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KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal conductivity k and inner temperature, T (^) 1.
FIND: The outer temperature of the wall, T 2.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties.
ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law,
q q q A = -k dT dx
A = kA
T T
cond (^) L x x
= = ′′ ⋅ ⋅ 1 −^2.
Solving for T 2 gives
T T
q L (^2 1) kA = − cond.
Substituting numerical values, find
T C -
3000W 0.025m (^2) 0.2W / m K 10m 2
×
⋅ ×
T 2 = 415 $^ C - 37.5 C$
T 2 = 378 $C. <
COMMENTS: Note direction of heat flow and fact that T 2 must be less than T (^) 1.
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.
FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from -15 to 38°C.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties, (4) Outside wall temperature is that of the ambient air.
ANALYSIS: From Fourier’s law, it is evident that the gradient, dT dx = − q ′′x k, is a constant, and
hence the temperature distribution is linear, if q′′x and k are each constant. The heat flux must be
constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T 2 = -15°C are
1 2 (^ )^2 x
dT T T 25 C^ 15 C q k k 1W m K 133.3W m dx L 0.30 m
−^ − −
$ $
. (1)
2 2 qx = qx ′′× A = 133.3 W m × 20 m = 2667 W. (2) <
Combining Eqs. (1) and (2), the heat rate q (^) x can be determined for the range of ambient temperature, - ≤ T 2 ≤ 38 °C, with different wall thermal conductivities, k.
-20 -10 0 10 20 30 40 Ambient air temperature, T2 (C)
500
1500
2500
3500
Heat loss, qx (W)
Wall thermal conductivity, k = 1.25 W/m.Kk = 1 W/m.K, concrete wall k = 0.75 W/m.K
For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearily from +2667 W to -867 W and is zero when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with increasing thermal conductivity.
COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane wall would not be linear.
KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed thickness.
FIND: Thermal conductivity, k, of the wood.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be determined from Fourier’s law, Eq. 1.2. Rearranging,
L W 0.05m k=q 40
x T 1 T 2 m 2 40-20 C
− $
k = 0.10 W / m K.⋅ <
COMMENTS: Note that the °C or K temperature units may be used interchangeably when
evaluating a temperature difference.
KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions.
FIND: Heat loss through window.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from Fourier’s law, Eq. 1.2.
T T
q k L W 15-5^ C q 1. m K 0.005m q 2800 W/m.
x
x
x
Since the heat flux is uniform over the surface, the heat loss (rate) is
q = qx A
q = 2800 W / m^2 3m^2
′′ ×
×
q = 8400 W. <
COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions.
KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures.
FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribed value.
SCHEMATIC:
ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 5
walls of area A = 4m 2 , (3) Steady-state conditions, (4) Constant properties.
ANALYSIS: Using Fourier’s law, Eq. 1.2, the heat rate is
q = q A = k
T
L
′′ ⋅ A (^) total
Solving for L and recognizing that A (^) total = 5×W 2 , find
L =
5 k T W q
∆^2
5 0.03 W/m K 35 - (^) ( -10) C (^) ( 4m ) L = 500 W
× ⋅ ^2
$
L = 0.054m = 54mm. <
COMMENTS: (^) The corners will cause local departures from one-dimensional conduction and a slightly larger heat loss.
KNOWN: Dimensions and thermal conductivity of food/beverage container. Inner and outer surface temperatures.
FIND: Heat flux through container wall and total heat load.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through bottom wall, (3) Uniform surface temperatures and one-dimensional conduction through remaining walls.
ANALYSIS: From Fourier’s law, Eq. 1.2, the heat flux is
T 2 T 1 0.023 W/m K^ (^20 2 ) C 2
q k 16.6 W/m L 0.025 m
−^ ⋅^ −
Since the flux is uniform over each of the five walls through which heat is transferred, the heat load is
q =^ q ′′^ ×^ A total =^ q ′′ H 2W^ ( 1 +^ 2W 2 )+^ W 1 ×W 2
q = 16.6 W/m^2 ^ 0.6m 1.6m + 1.2m + 0.8m × 0.6m =35.9 W
COMMENTS: The corners and edges of the container create local departures from one-
dimensional conduction, which increase the heat load. However, for H, W 1 , W 2 >> L, the effect is negligible.
KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil water. Rate of heat transfer to the pan.
FIND: Outer surface temperature of pan for an aluminum and a copper bottom.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction through bottom of pan.
ANALYSIS: From Fourier’s law, the rate of heat transfer by conduction through the bottom of the pan is
T 1 T 2
q kA L
Hence,
qL T 1 T 2 kA
where A = π D 2 / 4 = π( 0.2m )^2 / 4 =0.0314 m.^2
Aluminum :
600W 0.005 m T 1 110 C 110.40 C 240 W/m K 0.0314 m^2
Copper :
600W 0.005 m T 1 110 C 110.25 C 390 W/m K 0.0314 m^2
COMMENTS: Although the temperature drop across the bottom is slightly larger for aluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible for both materials. To a good approximation, the bottom may be considered isothermal at T ≈ 110 °C, which is a desirable feature of pots and pans.
KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one surface.
FIND: Temperature drop across the chip.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heat dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction in chip.
ANALYSIS: All of the electrical power dissipated at the back surface of the chip is transferred by conduction through the chip. Hence, from Fourier’s law,
P = q = kA
T
t
or
t P 0.001 m 4 W T =
kW 2 150 W/m K 0.005 m^2
⋅ ×
∆T = 1.1$ C. <
COMMENTS: For fixed P, the temperature drop across the chip decreases with increasing k and W, as well as with decreasing t.
KNOWN: Hand experiencing convection heat transfer with moving air and water.
FIND: Determine which condition feels colder. Contrast these results with a heat loss of 30 W/m^2 under normal room conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Temperature is uniform over the hand’s surface, (2) Convection coefficient is uniform over the hand, and (3) Negligible radiation exchange between hand and surroundings in the case of air flow.
ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss. The heat loss can be determined from Newton’s law of cooling, Eq. 1.3a, written as
q ′′ = h T ( s −T∞)
For the air stream:
qair ′′^ = 40 W m 2 ⋅ K 30 − − ( 5 ) K =1, 400 W m^2 <
For the water stream:
qwater ′′^ = 900 W m^2 ⋅ K 30 ( − 10 K) = 18,000 W m^2 <
COMMENTS: The heat loss for the hand in the water stream is an order of magnitude larger than when in the air stream for the given temperature and convection coefficient conditions. In contrast, the heat loss in a normal room environment is only 30 W/m^2 which is a factor of 400 times less than the loss in the air stream. In the room environment, the hand would feel comfortable; in the air and water streams, as you probably know from experience, the hand would feel uncomfortably cold since the heat loss is excessively high.
KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinder with an imbedded electrical heater for different air velocities.
FIND: (a) Determine the convection coefficient for each of the air velocity conditions and display the results graphically, and (b) Assuming that the convection coefficient depends upon air velocity as h = CV n, determine the parameters C and n.
SCHEMATIC:
V(m/s) 1 2 4 8 12 Pe ′ (W/m) 450 658 983 1507 1963 h (W/m^2 ⋅K) 22.0^ 32.2^ 48.1^ 73.8^ 96.
ASSUMPTIONS: (1) Temperature is uniform over the cylinder surface, (2) Negligible radiation exchange between the cylinder surface and the surroundings, (3) Steady-state conditions.
ANALYSIS: (a) From an overall energy balance on the cylinder, the power dissipated by the electrical heater is transferred by convection to the air stream. Using Newtons law of cooling on a per unit length basis,
Pe ′ = h ( π D )( Ts −T∞)
where Pe′^ is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/s
condition, using the data from the table above, find
h = 450 W m π× 0.025 m 300 − 40 $C = 22.0 W m 2 ⋅K <
Repeating the calculations, find the convection coefficients for the remaining conditions which are tabulated above and plotted below. Note that h is not linear with respect to the air velocity.
(b) To determine the (C,n) parameters, we plotted h vs. V on log-log coordinates. Choosing C = 22.12 W/m^2 ⋅K(s/m) n, assuring a match at V = 1, we can readily find the exponent n from the slope of the h vs. V curve. From the trials with n = 0.8, 0.6 and 0.5, we recognize that n = 0.6 is a reasonable
choice. Hence, C = 22.12 and n = 0.6. <
0 2 4 6 8 10 12 Air velocity, V (m/s)
20
40
60
80
100
Coefficient, h (W/m^2.K)
Data, smooth curve, 5-points
1 2 4 6 8 10 Air velocity, V (m/s)
10
20
40
60
80
100
Coefficient, h (W/m^2.K)
Data , smooth curve, 5 points h = C * V^n, C = 22.1, n = 0. n = 0. n = 0.
KNOWN: Dimensions of a cartridge heater. Heater power. Convection coefficients in air and water at a prescribed temperature.
FIND: Heater surface temperatures in water and air.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) All of the electrical power is transferred to the fluid by convection, (3) Negligible heat transfer from ends.
ANALYSIS: With P = q (^) conv, Newton’s law of cooling yields
P=hA T ( T ) h DL T( T )
P
T T.
h DL
s s
s
π
π
In water,
T C +
2000 W
5000 W / m K 0.02 m 0.200 m s =^2 ⋅ × × ×
π
Ts = 20 $^ C + 31.8 C = 51.8 C.$^ $ <
In air,
T C +
2000 W
50 W / m K 0.02 m 0.200 m s =^2 ⋅ × × ×
π
Ts = 20 $^ C + 3183 C = 3203 C.$^ $ <
COMMENTS: (1) Air is much less effective than water as a heat transfer fluid. Hence, the cartridge temperature is much higher in air, so high, in fact, that the cartridge would melt.
(2) In air, the high cartridge temperature would render radiation significant.
KNOWN: Length, diameter and calibration of a hot wire anemometer. Temperature of air stream. Current, voltage drop and surface temperature of wire for a particular application.
FIND: Air velocity
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from the wire by natural convection or radiation.
ANALYSIS: If all of the electric energy is transferred by convection to the air, the following equality must be satisfied
Pelec = EI = hA T ( s − T∞)
where A = π^ DL = π^ ( 0.0005m × 0.02m ) = 3.14 × 10 −^5 m.^2
Hence,
EI 5V 0.1A 2
h 318 W/m K A Ts T 3.14 10 5 m^2 50 C
×
− ∞ × − $
V = 6.25 × 10 −^ h = 6.25 × 10 − 318 W/m ⋅ K = 6.3 m/s <
COMMENTS: The convection coefficient is sufficiently large to render buoyancy (natural convection) and radiation effects negligible.
KNOWN: Length, diameter and maximum allowable surface temperature of a power transistor. Temperature and convection coefficient for air cooling.
FIND: Maximum allowable power dissipation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through base of transistor, (3) Negligible heat transfer by radiation from surface of transistor.
ANALYSIS: Subject to the foregoing assumptions, the power dissipated by the transistor is equivalent to the rate at which heat is transferred by convection to the air. Hence,
Pelec = q conv = hA T ( s − T∞)
where A = π( DL + D 2 / 4) = π ^ 0.012m × 0.01m + ( 0.012m) 2 / 4 = 4.90 × 10 −^4 m.^2
For a maximum allowable surface temperature of 85°C, the power is
( ) (^ )
P 100 W/m^2 K 4.90 10 4 m^2 85 25 C 2.94 W elec
= ⋅ × − − $^ = <
COMMENTS: (1) For the prescribed surface temperature and convection coefficient, radiation will be negligible relative to convection. However, conduction through the base could be significant, thereby permitting operation at a larger power.
(2) The local convection coefficient varies over the surface, and hot spots could exist if there are locations at which the local value of h is substantially smaller than the prescribed average value.
KNOWN: Air jet impingement is an effective means of cooling logic chips.
FIND: Procedure for measuring convection coefficients associated with a 10 mm × 10 mm chip.
SCHEMATIC:
ASSUMPTIONS: Steady-state conditions.
ANALYSIS: One approach would be to use the actual chip-substrate system, Case (a), to perform the measurements. In this case, the electric power dissipated in the chip would be transferred from the chip by radiation and conduction (to the substrate), as well as by convection to the jet. An energy balance for
the chip yields qelec = qconv + q cond + qrad. Hence, with q conv = hA T( s − T∞), where A = 100
mm^2 is the surface area of the chip,
elec cond rad s
q q q h A T T∞
(1)
While the electric power ( q (^) elec ) and the jet ( T∞ ) and surface ( Ts ) temperatures may be measured, losses
from the chip by conduction and radiation would have to be estimated. Unless the losses are negligible (an unlikely condition), the accuracy of the procedure could be compromised by uncertainties associated with determining the conduction and radiation losses.
A second approach, Case (b), could involve fabrication of a heater assembly for which the conduction and radiation losses are controlled and minimized. A 10 mm × 10 mm copper block (k ~ 400 W/m⋅K) could be inserted in a poorly conducting substrate (k < 0.1 W/m⋅K) and a patch heater could be applied to the back of the block and insulated from below. If conduction to both the substrate and insulation could thereby be rendered negligible, heat would be transferred almost exclusively through the block. If radiation were rendered negligible by applying a low emissivity coating (ε < 0.1) to the surface of the copper block, virtually all of the heat would be transferred by convection to the jet. Hence, q (^) cond
and q (^) rad may be neglected in equation (1), and the expression may be used to accurately determine h from the known (A) and measured ( q (^) elec , Ts , T∞ ) quantities.
COMMENTS: Since convection coefficients associated with gas flows are generally small, concurrent heat transfer by radiation and/or conduction must often be considered. However, jet impingement is one of the more effective means of transferring heat by convection and convection coefficients well in excess of 100 W/m^2 ⋅K may be achieved.
