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Solução dos exercícios com comentários e em inglês

Tipologia: Exercícios

2019

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Baixe Resolução capitulo 1 exercícios 1-20 Fundamentos de transferência de calor e massa Incropera e outras Exercícios em PDF para Calor e Transferência de Massa, somente na Docsity! PROBLEM 1.1 KNOWN: Thermal conductivity, thickness and temperature difference across a sheet of rigid extruded insulation. FIND: (a) The heat flux through a 2 m × 2 m sheet of the insulation, and (b) The heat rate through the sheet. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: From Equation 1.2 the heat flux is 1 2x T - TdTq = -k = k dx L ′′ Solving, "x W 10 Kq = 0.029 × m K 0.02 m⋅ x 2 Wq = 14.5 m ′′ < The heat rate is 2x x 2 Wq = q A = 14.5 × 4 m = 58 W m ′′ ⋅ < COMMENTS: (1) Be sure to keep in mind the important distinction between the heat flux (W/m2) and the heat rate (W). (2) The direction of heat flow is from hot to cold. (3) Note that a temperature difference may be expressed in kelvins or degrees Celsius. qcond A = 4 m2 T2T1 k = 0.029 ⋅ W m K x L = 20 mm T1 – T2 = 10˚C PROBLEM 1.2 KNOWN: Thickness and thermal conductivity of a wall. Heat flux applied to one face and temperatures of both surfaces. FIND: Whether steady-state conditions exist. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal energy generation. ANALYSIS: Under steady-state conditions an energy balance on the control volume shown is 2 in out cond 1 2( ) / 12 W/m K(50 C 30 C) / 0.01 m 24,000 W/mq q q k T T L′′ ′′ ′′= = = − = ⋅ ° − ° = Since the heat flux in at the left face is only 20 W/m2, the conditions are not steady state. < COMMENTS: If the same heat flux is maintained until steady-state conditions are reached, the steady-state temperature difference across the wall will be ΔT = 2/ 20 W/m 0.01 m /12 W/m K 0.0167 Kq L k′′ = × ⋅ = which is much smaller than the specified temperature difference of 20°C. q” = 20 W/m2 L = 10 mm k = 12 W/m·KT1 = 50°C T2 = 30°C q″cond PROBLEM 1.5 KNOWN: Thermal conductivity and thickness of a wall. Heat flux through wall. Steady-state conditions. FIND: Value of temperature gradient in K/m and in °C/m. SCHEMATIC: L = 20 mm k = 2.3 W/m·K q”x = 10 W/m2 x ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties. ANALYSIS: Under steady-state conditions, " 210 W/m 4.35 K/m 4.35 C/m 2.3 W/m K xdT q dx k = − = − = − = − ° ⋅ < Since the K units here represent a temperature difference, and since the temperature difference is the same in K and °C units, the temperature gradient value is the same in either units. COMMENTS: A negative value of temperature gradient means that temperature is decreasing with increasing x, corresponding to a positive heat flux in the x-direction. PROBLEM 1.6 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 0.05mk=q 40 T T m 40-20 C x 21 2 ′′ = − o k = 0.10 W / m K.⋅ < COMMENTS: Note that the °C or K temperature units may be used interchangeably when evaluating a temperature difference. PROBLEM 1.7 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 15-5 CWq 1.4 m K 0.005m q 2800 W/m . 1 2x x 2x − ′′ = ′′ = ⋅ ′′ = o Since the heat flux is uniform over the surface, the heat loss (rate) is q = qx A q = 2800 W / m2 3m2 ′′ × × q = 8400 W. < COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions. PROBLEM 1.9 KNOWN: Width, height, thickness and thermal conductivity of a single pane window and the air space of a double pane window. Representative winter surface temperatures of single pane and air space. FIND: Heat loss through single and double pane windows. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-state conditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy induced motion). ANALYSIS: From Fourier’s law, the heat losses are Single Pane: ( )T T 35 C21 2q k A 1.4 W/m K 2m 19,600 Wg g L 0.005m−= = ⋅ = o < Double Pane: ( )T T 25 C21 2q k A 0.024 2m 120 Wa a L 0.010 m−= = = o < COMMENTS: Losses associated with a single pane are unacceptable and would remain excessive, even if the thickness of the glass were doubled to match that of the air space. The principal advantage of the double pane construction resides with the low thermal conductivity of air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, use of the double pane construction would also increase the surface temperature of the glass exposed to the room (inside) air. PROBLEM 1.10 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 = 4m2, (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 Atotal′′ ⋅ Δ Solving for L and recognizing that Atotal = 5×W 2, find L = 5 k T W q 2Δ ( ) ( )5 0.03 W/m K 35 - -10 C 4m L = 500 W 2⎡ ⎤× ⋅ ⎣ ⎦ o L = 0.054m = 54mm. < COMMENTS: The corners will cause local departures from one-dimensional conduction and a slightly larger heat loss. PROBLEM 1.11 KNOWN: Heat flux at one face and air temperature and convection coefficient at other face of plane wall. Temperature of surface exposed to convection. FIND: If steady-state conditions exist. If not, whether the temperature is increasing or decreasing. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal energy generation. ANALYSIS: Conservation of energy for a control volume around the wall gives st in out g dE E E E dt = − +& & & [ ]st in in 2 2 2 ( ) ( ) 20 W/m 20 W/m K(50 C 30 C) 380 W/m s s dE q A hA T T q h T T A dt A A ∞ ∞ ′′ ′′= − − = − − = − ⋅ ° − ° = −⎡ ⎤⎣ ⎦ Since dEst/dt ≠ 0, the system is not at steady-state. < Since dEst/dt < 0, the stored energy is decreasing, therefore the wall temperature is decreasing. < COMMENTS: When the surface temperature of the face exposed to convection cools to 31°C, qin = qout and dEst/dt = 0 and the wall will have reached steady-state conditions. q” = 20 W/m2 Ts = 50°C h = 20 W/m2·K T∞ = 30°C Air q”conv PROBLEM 1.14 KNOWN: Expression for variable thermal conductivity of a wall. Constant heat flux. Temperature at x = 0. FIND: Expression for temperature gradient and temperature distribution. SCHEMATIC: ASSUMPTIONS: (1) One-dimensional conduction. ANALYSIS: The heat flux is given by Fourier’s law, and is known to be constant, therefore x dTq k constant dx ′′ = − = Solving for the temperature gradient and substituting the expression for k yields x xq qdT dx k ax b ′′ ′′ = − = − + < This expression can be integrated to find the temperature distribution, as follows: xqdT dx dx dx ax b ′′ = − +∫ ∫ Since xq constant′′ = , we can integrate the right hand side to find ( )xqT ln ax b c a ′′ = − + + where c is a constant of integration. Applying the known condition that T = T1 at x = 0, we can solve for c. Continued… q” x k = ax + b T1 PROBLEM 1.14 (Cont.) 1 x 1 x 1 T(x 0) T q ln b c T a qc T ln b a = = ′′ − + = ′′ = + Therefore, the temperature distribution is given by ( )x x1 q qT ln ax b T ln b a a ′′ ′′ = − + + + < x1 q bT lna ax b ′′ = + + < COMMENTS: Temperature distributions are not linear in many situations, such as when the thermal conductivity varies spatially or is a function of temperature. Non-linear temperature distributions may also evolve if internal energy generation occurs or non-steady conditions exist. PROBLEM 1.15 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 T1 2q kA L − = Hence, qLT T1 2 kA = + where ( )22 2A D / 4 0.2m / 4 0.0314 m .π π= = = Aluminum: ( ) ( ) 600W 0.005 m T 110 C 110.40 C1 2240 W/m K 0.0314 m = + = ⋅ o o < Copper: ( ) ( ) 600W 0.005 m T 110 C 110.24 C1 2390 W/m K 0.0314 m = + = ⋅ o o < 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. PROBLEM 1.18 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/m2 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 ( )sq h T T∞′′ = − For the air stream: ( )2 2airq 40 W m K 30 5 K 1,400 W m′′ ⎡ ⎤= ⋅ − − =⎣ ⎦ < For the water stream: ( )2 2waterq 900 W m K 30 10 K 18,000 W m′′ = ⋅ − = < 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/m2 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. PROBLEM 1.19 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 = CVn, 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/m2⋅K) 22.0 32.2 48.1 73.8 96.1 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 Newton’s law of cooling on a per unit length basis, ( )( )e sP h D T Tπ ∞′ = − where eP′ 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 ( ) 2h 450 W m 0.025m 300 40 C 22.0 W m Kπ= × − = ⋅o < 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/m2⋅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 C oe ffi ci en t, 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 C oe ffi ci en t, h (W /m ^2 .K ) Data , smooth curve, 5 points h = C * V^n, C = 22.1, n = 0.5 n = 0.6 n = 0.8 COMMENTS: Radiation may not be negligible, depending on surface emissivity. PROBLEM 1.20 KNOWN: Inner and outer surface temperatures of a wall. Inner and outer air temperatures and convection heat transfer coefficients. FIND: Heat flux from inner air to wall. Heat flux from wall to outer air. Heat flux from wall to inner air. Whether wall is under steady-state conditions. SCHEMATIC: ASSUMPTIONS: (1) Negligible radiation, (2) No internal energy generation. ANALYSIS: The heat fluxes can be calculated using Newton’s law of cooling. Convection from the inner air to the wall occurs in the positive x-direction: 2 2 x,i w i ,i s,iq h (T T ) 5 W/m K (20 C 16 C) 20 W/m− ∞′′ = − = ⋅ × ° − ° = < Convection from the wall to the outer air also occurs in the positive x-direction: 2 2 x,w o o s,o ,oq h (T T ) 20 W/m K (6 C 5 C) 20 W/m− ∞′′ = − = ⋅ × ° − ° = < From the wall to the inner air: 2 2 w i i s,i ,iq h (T T ) 5 W/m K (16 C 20 C) 20 W/m− ∞′′ = − = ⋅ × ° − ° = − < An energy balance on the wall gives st in out x,i w x,w o dE E E A(q q ) 0 dt − − ′′ ′′= − = − =& & Since dEst/dt = 0, the wall could be at steady-state and the spatially-averaged wall temperature is not changing. However, it is possible that stored energy is increasing in one part of the wall and decreasing in another, therefore we cannot tell if the wall is at steady-state or not. If we found dEst/dt ≠ 0, we would know the wall was not at steady-state. < COMMENTS: The heat flux from the wall to the inner air is equal and opposite to the heat flux from the inner air to the wall. Air T∞,o= 5°C ho = 20 W/m2·K Ts,i = 16°C Ts,o = 6°C x q″x,w-o q″x,i-w Air T∞,i= 20°C hi = 5 W/m2·K Inner surface Outer surface dEst/dt