

1. 785 cm^2 191 W/cm

150 W

s

s A

Q

q

( b ) The heat flux on the surface of glass bulb is

A =π D^2 =π( 8 cm)^2 = 201. 1 cm^2

s

= = =^2 = 7500 W/m^2

1. 1 cm^2 0.^75 W/cm

150 W

s

s (^) A

Q

q

( c ) The amount and cost of electrical energy consumed during a one-year period is

Q &

Lamp 150 W

= $35.04/yr

= ∆ = × =

AnnualCost=(438kWh/yr)($0.08/ kWh)

Electricity Consumption Q & t ( 0. 15 kW)(365 8 h/yr) 438 kWh/yr

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1-23 Water is heated in an insulated tube by an electric resistance heater. The mass flow rate of water through the heater is to be determined.

Assumptions 1 Water is an incompressible substance with a constant specific heat. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Heat loss from the insulated tube is negligible.

Properties The specific heat of water at room temperature is cp = 4.18 kJ/kg·°C.

Analysis We take the tube as the system. This is a control volume since mass crosses the system boundary during the process. We observe that this is a steady-flow process since there is no change with time at any point and thus ∆ m (^) CV = 0 and∆ E CV= 0 , there is only one inlet and one exit and thus m &^1 (^) = m & 2 = m &, and the tube is insulated. The energy

balance for this steady-flow system can be expressed in the rate form as

(since ke pe 0)

e,in 2 1

e,in 1 2

potential,etc.energies

Rateofchangeininternal,kinetic,

0 (steady) system byheat,work,andmass

Rateofnetenergy transfer

W mc T T

W mh mh

E E E E E

p

in out in out

& & Ê

15 °C

WATER

60 °C

5 kW Thus,

= 0.0266 kg/s ⋅° − °

(4.18kJ/kg C)( 60 15 )C

5 kJ/s ( 2 1 )

e,in c T T

W

m p

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1-25 Liquid water is to be heated in an electric teapot. The heating time is to be determined.

Assumptions 1 Heat loss from the teapot is negligible. 2 Constant properties can be used for both the teapot and the water.

Properties The average specific heats are given to be 0.7 kJ/kg·K for the teapot and 4.18 kJ/kg·K for water.

Analysis We take the teapot and the water in it as the system, which is a closed system (fixed mass). The energy balance in this case can be expressed as

in system water teapot

potential, etc.energies

Changeininternal,kinetic,

system

by heat,work,andmass

Netenergy transfer

E U U U

E (^) in Eout E

Then the amount of energy needed to raise the temperature of water and the teapot from 15°C to 95°C is

1. 3 kJ

(1.2kg)(4.18kJ/kg C)(95 15 )C (0.5kg)(0.7kJ/kg C)(95 15 )C

in (^ )water ( )teapot

E = mc ∆ T + mc ∆ T

The 1200-W electric heating unit will supply energy at a rate of 1.2 kW or 1.2 kJ per second. Therefore, the time needed for this heater to supply 429.3 kJ of heat is determined from

∆ = = = = 358 s= 6.0 min

1. 2 kJ/s

2. 3 kJ Rateofenergy transfer

Totalenergy transferred transfer

in E

E

t &

Discussion In reality, it will take more than 6 minutes to accomplish this heating process since some heat loss is inevitable during heating. Also, the specific heat units kJ/kg · °C and kJ/kg · K are equivalent, and can be interchanged.

1-26 It is observed that the air temperature in a room heated by electric baseboard heaters remains constant even though the heater operates continuously when the heat losses from the room amount to 9000 kJ/h. The power rating of the heater is to be determined.

Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - 141 °C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 The temperature of the room remains constant during this process.

Analysis We take the room as the system. The energy balance in this case reduces to

We

AIR

ein out

ein out

in out

W Q

W Q U

E E E

,

,

potential, etc.energies

Changeininternal,kinetic,

system by heat,work,andmass

Netenergy transfer

0

since ∆ U = mc v∆ T = 0 for isothermal processes of ideal gases. Thus,

⎟= 2.5 kW ⎠

3600 kJ/h

1 kW W & e (^) , in Q & out 9000 kJ/h

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1-27 A room is heated by an electrical resistance heater placed in a short duct in the room in 15 min while the room is losing heat to the outside, and a 300-W fan circulates the air steadily through the heater duct. The power rating of the electric heater and the temperature rise of air in the duct are to be determined.

Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - 141 °C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications. 3 Heat loss from the duct is negligible. 4 The house is air-tight and thus no air is leaking in or out of the room.

Properties The gas constant of air is R = 0.287 kPa.m^3 /kg.K (Table A-1). Also, cp = 1.007 kJ/kg·K for air at room temperature (Table A-15) and c v = cp – R = 0.720 kJ/kg·K.

Analysis ( a ) We first take the air in the room as the system. This is a constant volume closed system since no mass crosses the system boundary. The energy balance for the room can be expressed as

( (^) e,in fan,in out) ( 2 1 ) ( 2 1 )

e,in fan,in out

potential,etc.energies

Changeininternal,kinetic,

system byheat,work,andmass Netenergy transfer

W W Q t mu u mc T T

W W Q U

E E E

v

in out

5 × 6 ×8 m^3

We

300 W

200 kJ/min

The total mass of air in the room is

284.6 kg (0.287kPam/kgK)( 288 K)

( 98 kPa)( 240 m )

5 6 8 m 240 m

3

3

1

1

3 3

= × × =

RT

P

m

V

V

Then the power rating of the electric heater is determined to be

= − + ⋅° − ° × = 4.93 kW

(200/60kJ/s) (0.3kJ/s) (284.6kg)(0.720kJ/kg C)( 25 15 C)/(18 60 s)

W &e, in Q &out W^ &fan,in mc v( T 2 T 1 )/ t

( b ) The temperature rise that the air experiences each time it passes through the heater is determined by applying the energy balance to the duct,

W W m h mc T

W W mh Q mh

E E

p

in out

e,in fan, in

2

0 e,in fan,in 1 out (since ke pe 0)

Ê

Thus,

= 6.2 ° C

(50/60kg/s)(1.007kJ/kg K)

e,in fan,in (4.93 0.3)kJ/s mc p

W W

T

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1-29 Air is moved through the resistance heaters in a 900-W hair dryer by a fan. The volume flow rate of air at the inlet and the velocity of the air at the exit are to be determined.

Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - 141 °C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at room temperature can be used for air. 4 The power consumed by the fan and the heat losses through the walls of the hair dryer are negligible.

Properties The gas constant of air is R = 0.287 kPa.m^3 /kg.K (Table A-1). Also, cp = 1.007 kJ/kg·K for air at room temperature (Table A-15).

Analysis ( a ) We take the hair dryer as the system. This is a control volume since mass crosses the system boundary during the process. We observe that this is a steady-flow process since there is no change with time at any point and thus ∆ m (^) CV = 0 and∆ E CV= 0 , and there is only one inlet and one exit and thus m &^1 (^) = m & 2 = m &. The energy balance for this

steady-flow system can be expressed in the rate form as

(since ke pe 0)

e,in 2 1

2

0 1

0 e,in fan,in

potential, etc.energies

Rateofchangeininternal,kinetic,

0 (steady) system by heat,work,andmass

Rateofnetenergy transfer

W mc T T

W W mh Q mh

E E E E E

p

out

in out in out

Ê Ê

Ê

T 2 = 50°C

A 2 = 60 cm^2

P 1 = 100 kPa T 1 = 25°C

W e = 900 W

Thus,

( )

0.03575 kg/s (1.007kJ/kg C)( 50 25 )C

0.9kJ/s

2 1

e,in

c T T

W

m p

Then,

= = = 0.0306 m^3 /s

(0.03575kg/s)(0.8553m/kg )

0.8553m/kg 100 kPa

(0.287kPa m/kg K)( 298 K)

3 1 1

3 3

1

1 1

V v

v

m

P

RT

( b ) The exit velocity of air is determined from the conservation of mass equation,

= 5.52 m/s ×

− 4 2

3

2

2 2 2 2 2

3

3

2

2 2

60 10 m

1 (0.03575kg/s)(0.9270m/kg)

0.9270m/kg 100 kPa

(0.287kPa m/kgK)( 323 K)

A

m m AV V

P

RT

v v

v

& &

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1-30 The ducts of an air heating system pass through an unheated area, resulting in a temperature drop of the air in the duct. The rate of heat loss from the air to the cold environment is to be determined.

Assumptions 1 Air is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of - 141 °C and 3.77 MPa. 2 The kinetic and potential energy changes are negligible, ∆ke ≅ ∆pe ≅ 0. 3 Constant specific heats at room temperature can be used for air. This assumption results in negligible error in heating and air-conditioning applications.

Properties The specific heat of air at room temperature is cp = 1.007 kJ/kg·°C (Table A-15).

Analysis We take the heating duct as the system. This is a control volume since mass crosses the system boundary during the process. We observe that this is a steady-flow process since there is no change with time at any point and thus ∆ m (^) CV = 0 and∆ E CV= 0. Also, there is only one inlet and one exit and thus m &^1 (^) = m & 2 = m &. The energy balance for this

steady-flow system can be expressed in the rate form as

(since ke pe 0)

out 1 2

1 2

potential, etc.energies

Rateofchangeininternal,kinetic,

0 (steady) system by heat,work,andmass

Rateofnetenergy transfer

Q mc T T

mh Q mh

E E E E E

p

out

in out in out

& & Ê

90 kg/min AIR

Q

Substituting,

Q &out (^) = m & cp ∆ T = ( 90 kg/min)(1.007 kJ/kg⋅°C)( 3 °C) = 272 kJ/min

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1-32 A classroom is to be air-conditioned using window air-conditioning units. The cooling load is due to people, lights, and heat transfer through the walls and the windows. The number of 5-kW window air conditioning units required is to be determined.

Assumptions There are no heat dissipating equipment (such as computers, TVs, or ranges) in the room.

Analysis The total cooling load of the room is determined from

Q &^ cooling (^) = Q &lights+ Q &people+ Q &heat gain

Room 50 people 10 bulbs

12 ,000 kJ/h

where

Qcool ·

12,000kJ/h 3 .33 kW

50 360 kJ/h 18 , 000 kJ/h 5 kW

10 100 W 1 kW

heatgain

people

lights

= × = =

= × =

Q

Q

Q

Substituting,

Q &cooling = 1 + 5 + 3. 33 = 9 .33 kW

Thus the number of air-conditioning units required is

1. 87 2 units 5 kW/unit

9.33 kW = ⎯⎯→

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Heat Transfer Mechanisms

1-33C Diamond is a better heat conductor.

1-34C The thermal conductivity of a material is the rate of heat transfer through a unit thickness of the material per unit area and per unit temperature difference. The thermal conductivity of a material is a measure of how fast heat will be conducted in that material.

1-35C The mechanisms of heat transfer are conduction, convection and radiation. Conduction is the transfer of energy from the more energetic particles of a substance to the adjacent less energetic ones as a result of interactions between the particles. Convection is the mode of energy transfer between a solid surface and the adjacent liquid or gas which is in motion, and it involves combined effects of conduction and fluid motion. Radiation is energy emitted by matter in the form of electromagnetic waves (or photons) as a result of the changes in the electronic configurations of the atoms or molecules.

1-36C In solids, conduction is due to the combination of the vibrations of the molecules in a lattice and the energy transport by free electrons. In gases and liquids, it is due to the collisions of the molecules during their random motion.

1-37C The parameters that effect the rate of heat conduction through a windowless wall are the geometry and surface area of wall, its thickness, the material of the wall, and the temperature difference across the wall.

1-38C Conduction is expressed by Fourier's law of conduction as dx

dT Q^ &^ cond=− kA where dT/dx is the temperature gradient,

k is the thermal conductivity, and A is the area which is normal to the direction of heat transfer.

Convection is expressed by Newton's law of cooling as where h is the convection heat

transfer coefficient, A

Q^ &conv (^) = hAs ( Ts − T ∞ ) s is the surface area through which convection heat transfer takes place,^ Ts is the surface temperature and T ∞ is the temperature of the fluid sufficiently far from the surface.

Radiation is expressed by Stefan-Boltzman law as Q & rad =εσ As ( T s^4 − T surr^4 )where ε is the emissivity of surface, As

is the surface area, Ts is the surface temperature, T surr is the average surrounding surface temperature and

σ= 5. 67 × 10 −^8 W/m^2 ⋅K^4 is the Stefan-Boltzman constant.

1-39C Convection involves fluid motion, conduction does not. In a solid we can have only conduction.

1-40C No. It is purely by radiation.

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1-49C Superinsulations are obtained by using layers of highly reflective sheets separated by glass fibers in an evacuated space. Radiation heat transfer between two surfaces is inversely proportional to the number of sheets used and thus heat loss by radiation will be very low by using this highly reflective sheets. At the same time, evacuating the space between the layers forms a vacuum under 0.000001 atm pressure which minimize conduction or convection through the air space between the layers.

1-50C Most ordinary insulations are obtained by mixing fibers, powders, or flakes of insulating materials with air. Heat transfer through such insulations is by conduction through the solid material, and conduction or convection through the air space as well as radiation. Such systems are characterized by apparent thermal conductivity instead of the ordinary thermal conductivity in order to incorporate these convection and radiation effects.

1-51C The thermal conductivity of an alloy of two metals will most likely be less than the thermal conductivities of both metals.

1-52 Two surfaces of a flat plate are maintained at specified temperatures, and the rate of heat transfer through the plate is measured. The thermal conductivity of the plate material is to be determined.

Assumptions 1 Steady operating conditions exist since the surface temperatures of the plate Plate remain constant at the specified values. 2 Heat transfer through the plate is one- dimensional. 3 Thermal properties of the plate are constant. Q Analysis The thermal conductivity is determined directly from the steady one-dimensional heat conduction relation to be

= 0.125 W/m ⋅° C − °

(80 0)C

( 500 W/m )(0.02m) ( )

( / )^2

1 2

1 2 T T

Q AL

k L

T T

Q kA

& 80 °C 0 °C

1-53 Four power transistors are mounted on a thin vertical aluminum plate that is cooled by a fan. The temperature of the aluminum plate is to be determined.

Assumptions 1 Steady operating conditions exist. 2 The entire plate is nearly isothermal. 3 Thermal properties of the wall are constant. 4 The exposed surface area of the transistor can be taken to be equal to its base area. 5 Heat transfer by radiation is disregarded. 6 The convection heat transfer coefficient is constant and uniform over the surface.

Analysis The total rate of heat dissipation from the aluminum plate and the total heat transfer area are

( 0. 22 m)(0.22m) 0. 0484 m^2

4 12 W 48 W

= × =

A s

Q &

12 W

Ts Disregarding any radiation effects, the temperature of the aluminum plate is determined to be

= 64.7 ° C

( 25 W/m C)( 0. 0484 m )

48 W

( ) 25 C 2 2

s

s s s hA

Q

Q hA T T T T

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1-54 The convection heat transfer coefficient heat transfer between the surface of a pipe carrying superheated vapor and the surrounding is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Heat transfer by radiation is not considered. 3 Rate of heat loss from the vapor in the pipe is equal to the heat transfer rate by convection between pipe surface and the surrounding.

Properties The specific heat of vapor is given to be 2190 J/kg · °C.

Analysis The surface area of the pipe is

A =π DL =π( 0. 05 m)( 10 m)= 1. 571 m^2

s

The rate of heat loss from the vapor in the pipe can be determined from

19710 W

( 0. 3 kg/s)( 2190 J/kg C)( 30 ) C 19710 J/s

loss (^ in out)

Q & = m & cpT − T

With the rate of heat loss from the vapor in the pipe assumed equal to the heat transfer rate by convection, the heat transfer coefficient can be determined using the Newton’s law of cooling:

Q &loss (^) = Q^ &conv= hAs ( Ts − T ∞ )

Rearranging, the heat transfer coefficient is determined to be

= 157 W/m^2 ⋅° C − °

∞ (^1.^571 m )(^10020 ) C

19710 W

( )^2

loss A T T

Q

h s s

Discussion By insulating the pipe surface, heat loss from the vapor in the pipe can be reduced.

1-55 An electrical resistor with a uniform temperature of 90 °C is in a room at 20 °C. The heat transfer coefficient by convection is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Radiation heat transfer is negligible. 3 No hot spot exists on the resistor.

Analysis The total heat transfer area of the resistor is

A = 2 ( π D^2 / 4 )+π DL = 2 π( 0. 025 m)^2 / 4 +π( 0. 025 m)( 0. 15 m)= 0. 01276 m^2

s

The electrical energy converted to thermal energy is transferred by convection:

Q &conv = IV =( 5 A)( 6 V)= 30 W

From Newton’s law of cooling, the heat transfer by convection is given as

Q^ &conv (^) = hAs ( Ts − T ∞ )

Rearranging, the heat transfer coefficient is determined to be

= 33.6 W/m^2 ⋅° C − °

∞ (^0.^01276 m )(^9020 ) C

30 W

( )^2

conv A T T

Q

h s s

Discussion By comparing the magnitude of the heat transfer coefficient determined here with the values presented in Table 1-5, one can conclude that it is likely that forced convection is taking place rather than free convection.