Fundamentals of Heat and Mass Transfer - CH01 (41-73).pdf

PROBLEM 1.41

KNOWN: Hot plate-type wafer thermal processing tool based upon heat transfer modes by conduction through gas within the gap and by radiation exchange across gap. FIND: (a) Radiative and conduction heat fluxes across gap for specified hot plate and wafer temperatures and gap separation; initial time rate of change in wafer temperature for each mode, and (b) heat fluxes and initial temperature-time change for gap separations of 0.2, 0.5 and 1.0 mm for hot plate temperatures 300 Th 1300 C. Comment on the relative importance of the modes and the influence of the gap distance. Under what conditions could a wafer be heated to 900 C in less than 10 seconds? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions for flux calculations, (2) Diameter of hot plate and wafer much larger than gap spacing, approximating plane, infinite planes, (3) One-dimensional conduction through gas, (4) Hot plate and wafer are blackbodies, (5) Negligible heat losses from wafer backside, and (6) Wafer temperature is uniform at the onset of heating. PROPERTIES: Wafer: = 2700 kg/m , c = 875 J/kgK; Gas in gap: k = 0.0436 W/mK. ANALYSIS: (a) The radiative heat flux between the hot plate and wafer for Th = 600 C and Tw = 20 C follows from the rate equation,

4 4 q = Th - Tw = 5.67 10-8 W / m 2 K 4 rad

3

(

)

((600 + 273)

4

- ( 20 + 273)

4

)K

4

= 32.5 kW / m2

The conduction heat flux through the gas in the gap with L = 0.2 mm follows from Fourier's law,

(600 - 20) K = 126 kW / m2 Th - Tw q = 0.0436 W / m K cond = k L 0.0002 m

The initial time rate of change of the wafer can be determined from an energy balance on the wafer at the instant of time the heating process begins,

Ein - E = E out st

dT E = c d w st dt i

where E = 0 and E = q or q . Substituting numerical values, find out in rad cond

dTw dt dTw dt

q 32.5 103 W / m 2 = rad = = 17.6 K / s i,rad cd 2700 kg / m3 875 J / kg K 0.00078 m q = cond = 68.4 K / s cd i,cond

Continued .

PROBLEM 1.41 (Cont.)

(b) Using the foregoing equations, the heat fluxes and initial rate of temperature change for each mode can be calculated for selected gap separations L and range of hot plate temperatures Th with Tw = 20 C.

400

Initial rate of change, dTw/dt (K.s -1)

200

300 Heat flux (kW/m 2)

150

200

100

100

50

0 300 500 700 900 1100 1300

0 300 500 700 900 1100 1300

Hot plate temperature, Th (C)

Hot plate temperature, Th (C) q'rad q'cond, L = 1.0 mm q'cond, L = 0.5 mm q'cond, L = 0.2 mm

q'rad q'cond, L = 1.0 m m q'cond, L = 0.5 m m q'cond, L = 0.2 m m

In the left-hand graph, the conduction heat flux increases linearly with Th and inversely with L as expected. The radiative heat flux is independent of L and highly non-linear with Th, but does not approach that for the highest conduction heat rate until Th approaches 1200 C. The general trends for the initial temperature-time change, (dTw/dt)i, follow those for the heat fluxes. To reach 900 C in 10 s requires an average temperature-time change rate of 90 K/s. Recognizing that (dTw/dt) will decrease with increasing Tw, this rate could be met only with a very high Th and the smallest L.

PROBLEM 1.42

KNOWN: Silicon wafer, radiantly heated by lamps, experiencing an annealing process with known backside temperature. FIND: Whether temperature difference across the wafer thickness is less than 2 C in order to avoid damaging the wafer. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wafer, (3) Radiation exchange between upper surface of wafer and surroundings is between a small object and a large enclosure, and (4) Vacuum condition in chamber, no convection. PROPERTIES: Wafer: k = 30 W/mK, = = 0.65. ANALYSIS: Perform a surface energy balance on the upper surface of the wafer to determine Tw,u . The processes include the absorbed radiant flux from the lamps, radiation exchange with the chamber walls, and conduction through the wafer.

E - E = 0 in out

q - q - q = 0 s rad cd

4 4 q - Tw,u - Tsur - k s

(

)

Tw,u - Tw, L

=0

4 0.65 3.0 105 W / m 2 - 0.65 5.67 10-8 W / m 2 K 4 Tw,u - ( 27 + 273)

(

4

)K

4

-30W / m K Tw,u - (997 + 273) K / 0.00078 m = 0 Tw,u = 1273K = 1000 C

COMMENTS: (1) The temperature difference for this steady-state operating condition, Tw,u - Tw,l , is larger than 2 C. Warping of the wafer and inducing slip planes in the crystal structure could occur. (2) The radiation exchange rate equation requires that temperature must be expressed in kelvin units. Why is it permissible to use kelvin or Celsius temperature units in the conduction rate equation? (3) Note how the surface energy balance, Eq. 1.12, is represented schematically. It is essential to show the control surfaces, and then identify the rate processes associated with the surfaces. Make sure the directions (in or out) of the process are consistent with the energy balance equation.

PROBLEM 1.43

KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces exposed to hot and cool zones, respectively. FIND: (a) Initial rate of change of the wafer temperature corresponding to the wafer temperature Tw,i = 300 K, and (b) Steady-state temperature reached if the wafer remains in this position. How significant is convection for this situation? Sketch how you'd expect the wafer temperature to vary as a function of vertical distance. SCHEMATIC:

ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Transient conditions when wafer is initially positioned, (3) Hot and cool zones have uniform temperatures, (3) Radiation exchange is between small surface (wafer) and large enclosure (chamber, hot or cold zone), and (4) Negligible heat losses from wafer to mounting pin holder. ANALYSIS: The energy balance on the wafer illustrated in the schematic above includes convection from the upper (u) and lower (l) surfaces with the ambient gas, radiation exchange with the hot- and cool-zone (chamber) surroundings, and the rate of energy storage term for the transient condition.

E - E = E in out st

q rad,h + q rad,c - q cv,u - q = cd cv,l

d Tw dt d Tw dt

4 4 4 4 Tsur,h - Tw + Tsur,c - Tw - h u (Tw - T ) - h l (Tw - T ) = cd

(

) (

(

)

(a) For the initial condition, the time rate of temperature change of the wafer is determined using the energy balance above with Tw = Tw,i = 300 K, 0.65 5.67 10-8 W / m 2 K 4 15004 - 3004 K 4 + 0.65 5.67 10-8 W / m 2 K 4 3304 - 3004 K 4

)

(

)

-8 W / m 2 K (300 - 700 ) K - 4 W / m 2 K (300 - 700 ) K = 2700 kg / m3 875 J / kg K 0.00078 m ( d Tw / dt )i

(d Tw / dt )i = 104 K / s

(b) For the steady-state condition, the energy storage term is zero, and the energy balance can be solved for the steady-state wafer temperature, Tw = Tw,ss .

Continued .

PROBLEM 1.43 (Cont.)

4 4 0.65 15004 - Tw,ss K 4 + 0.65 3304 - Tw,ss K 4

(

)

(

)

-8 W / m 2 K Tw,ss - 700 K - 4 W / m 2 K Tw,ss - 700 K = 0 Tw,ss = 1251 K

(

)

(

)

To determine the relative importance of the convection processes, re-solve the energy balance above ignoring those processes to find ( d Tw / dt )i = 101 K / s and Tw,ss = 1262 K. We conclude that the radiation exchange processes control the initial time rate of temperature change and the steady-state temperature. If the wafer were elevated above the present operating position, its temperature would increase, since the lower surface would begin to experience radiant exchange with progressively more of the hot zone chamber. Conversely, by lowering the wafer, the upper surface would experience less radiant exchange with the hot zone chamber, and its temperature would decrease. The temperature-distance trend might appear as shown in the sketch.

PROBLEM 1.44 KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive wastes. Surface convection conditions. FIND: Total energy generation rate and surface temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature drop across thin container wall. ANALYSIS: The rate of energy generation is

r 2 Eg = qdV=qo o 1- ( r/ro ) 2 rLdr 0 = 2 Lq r 2 / 2 - r 2 / 4 o o Eg o

(

)

or per unit length, q r E = o o . g 2

2

Performing an energy balance for a control surface about the container yields, at an instant, E - E = 0 g out and substituting for the convection heat rate per unit length, 2 qo ro = h ( 2 ro )( Ts - T ) 2 Ts = T + q o ro . 4h

COMMENTS: The temperature within the radioactive wastes increases with decreasing r from Ts at ro to a maximum value at the centerline.

PROBLEM 1.45

KNOWN: Rod of prescribed diameter experiencing electrical dissipation from passage of electrical current and convection under different air velocity conditions. See Example 1.3. FIND: Rod temperature as a function of the electrical current for 0 I 10 A with convection 2 coefficients of 50, 100 and 250 W/m K. Will variations in the surface emissivity have a significant effect on the rod temperature? SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform rod temperature, (3) Radiation exchange between the outer surface of the rod and the surroundings is between a small surface and large enclosure. ANALYSIS: The energy balance on the rod for steady-state conditions has the form,

q conv + q = E rad gen

4 Dh (T - T ) + D T 4 - Tsur = I 2R e Using this equation in the Workspace of IHT, the rod temperature is calculated and plotted as a function of current for selected convection coefficients.

150

(

)

125 R o d te m p e ra tu re , T (C )

100

75

50

25

0 0 2 4 6 8 10

C u rre n t, I (a m p e re s ) h = 5 0 W /m 2 .K h = 1 0 0 W /m 2 .K h = 2 5 0 W /m 2 .K

COMMENTS: (1) For forced convection over the cylinder, the convection heat transfer coefficient is 0.6 dependent upon air velocity approximately as h V . Hence, to achieve a 5-fold change in the 2 convection coefficient (from 50 to 250 W/m K), the air velocity must be changed by a factor of nearly 15. Continued .

PROBLEM 1.45 (Cont.)

(2) For the condition of I = 4 A with h = 50 W/m K with T = 63.5 C, the convection and radiation exchange rates per unit length are, respectively, q = 5.7 W / m and q = 0.67 W / m. We conclude cv rad that convection is the dominate heat transfer mode and that changes in surface emissivity could have 2 only a minor effect. Will this also be the case if h = 100 or 250 W/m K? (3) What would happen to the rod temperature if there was a "loss of coolant" condition where the air flow would cease? (4) The Workspace for the IHT program to calculate the heat losses and perform the parametric analysis to generate the graph is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results. It is also good practice to show plots in customary units, that is, the units used to prescribe the problem. As such the graph of the rod temperature is shown above with Celsius units, even though the calculations require temperatures in kelvins.

2

// Energy balance; from Ex. 1.3, Comment 1 -q'cv - q'rad + Edot'g = 0 q'cv = pi*D*h*(T - Tinf) q'rad = pi*D*eps*sigma*(T 4 - Tsur 4) sigma = 5.67e-8 // The generation term has the form Edot'g = I 2*R'e qdot = I 2*R'e / (pi*D 2/4) // Input parameters D = 0.001 Tsur = 300 T C = T 273 eps = 0.8 Tinf = 300 h = 100 //h = 50 //h = 250 I = 5.2 //I = 4 R'e = 0.4

// Representing temperature in Celsius units using C subscript

// Values of coefficient for parameter study // For graph, sweep over range from 0 to 10 A // For evaluation of heat rates with h = 50 W/m 2.K

/* Base case results: I = 5.2 A with h = 100 W/m 2.K, find T = 60 C (Comment 2 case). Edot'g T T C q'cv q'rad qdot D I R'e Tinf Tsur eps h sigma 10.82 332.6 59.55 10.23 0.5886 1.377E7 0.001 5.2 0.4 300 300 0.8 100 5.67E-8 */ /* Results: I = 4 A with h = 50 W/m 2.K, find q'cv = 5.7 W/m and q'rad = 0.67 W/m Edot'g T T C q'cv q'rad qdot D I R'e Tinf Tsur eps h sigma 6.4 336.5 63.47 5.728 0.6721 8.149E6 0.001 4 0.4 300 300 0.8 50 5.67E-8 */

PROBLEM 1.46

KNOWN: Long bus bar of prescribed diameter and ambient air and surroundings temperatures. Relations for the electrical resistivity and free convection coefficient as a function of temperature. FIND: (a) Current carrying capacity of the bus bar if its surface temperature is not to exceed 65 C; compare relative importance of convection and radiation exchange heat rates, and (b) Show graphically the operating temperature of the bus bar as a function of current for the range 100 I 5000 A for bus-bar diameters of 10, 20 and 40 mm. Plot the ratio of the heat transfer by convection to the total heat transfer for these conditions. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Bus bar and conduit are very long in direction normal to page, (3) Uniform bus-bar temperature, (4) Radiation exchange between the outer surface of the bus bar and the conduit is between a small surface and a large enclosure. PROPERTIES: Bus-bar material, e = e,o [1 + ( T - To )] , e,o = 0.0171 m, To = 25 C,

= 0.00396 K -1.

ANALYSIS: An energy balance on the bus-bar for a unit length as shown in the schematic above has the form

E - E + E = 0 in out gen

2 -q - q rad conv + I R = 0 e 4 - D T 4 - Tsur - h D ( T - T ) + I2 e / Ac = 0 where R = e / A c and A c = D 2 / 4. Using the relations for e ( T ) and h ( T, D ) , and substituting e numerical values with T = 65 C, find q = 0.85 ( 0.020m ) 5.67 10-8 W / m 2 K 4 [65 + 273] - [30 + 273] rad

4

(

)

(

4

)K

4

= 223 W / m

2 q conv = 7.83W / m K ( 0.020m )( 65 - 30 ) K = 17.2 W / m -0.25 where h = 1.21W m -1.75 K -1.25 ( 0.020m ) (65 - 30 )0.25 = 7.83 W / m2 K

I2 R = I2 198.2 10-6 m / (0.020 ) m 2 / 4 = 6.31 10-5 I2 W / m e

2 where

(

)

e = 0.0171 10-6 m 1 + 0.00396 K -1 (65 - 25) K = 198.2 m

I = 1950 A q / ( q + q ) = q / qtot = 0.072 cv cv rad cv

The maximum allowable current capacity and the ratio of the convection to total heat transfer rate are

For this operating condition, convection heat transfer is only 7.2% of the total heat transfer. (b) Using these equations in the Workspace of IHT, the bus-bar operating temperature is calculated and plotted as a function of the current for the range 100 I 5000 A for diameters of 10, 20 and 40 mm. Also shown below is the corresponding graph of the ratio (expressed in percentage units) of the heat transfer by convection to the total heat transfer, q / q tot . cv Continued .

PROBLEM 1.46 (Cont.)

100 Ba r te m p e ra tu re , Ts (C ) 80 60 40 20 0 1000 2000 3000 4000 5000 C u rre n t, I (A) D = 10 m m D = 20 m m D = 40 m m

13 Ratio q'cv / q'tot, (%) 11 9 7 5 3 1 20 40 60 80 100 Bus bar temperature, T (C) D = 10 mm D = 20 mm D = 40 mm

COMMENTS: (1) The trade-off between current-carrying capacity, operating temperature and bar diameter is shown in the first graph. If the surface temperature is not to exceed 65 C, the maximum current capacities for the 10, 20 and 40-mm diameter bus bars are 960, 1950, and 4000 A, respectively. (2) From the second graph with q / q tot vs. T, note that the convection heat transfer rate is always a cv small fraction of the total heat transfer. That is, radiation is the dominant mode of heat transfer. Note also that the convection contribution increases with increasing diameter. (3) The Workspace for the IHT program to perform the parametric analysis and generate the graphs is shown below. It is good practice to provide commentary with the code making your solution logic clear, and to summarize the results.

/* Results: base-case conditions, Part (a) I R'e cvovertot hbar q'cv Tsur C eps 1950 6.309E-5 7.171 7.826 17.21 30 0.85 */ q'rad 222.8 rhoe D Tinf C 30 Ts C 65

1.982E-8 0.02

// Energy balance, on a per unit length basis; steady-state conditions // Edot'in - Edot'out + Edot'gen = 0 -q'cv - q'rad + Edot'gen = 0 q'cv = hbar * P * (Ts - Tinf) P = pi * D q'rad = eps * sigma * (Ts 4 - Tsur 4) sigma = 5.67e-8 Edot'gen = I 2 * R'e R'e = rhoe / Ac rhoe = rhoeo * (1 + alpha * (Ts - To) ) To = 25 + 273 Ac = pi * D 2 / 4 // Convection coefficient hbar = 1.21 * (D -0.25) * (Ts - Tinf) 0.25 // Convection vs. total heat rates cvovertot = q'cv / (q'cv + q'rad) * 100 // Input parameters D = 0.020 // D = 0.010 // D = 0.040 // I = 1950 rhoeo = 0.01711e-6 alpha = 0.00396 Tinf C = 30 Tinf = Tinf C + 273 Ts C = 65 Ts = Ts C + 273 Tsur C = 30 Tsur = Tsur C + 273 eps = 0.85

// Compact convection coeff. correlation

// Values of diameter for parameter study // Base case condition unknown

// Base case condition to determine current

PROBLEM 1.47

KNOWN: Elapsed times corresponding to a temperature change from 15 to 14 C for a reference sphere and test sphere of unknown composition suddenly immersed in a stirred water-ice mixture. Mass and specific heat of reference sphere. FIND: Specific heat of the test sphere of known mass. SCHEMATIC:

ASSUMPTIONS: (1) Spheres are of equal diameter, (2) Spheres experience temperature change from 15 to 14 C, (3) Spheres experience same convection heat transfer rate when the time rates of surface temperature are observed, (4) At any time, the temperatures of the spheres are uniform, (5) Negligible heat loss through the thermocouple wires. PROPERTIES: Reference-grade sphere material: cr = 447 J/kg K. ANALYSIS: Apply the conservation of energy requirement at an instant of time, Eq. 1.11a, after a sphere has been immersed in the ice-water mixture at T.

Ein - E out = Est -q conv = Mc dT dt

where q conv = h A s ( T - T ). Since the temperatures of the spheres are uniform, the change in

energy storage term can be represented with the time rate of temperature change, dT/dt. The convection heat rates are equal at this instant of time, and hence the change in energy storage terms for the reference (r) and test (t) spheres must be equal.

M r cr

dT dT = M t ct dt r dt t

Approximating the instantaneous differential change, dT/dt, by the difference change over a short period of time, T/t, the specific heat of the test sphere can be calculated.

0.515 kg 447 J / kg K c t = 132 J / kg K

(15 - 14 ) K

6.35s

= 1.263kg c t

(15 - 14 ) K

4.59s

COMMENTS: Why was it important to perform the experiments with the reference and test spheres over the same temperature range (from 15 to 14 C)? Why does the analysis require that the spheres have uniform temperatures at all times?

PROBLEM 1.48

KNOWN: Inner surface heating and new environmental conditions associated with a spherical shell of prescribed dimensions and material. FIND: (a) Governing equation for variation of wall temperature with time. Initial rate of temperature change, (b) Steady-state wall temperature, (c) Effect of convection coefficient on canister temperature. SCHEMATIC:

ASSUMPTIONS: (1) Negligible temperature gradients in wall, (2) Constant properties, (3) Uniform, time-independent heat flux at inner surface. PROPERTIES: Table A.1, Stainless Steel, AISI 302: = 8055 kg/m3, c p = 510 J/kgK.

ANALYSIS: (a) Performing an energy balance on the shell at an instant of time, Ein - E out = Est . Identifying relevant processes and solving for dT/dt, 4 dT 2 3 q 4 ri2 -; ro ( T - T ) = ro - ri3 cp i 3 dt

(

) ( (

)

(

)

dT 3 q r 2 - hr 2 ( T - T ) . = o 3 - r3 i i dt c r p o i

)

Substituting numerical values for the initial condition, find

dT dt i

W W 3 105 (0.5m )2 - 500 2 (0.6m )2 (500 - 300 ) K m2 m K = kg J 8055 510 (0.6 )3 - (0.5 )3 m3 3 kg K m

(b) Under steady-state conditions with Est = 0, it follows that 2 q 4 ri2 =; ro ( T - T ) i

dT = -0.089 K/s . dt i

(

) (

)

Continued .

PROBLEM 1.48 (Cont.)

2 q ri 105 W/m2 0.5m i T = T + = 300K + = 439K h ro 500W/m 2 K 0.6m 2

(c) Parametric calculations were performed using the IHT First Law Model for an Isothermal Hollow Sphere. As shown below, there is a sharp increase in temperature with decreasing values of h 1000 W/m2K. For T 380 K, boiling will occur at the canister surface, and for T 410 K a condition known as film boiling (Chapter 10) will occur. The condition corresponds to a precipitous reduction in h and increase in T.

1000 900 Temperature, T(K) 800 700 600 500 400 300 100 400 800 2000 6000 10000

Convection coefficient, h(W/m 2.K)

Although the canister remains well below the melting point of stainless steel for h = 100 W/m2K, boiling should be avoided, in which case the convection coefficient should be maintained at h 1000 W/m2K. COMMENTS: The governing equation of part (a) is a first order, nonhomogenous differential equation with constant coefficients. Its solution is = (S/R ) 1 - e-Rt + i e-Rt , where T - T ,

) ( 3 2 3 S 3qi ri2 / cp ( ro - ri3 ) , R = 3hro / cp ( ro - ri3 ) . Note results for t and for S = 0.

PROBLEM 1.49

KNOWN: Boiling point and latent heat of liquid oxygen. Diameter and emissivity of container. Free convection coefficient and temperature of surrounding air and walls. FIND: Mass evaporation rate. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature of container outer surface equals boiling point of oxygen. ANALYSIS: (a) Applying an energy balance to a control surface about the container, it follows that, at any instant, Ein - Eout = 0 or qconv + q rad - q evap = 0 . The evaporative heat loss is equal to the product of the mass rate of vapor production and the heat of vaporization. Hence, h ( T - T ) + T 4 - T 4 A - m (1) s sur s s evap h fg = 0

(

h ( T - T ) + T 4 - T 4 D2 s sur s mevap = h fg

(

)

)

10 W m 2 K ( 298 - 263) K + 0.2 5.67 10-8 W m 2 K 4 2984 - 2634 K 4 (0.5 m )2 mevap = mevap =

(

)

(350 + 35.2 ) W / m2

214 kJ kg

(0.785 m2 ) = 1.41 10-3 kg s .

214 kJ kg

(b) Using the energy balance, Eq. (1), the mass rate of vapor production can be determined for the range of emissivity 0.2 to 0.94. The effect of increasing emissivity is to increase the heat rate into the container and, hence, increase the vapor production rate.

1.9 Evaporation rate, mdot*1000 (kg/s)

1.8

1.7

1.6

1.5

1.4 0.2 0.4 0.6 Surface emissivity, eps 0.8 1

COMMENTS: To reduce the loss of oxygen due to vapor production, insulation should be applied to the outer surface of the container, in order to reduce qconv and qrad. Note from the calculations in part (a), that heat transfer by convection is greater than by radiation exchange.

PROBLEM 1.50

KNOWN: Frost formation of 2-mm thickness on a freezer compartment. Surface exposed to convection process with ambient air. FIND: Time required for the frost to melt, tm. SCHEMATIC:

ASSUMPTIONS: (1) Frost is isothermal at the fusion temperature, Tf, (2) The water melt falls away from the exposed surface, (3) Negligible radiation exchange at the exposed surface, and (4) Backside surface of frost formation is adiabatic. PROPERTIES: Frost, f = 770 kg / m3 , h sf = 334 kJ / kg. ANALYSIS: The time tm required to melt a 2-mm thick frost layer may be determined by applying an energy balance, Eq 1.11b, over the differential time interval dt and to a differential control volume extending inward from the surface of the layer dx. From the schematic above, the energy in is the convection heat flux over the time period dt and the change in energy storage is the latent energy change within the control volume, Asdx.

Ein - E out = Est q As dt = dUat conv

h As ( T - Tf ) dt = - f As h sf dx

Integrating both sides of the equation and defining appropriate limits, find t 0 h ( T - Tf ) m dt = - f hsf dx 0 xo

tm = tm =

f h sf x o h ( T - Tf )

700 kg / m3 334 103 J / kg 0.002m 2 W / m 2 K ( 20 - 0 ) K

= 11, 690 s = 3.2 hour

melting the frost layer ( x o h sf ). This equality is directly comparable to the derived expression above for tm. (2) Explain why the energy storage term in the analysis has a negative sign, and the limits of integration are as shown. Hint: Recall from the formulation of Eq. 1.11b, that the storage term represents the change between the final and initial states.

COMMENTS: (1) The energy balance could be formulated intuitively by recognizing that the total heat in by convection during the time interval t m ( q t m ) must be equal to the total latent energy for cv

PROBLEM 1.51

KNOWN: Vertical slab of Woods metal initially at its fusion temperature, Tf, joined to a substrate. Exposed surface is irradiated with laser source, G W / m 2 .

FIND: Instantaneous rate of melting per unit area, m (kg/sm ), and the material removed in a m period of 2 s, (a) Neglecting heat transfer from the irradiated surface by convection and radiation exchange, and (b) Allowing for convection and radiation exchange.

(

)

2

SCHEMATIC:

ASSUMPTIONS: (1) Woods metal slab is isothermal at the fusion temperature, Tf, and (2) The melt runs off the irradiated surface. ANALYSIS: (a) The instantaneous rate of melting per unit area may be determined by applying an energy balance, Eq 1.11a, on the metal slab at an instant of time neglecting convection and radiation exchange from the irradiated surface.

E - E = E in out st

G =

d dM ( -Mh sf ) = -hsf dt dt

where dM / dt = m is the time rate of change of mass in the control volume. Substituting values, m m m 0.4 5000 W / m 2 = -33, 000 J / kg m m = -60.6 10-3 kg / s m 2 The material removed in a 2s period per unit area is M = m t = 121 g / m 2 2s m (b) The energy balance considering convection and radiation exchange with the surroundings yields

m G - q - q = -h sf m cv rad q = h ( Tf - T ) = 15 W / m 2 K ( 72 - 20 ) K = 780 W / m 2 cv

4 q = Tf4 - T = 0.4 5.67 10-8 W / m 2 K [72 + 273] - [20 + 273] rad 4

(

)

(

4

)K

4

= 154 W / m 2

m m = -32.3 10-3 kg / s m 2

M 2s = 64 g / m 2

COMMENTS: (1) The effects of heat transfer by convection and radiation reduce the estimate for the material removal rate by a factor of two. The heat transfer by convection is nearly 5 times larger than by radiation exchange. (2) Suppose the work piece were horizontal, rather than vertical, and the melt puddled on the surface rather than ran off. How would this affect the analysis? (3) Lasers are common heating sources for metals processing, including the present application of melting (heat transfer with phase change), as well as for heating work pieces during milling and turning (laser-assisted machining).

PROBLEM 1.52 KNOWN: Hot formed paper egg carton of prescribed mass, surface area and water content exposed to infrared heater providing known radiant flux. FIND: Whether water content can be reduced from 75% to 65% by weight during the 18s period carton is on conveyor. SCHEMATIC:

ASSUMPTIONS: (1) All the radiant flux from the heater bank is absorbed by the carton, (2) Negligible heat loss from carton by convection and radiation, (3) Negligible mass loss occurs from bottom side. PROPERTIES: Water (given): hfg = 2400 kJ/kg. ANALYSIS: Define a control surface about the carton, and write the conservation of energy requirement for an interval of time, t,

E in - E o ut = E st = 0

where Ein is due to the absorbed radiant flux, q , from the h heater and Eout is the energy leaving due to evaporation of water from the carton. Hence. q A s t = M h fg . h For the prescribed radiant flux q , h q A st 5000 W / m2 0.0625 m2 18s M = h = = 0.00234 kg. h fg 2400 kJ / kg The chief engineer's requirement was to remove 10% of the water content, or M req = M 0.10 = 0.220 kg 0.10 = 0.022 kg which is nearly an order of magnitude larger than the evaporative loss. Considering heat losses by convection and radiation, the actual water removal from the carton will be less than M. Hence, the purchase should not be recommended, since the desired water removal cannot be achieved.

PROBLEM 1.53 KNOWN: Average heat sink temperature when total dissipation is 20 W with prescribed air and surroundings temperature, sink surface area and emissivity. FIND: Sink temperature when dissipation is 30 W. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) All dissipated power in devices is transferred to the sink, (3) Sink is isothermal, (4) Surroundings and air temperature remain the same for both power levels, (5) Convection coefficient is the same for both power levels, (6) Heat sink is a small surface within a large enclosure, the surroundings. ANALYSIS: Define a control volume around the heat sink. Power dissipated within the devices is transferred into the sink, while the sink loses heat to the ambient air and surroundings by convection and radiation exchange, respectively. Ein - Eout = 0 (1) 4 4 Pe - hAs ( Ts - T ) - As Ts - Tsur = 0.

(

)

Consider the situation when Pe = 20 W for which Ts = 42 C; find the value of h.

4 4 h= Pe / As - Ts - Tsur / ( Ts - T ) h= 20 W/0.045 m 2 - 0.8 5.67 10-8 W/m 2 K 4 3154 - 3004 K 4 / (315 - 300 ) K 2 h = 24.4 W / m K.

(

)

(

)

For the situation when Pe = 30 W, using this value for h with Eq. (1), obtain 30 W - 24.4 W/m2 K 0.045 m 2 (Ts - 300 ) K

4 -0.045 m 2 0.8 5.67 10-8 W/m 2 K 4 Ts - 3004 K 4 = 0 4 30 = 1.098 (Ts - 300 ) + 2.041 10-9 Ts - 3004 .

By trial-and-error, find Ts 322 K = 49 C.

(

)

(

)

COMMENTS: (1) It is good practice to express all temperatures in kelvin units when using energy balances involving radiation exchange. (2) Note that we have assumed As is the same for the convection and radiation processes. Since not all portions of the fins are completely exposed to the surroundings, As,rad is less than As,conv = As. (3) Is the assumption that the heat sink is isothermal reasonable?

PROBLEM 1.54

KNOWN: Number and power dissipation of PCBs in a computer console. Convection coefficient associated with heat transfer from individual components in a board. Inlet temperature of cooling air and fan power requirement. Maximum allowable temperature rise of air. Heat flux from component most susceptible to thermal failure. FIND: (a) Minimum allowable volumetric flow rate of air, (b) Preferred location and corresponding surface temperature of most thermally sensitive component.

SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and kinetic energy changes of air flow, (4) Negligible heat transfer from console to ambient air, (5) Uniform convection coefficient for all components. ANALYSIS: (a) For a control surface about the air space in the console, conservation of energy for an open system, Eq. (1.11e), reduces to

m ( u + pv ) - m ( u + pv ) + q - W = 0 i o where u + pv = i, q = 5Pb , and W = - Pf . Hence, with m (ii - io ) = mcp (Ti - To ) , mcp (To - Ti ) = 5 Pb + Pf

For a maximum allowable temperature rise of 15 C, the required mass flow rate is

m=

5 Pb + Pf 5 20 W + 25 W = = 8.28 10-3 kg/s cp ( To - Ti ) 1007 J/kg K 15 $C

(

)

The corresponding volumetric flow rate is

=

m 8.28 10-3 kg/s = = 7.13 10-3 m3 / s 3 1.161 kg/m

(b) The component which is most susceptible to thermal failure should be mounted at the bottom of one of the PCBs, where the air is coolest. From the corresponding form of Newton's law of cooling, q = h ( Ts - Ti ) , the surface temperature is

4 2 q $ C + 1 10 W/m = 70$ C Ts = Ti + = 20; W/m2 K

COMMENTS: (1) Although the mass flow rate is invariant, the volumetric flow rate increases as the air is heated in its passage through the console, causing a reduction in the density. However, for the prescribed temperature rise, the change in , and hence the effect on , is small. (2) If the thermally sensitive component were located at the top of a PCB, it would be exposed to warmer air (To = 35 C) and the surface temperature would be Ts = 85 C.

PROBLEM 1.55 KNOWN: Top surface of car roof absorbs solar flux, qS,abs , and experiences for case (a): convection with air at T and for case (b): the same convection process and radiation emission from the roof.

FIND: Temperature of the plate, Ts , for the two cases. Effect of airflow on roof temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer to auto interior, (3) Negligible radiation from atmosphere. ANALYSIS: (a) Apply an energy balance to the control surfaces shown on the schematic. For an instant of time, Ein - E out = 0. Neglecting radiation emission, the relevant processes are convection

between the plate and the air, q conv , and the absorbed solar flux, qS,abs . Considering the roof to have an area As , qS,abs As - hAs ( Ts - T ) = 0 Ts = T + qS,abs /h Ts = 20 C + 800W/m 2 12W/m 2 K = 20 C + 66.7 C = 86.7 C

(b) With radiation emission from the surface, the energy balance has the form

qS,abs As - q conv - E As = 0 qS,abs As - hAs ( Ts - T ) - As Ts4 = 0 .

Substituting numerical values, with temperature in absolute units (K),

800

W m2

- 12

W m2 K

(Ts - 293K ) - 0.8 5.67 10-8

W T4 = 0 2 K4 s m

4 12Ts + 4.536 10-8 Ts = 4316 It follows that Ts = 320 K = 47 C.

Continued.

PROBLEM 1.55 (Cont.)

(c) Parametric calculations were performed using the IHT First Law Model for an Isothermal Plane Wall. As shown below, the roof temperature depends strongly on the velocity of the auto relative to the ambient air. For a convection coefficient of h = 40 W/m2K, which would be typical for a velocity of 55 mph, the roof temperature would exceed the ambient temperature by less than 10 C.

360 350 Temperature, Ts(K) 340 330 320 310 300 290 0 20 40 60 80 100 120 140 160 180 200 Convection coefficient, h(W/m 2.K)

rate equation must be h ( Ts - T ) and not h ( T - Ts ) .

COMMENTS: By considering radiation emission, Ts decreases, as expected. Note the manner in which q is formulated using Newton's law of cooling; since q is shown leaving the control surface, the conv conv

PROBLEM 1.56

KNOWN: Detector and heater attached to cold finger immersed in liquid nitrogen. Detector surface of = 0.9 is exposed to large vacuum enclosure maintained at 300 K. FIND: (a) Temperature of detector when no power is supplied to heater, (b) Heater power (W) required to maintain detector at 195 K, (c) Effect of finger thermal conductivity on heater power. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction through cold finger, (3) Detector and heater are very thin and isothermal at Ts , (4) Detector surface is small compared to enclosure surface. PROPERTIES: Cold finger (given): k = 10 W/mK. ANALYSIS: Define a control volume about detector and heater and apply conservation of energy requirement on a rate basis, Eq. 1.11a,

Ein - Eout = 0

where

(1)

Ein = q rad + q elec ;

E out = q cond

(2,3)

Combining Eqs. (2,3) with (1), and using the appropriate rate equations, 4 4 As Tsur - Ts + q elec = kAs (Ts - TL )/L . (a) Where q elec = 0, substituting numerical values 4 0.9 5.67 10-8 W/m 2 K 4 3004 - Ts K 4 = 10W/m K (Ts - 77 ) K/0.050 m

(

)

(4)

5.103 10-8 3004 - Ts4 = 200 (Ts - 77 ) Ts = 79.1K

(

)

(

)

Continued.

PROBLEM 1.56 (Cont.)

(b) When Ts = 195 K, Eq. (4) yields

0.9 [ ( 0.005 m ) / 4] 5.67 10-8 W/m 2 K 4 3004 - 1954 K 4 + q elec

2

(

)

= 10W/m K [ (0.005 m ) /4] (195 - 77 ) K / 0.050 m

2

qelec = 0.457 W = 457 mW

(c) Calculations were performed using the First Law Model for a Nonisothermal Plane Wall. With net radiative transfer to the detector fixed by the prescribed values of Ts and Tsur , Eq. (4) indicates that q elec increases linearly with increasing k.

19 Heater power, qelec(W) 17 15 13 11 9 7 5 3 1 -1 0 100 200 300 400 Thermal conductivity, k(W/m.K)

Heat transfer by conduction through the finger material increases with its thermal conductivity. Note that, for k = 0.1 W/mK, q elec = -2 mW, where the minus sign implies the need for a heat sink, rather than a heat source, to maintain the detector at 195 K. In this case q rad exceeds q cond , and a heat sink would be needed to dispose of the difference. A conductivity of k = 0.114 W/mK yields a precise balance between q rad and q cond . Hence to circumvent heaving to use a heat sink, while minimizing the heater power requirement, k should exceed, but remain as close as possible to the value of 0.114 W/mK. Using a graphite fiber composite, with the fibers oriented normal to the direction of conduction, Table A.2 indicates a value of k 0.54 W/mK at an average finger temperature of T = 136 K. For this value, q elec = 18 mW COMMENTS: The heater power requirement could be further reduced by decreasing .

PROBLEM 1.57

KNOWN: Conditions at opposite sides of a furnace wall of prescribed thickness, thermal conductivity and surface emissivity. FIND: Effect of wall thickness and outer convection coefficient on surface temperatures. Recommended values of L and; . SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction, (3) Negligible radiation exchange at surface 1, (4) Surface 2 is exposed to large surroundings. ANALYSIS: The unknown temperatures may be obtained by simultaneously solving energy balance equations for the two surface. At surface 1,

q conv,1 = q cond

h1 T,1 - T1 = k (T1 - T2 )/L

At surface 2,

(

)

(1)

q cond = q conv + q rad

4 4 k ( T1 - T2 )/L =; T2 - T,2 + T2 - Tsur

(

)

(

)

(2)

Using the IHT First Law Model for a Nonisothermal Plane Wall, we obtain

Surface temperature, T(K)

1700 1500 1300 1100 900 700 500 300 0 0.1 0.2 0.3 0.4 0.5 Wall thickness, L(m) Inner surface temperature, T1(K) Outer surface temperature, T2(K)

Continued .

PROBLEM 1.57 (Cont.)

Both q cond and T2 decrease with increasing wall thickness, and for the prescribed value of; = 10 2 W/m K, a value of L 0.275 m is needed to maintain T2 373 K = 100 C. Note that inner surface temperature T1 , and hence the temperature difference, T1 - T2 , increases with increasing L.

Performing the calculations for the prescribed range of; , we obtain

Surface temperature, T(K) 1700 1500 1300 1100 900 700 500 300 0 10 20 30 40 50 Convection coefficient, h2(W/m 2.K) Inner surface temperature, T(K) Outer surface temperature, T(K)

For the prescribed value of L = 0.15 m, a value of; 24 W/m2K is needed to maintain T2 373 K. The variation has a negligible effect on T1 , causing it to decrease slightly with increasing; , but does have a strong influence on T2 . COMMENTS: If one wishes to avoid use of active (forced convection) cooling on side 2, reliance will have to be placed on free convection, for which; 5 W/m2K. The minimum wall thickness would then be L = 0.40 m.

PROBLEM 1.58

KNOWN: Furnace wall with inner surface temperature T1 = 352 C and prescribed thermal conductivity experiencing convection and radiation exchange on outer surface. See Example 1.5. FIND: (a) Outer surface temperature T2 resulting from decreasing the wall thermal conductivity k or increasing the convection coefficient h by a factor of two; benefit of applying a low emissivity coating ( 0.8); comment on the effectiveness of these strategies to reduce risk of burn injury when 2 T2 65 C; and (b) Calculate and plot T2 as a function of h for the range 20 h 100 W/m K for three materials with k = 0.3, 0.6, and 1.2 W/mK; what conditions will provide for safe outer surface temperatures. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Radiation exchange is between small surface and large enclosure, (4) Inner surface temperature remains constant for all conditions. ANALYSIS: (a) The surface (x = L) energy balance is

T -T 4 4 k 1 2 = h ( T2 - T ) + T2 - Tsur L

(

)

h W / m2 K

20 20 40 40 20

With T1 = 352 C, the effects of parameters h, k and on the outer surface temperature are calculated and tabulated below. Conditions Example 1.5 Decrease k by Increase h by 2 Change k and h Decrease

k (W / m K )

1.2 0.6 1.2 0.6 1.2

(

)

0.8 0.8 0.8 0.8 0.1

T2 ( C )

100 69 73 51 115

(b) Using the energy balance relation in the Workspace of IHT, the outer surface temperature can be calculated and plotted as a function of the convection coefficient for selected values of the wall thermal conductivity. Continued .

PROBLEM 1.58 (Cont.)

O u te r su rfa ce te m p e ra tu re , T2 (C ) 100 80 60 40 20 20 40 60 80 100 C o n ve ctio n co e fficie n t, h (W /m 2 .K ) k = 1 .2 W /m .K k = 0 .6 W /m .K k = 0 .3 W /m .K

COMMENTS: (1) From the parameter study of part (a), note that decreasing the thermal conductivity is more effective in reducing T2 than is increasing the convection coefficient. Only if both changes are made will T2 be in the safe range. (2) From part (a), note that applying a low emissivity coating is not beneficial. Did you suspect that before you did the analysis? Give a physical explanation for this result. (3) From the parameter study graph we conclude that safe wall conditions (T2 65 C) can be 2 maintained for these conditions: with k = 1.2 W/mK when h 55 W/m K; with k = 0.6 W/mK 2 when h 25 W/m K; and with k = 0.3 W/mK when h 20 W/mK.

PROBLEM 1.59 KNOWN: Inner surface temperature, thickness and thermal conductivity of insulation exposed at its outer surface to air of prescribed temperature and convection coefficient. FIND: Outer surface temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in the insulation, (3) Negligible radiation exchange between outer surface and surroundings. ANALYSIS: From an energy balance at the outer surface at an instant of time, q cond = q conv . Using the appropriate rate equations, k

(T1 - T2 ) = h

L

(T2 - T ).

Solving for T2, find 0.1 W/m K W k 400 C + 500 35 C T1 + h T 2 K 0.025m m T2 = L = k W 0.1 W/m K h+ 500 + L 0.025m m2 K T2 = 37.9 C.

(

)

( )

COMMENTS: If the temperature of the surroundings is approximately that of the air, radiation exchange between the outer surface and the surroundings will be negligible, since T2 is small. In this case convection makes the dominant contribution to heat transfer from the outer surface, and assumption (3) is excellent.

PROBLEM 1.60

KNOWN: Thickness and thermal conductivity, k, of an oven wall. Temperature and emissivity, , of front surface. Temperature and convection coefficient, h, of air. Temperature of large surroundings. FIND: (a) Temperature of back surface, (b) Effect of variations in k, h and . SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Radiation exchange with large surroundings. ANALYSIS: (a) Applying an energy balance, Eq. 1.13, at an instant of time to the front surface and substituting the appropriate rate equations, Eqs. 1.2, 1.3a and 1.7, find

T -T 4 4 k 1 2 = h ( T2 - T ) + T2 - Tsur . L

Substituting numerical values, find

T1 - T2 =

(

)

W W -8 ( 400 K )4 - (300 K )4 = 200 K . 20 2 100 K + 0.8 5.67 10 0.7 W/m K m K m2 K 4

0.05 m

Since T2 = 400 K, it follows that T1 = 600 K.

(b) Parametric effects may be evaluated by using the IHT First Law Model for a Nonisothermal Plane Wall. Changes in k strongly influence conditions for k 20 W/mK, but have a negligible effect for larger values, as T2 approaches T1 and the heat fluxes approach the corresponding limiting values

10000

600

Heat flux, q'(W/m 2)

8000 6000 4000 2000 0 0 100 200 300 400 Thermal conductivity, k(W/m.K)

Temperature, T2(K)

500

400

300 0 100 200 300 400 Thermal conductivity, k(W/m.K)

Conduction heat flux, q'cond(W/m 2) Convection heat flux, q'conv(W/m 2) Radiation heat flux, q'rad(W/m 2)

PROBLEM 1.60 (Cont.)

The implication is that, for k 20 W/mK, heat transfer by conduction in the wall is extremely efficient relative to heat transfer by convection and radiation, which become the limiting heat transfer processes. Larger fluxes could be obtained by increasing and h and/or by decreasing T and Tsur . With increasing h, the front surface is cooled more effectively ( T2 decreases), and although q rad decreases, the reduction is exceeded by the increase in q conv . With a reduction in T2 and fixed values

of k and L, q cond must also increase.

30000

600

Heat flux, q'(W/m 2)

20000

Temperature, T2(K)

10000

500

0 0

400 0 100 Convection coefficient, h(W/m 2.K) 200

100 Convection coefficient, h(W/m 2.K)

200

Conduction heat flux, q'cond(W/m 2) Convection heat flux, q'conv(W/m 2) Radiation heat flux, q'rad(W/m 2)

The surface temperature also decreases with increasing , and the increase in q exceeds the reduction rad in q , allowing q to increase with . conv cond

10000 Heat flux, q'(W/m 2)

575

8000 6000 4000 2000 0

570 Temperature, T2(K)

565

560

555

0

0.2

0.4

0.6

0.8

1

Emissivity

550 0 0.2 0.4 0.6 0.8 1 Emissivity

Conduction heat flux, q'cond(W/m 2) Convection heat flux, q'conv(W/m 2) Radiation heat flux, q'rad(W/m 2)

COMMENTS: Conservation of energy, of course, dictates that, irrespective of the prescribed conditions, q cond = q conv + q . rad

PROBLEM 1.61

KNOWN: Temperatures at 10 mm and 20 mm from the surface and in the adjoining airflow for a thick steel casting. FIND: Surface convection coefficient, h. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in the x-direction, (3) Constant properties, (4) Negligible generation. ANALYSIS: From a surface energy balance, it follows that

q cond = q conv

where the convection rate equation has the form

q conv = h ( T - T0 ) ,

and q cond can be evaluated from the temperatures prescribed at surfaces 1 and 2. That is, from Fourier's law,

T1 - T2 q cond = k x 2 - x1

(50 - 40 ) C = 15, 000 W/m2 . W q cond = 15 m K ( 20-10 ) 10-3 m

Since the temperature gradient in the solid must be linear for the prescribed conditions, it follows that

T0 = 60 C.

Hence, the convection coefficient is

h= h=

q cond T - T0

15,000 W / m2 40 C

= 375 W / m2 K.

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

PROBLEM 1.62

KNOWN: Duct wall of prescribed thickness and thermal conductivity experiences prescribed heat flux q at outer surface and convection at inner surface with known heat transfer coefficient. o FIND: (a) Heat flux at outer surface required to maintain inner surface of duct at Ti = 85 C, (b) Temperature of outer surface, To , (c) Effect of h on To and q . o SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wall, (3) Constant properties, (4) Backside of heater perfectly insulated, (5) Negligible radiation.

ANALYSIS: (a) By performing an energy balance on the wall, recognize that balance on the top surface, it follows that

q cond = q conv = q . Hence, using the convection rate equation, o

q = q o cond . From an energy

2 2 q = q o conv = h ( Ti - T ) = 100 W / m K (85 - 30 ) C = 5500W /m . (b) Considering the duct wall and applying Fourier's Law,

q = k o

T -T T =k o i L X

q L 5500 W/m 2 0.010 m = (85 + 2.8 ) C = 87.8 C . To = Ti + o = 85 C + k 20 W/m K (c) For Ti = 85 C, the desired results may be obtained by simultaneously solving the energy balance equations T - Ti T - Ti q = k o k o = h ( Ti - T ) and o L L Using the IHT First Law Model for a Nonisothermal Plane Wall, the following results are obtained.

12000 Heat flux, q'o(W/m 2) 10000 8000 6000 4000 2000 0 0 40 80 120 160 200 Convection coefficient, h(W/m 2.K)

Surface temperature, To(C) 91 90 89 88 87 86 85 0 40 80 120 160 200 Convection coefficient, h(W/m 2.K)

Since q conv increases linearly with increasing h, the applied heat flux q must be balanced by an o increase in q , which, with fixed k, Ti and L, necessitates an increase in To . cond COMMENTS: The temperature difference across the wall is small, amounting to a maximum value of (To - Ti ) = 5.5 C for h = 200 W/m2K. If the wall were thinner (L 10 mm) or made from a material with higher conductivity (k 20 W/mK), this difference would be reduced.

PROBLEM 1.63 KNOWN: Dimensions, average surface temperature and emissivity of heating duct. Duct air inlet temperature and velocity. Temperature of ambient air and surroundings. Convection coefficient. FIND: (a) Heat loss from duct, (b) Air outlet temperature. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and kinetic energy changes of air flow, (4) Radiation exchange between a small surface and a large enclosure. ANALYSIS: (a) Heat transfer from the surface of the duct to the ambient air and the surroundings is given by Eq. (1.10) 4 4 q = hAs ( Ts - T ) + As Ts - Tsur

(

)

where As = L (2W + 2H) = 15 m (0.7 m + 0.5 m) = 16.5 m . Hence,

2

q = 4 W/m2 K 16.5 m2 45$ C + 0.5 16.5 m2 5.67 10-8 W/m2 K 4 3234 - 2784 K 4 q = qconv + q rad = 2970 W + 2298 W = 5268 W (b) With i = u + pv, W = 0 and the third assumption, Eq. (1.11e) yields, m (ii - io ) = mcp (Ti - To ) = q where the sign on q has been reversed to reflect the fact that heat transfer is from the system. With m = VAc = 1.10 kg/m3 4 m/s (0.35m 0.20m ) = 0.308 kg/s, the outlet temperature is q 5268 W To = Ti - = 58$ C - = 41$ C mcp 0.308 kg/s 1008 J/kg K

( )

(

)

COMMENTS: The temperature drop of the air is large and unacceptable, unless the intent is to use the duct to heat the basement. If not, the duct should be insulated to insure maximum delivery of thermal energy to the intended space(s).

PROBLEM 1.64

KNOWN: Uninsulated pipe of prescribed diameter, emissivity, and surface temperature in a room with fixed wall and air temperatures. See Example 1.2. FIND: (a) Which option to reduce heat loss to the room is more effective: reduce by a factor of two 2 the convection coefficient (from 15 to 7.5 W/m K) or the emissivity (from 0.8 to 0.4) and (b) Show 2 graphically the heat loss as a function of the convection coefficient for the range 5 h 20 W/m K for emissivities of 0.2, 0.4 and 0.8. Comment on the relative efficacy of reducing heat losses associated with the convection and radiation processes. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between pi