PROBLEM 1.5. ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties.

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1 PROBLEM 1.5 KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions. FIND: Heat loss through window. ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, () Steady-state conditions, (3) Constant properties. ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from Fourier s law, Eq. 1.. T T q k 1 x = L $ W ( 15-5) C q x = 1.4 m K 0.005m q 800 W/m x =. Since the heat flux is uniform over the surface, the heat loss (rate) is q = q x A q = 800 W / m 3m q = 8400 W. < COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions.

2 PROBLEM 1.15 KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required to maintain a specified surface temperature for water and air flows. FIND: Convection coefficients for the water and air flow convection processes, h w and h a, respectively. ASSUMPTIONS: (1) Flow is cross-wise over cylinder which is very long in the direction normal to flow. ANALYSIS: The convection heat rate from the cylinder per unit length of the cylinder has the form ( π ) ( ) q = h D Ts T and solving for the heat transfer convection coefficient, find q h =. π D ( Ts T ) Substituting numerical values for the water and air situations: Water Air W/m h = = 4,570 W/m w K < π $ 0.030m 90-5 C ( ) 400 W/m h = 65 W/m a = K. < $ π 0.030m 90-5 C ( ) COMMENTS: Note that the air velocity is 10 times that of the water flow, yet h w 70 h a. These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases. See Table 1.1.

3 PROBLEM 1.8 KNOWN: Length, diameter, surface temperature and emissivity of steam line. Temperature and convection coefficient associated with ambient air. Efficiency and fuel cost for gas fired furnace. FIND: (a) Rate of heat loss, (b) Annual cost of heat loss. ASSUMPTIONS: (1) Steam line operates continuously throughout year, () Net radiation transfer is between small surface (steam line) and large enclosure (plant walls). ANALYSIS: (a) From Eqs. (1.3a) and (1.7), the heat loss is 4 4 ( ) εσ ( ) q = qconv + qrad = A h Ts T + Ts T sur where π π ( ) Hence, A = DL = 0.1m 5m = 7.85m. q = 7.85m 10 W/m K K W/m K K 4 q = 7.85m 1, , 095 w/m = W = 18, 405 W < (b) The annual energy loss is ( ) ( ) ( ) ( ) 11 E qt 18, 405 W 3600 s/h 4h/d 365 d/y J = = = With a furnace energy consumption of is E E/ f = ηf = J, the annual cost of the loss C = C E 0.01 $/MJ g f = MJ = $6450 < COMMENTS: The heat loss and related costs are unacceptable and should be reduced by insulating the steam line.

4 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. ASSUMPTIONS: (1) Steady-state conditions, () Negligible temperature drop across thin container wall. ANALYSIS: The rate of energy generation is E r qdv=q o g = 1-( r/r ) o o π rldr 0 E Lq r / r g = π o o o /4 or per unit length, πq r E g = oo. ( ) Performing an energy balance for a control surface about the container yields, at an instant, E E = g out 0 and substituting for the convection heat rate per unit length, πq r oo = h( π ro)( Ts T ) T s q r = T + 4h oo. < COMMENTS: The temperature within the radioactive wastes increases with decreasing r from T s at r o to a maximum value at the centerline. <

5 PROBLEM 1.7(a) KNOWN: Solar radiation is incident on an asphalt paving. FIND: Relevant heat transfer processes. The relevant processes shown on the schematic include: q S Incident solar radiation, a large portion of which q S,abs, is absorbed by the asphalt surface, q rad q conv q cond Radiation emitted by the surface to the air, Convection heat transfer from the surface to the air, and Conduction heat transfer from the surface into the asphalt. Applying the surface energy balance, Eq. 1.1, qs,abs q rad qconv = qcond. COMMENTS: (1) q cond and q conv could be evaluated from Eqs. 1.1 and 1.3, respectively. () It has been assumed that the pavement surface temperature is higher than that of the underlying pavement and the air, in which case heat transfer by conduction and convection are from the surface. (3) For simplicity, radiation incident on the pavement due to atmospheric emission has been ignored (see Section 1.8 for a discussion). Eq. 1.6 may then be used for the absorbed solar irradiation and Eq. 1.5 may be used to obtain the emitted radiation q rad. (4) With the rate equations, the energy balance becomes 4 dt q S,abs εσts h( Ts T ) = k. dx s

6 PROBLEM 1.7(b) KNOWN: Physical mechanism for microwave heating. FIND: Comparison of (a) cooking in a microwave oven with a conventional radiant or convection oven and (b) a microwave clothes dryer with a conventional dryer. (a) Microwave cooking occurs as a result of volumetric thermal energy generation throughout the food, without heating of the food container or the oven wall. Conventional cooking relies on radiant heat transfer from the oven walls and/or convection heat transfer from the air space to the surface of the food and subsequent heat transfer by conduction to the core of the food. Microwave cooking is more efficient and is achieved in less time. (b) In a microwave dryer, the microwave radiation would heat the water, but not the fabric, directly (the fabric would be heated indirectly by energy transfer from the water). By heating the water, energy would go directly into evaporation, unlike a conventional dryer where the walls and air are first heated electrically or by a gas heater, and thermal energy is subsequently transferred to the wet clothes. The microwave dryer would still require a rotating drum and air flow to remove the water vapor, but is able to operate more efficiently and at lower temperatures. For a more detailed description of microwave drying, see Mechanical Engineering, March 1993, page 10.

7 PROBLEM 1.7(c) KNOWN: Surface temperature of exposed arm exceeds that of the room air and walls. FIND: Relevant heat transfer processes. Neglecting evaporation from the surface of the skin, the only relevant heat transfer processes are: q conv q rad Convection heat transfer from the skin to the room air, and Net radiation exchange between the surface of the skin and the surroundings (walls of the room). You are not imagining things. Even though the room air is maintained at a fixed temperature (T = 15 C), the inner surface temperature of the outside walls, T sur, will decrease with decreasing outside air temperature. Upon exposure to these walls, body heat loss will be larger due to increased q rad. COMMENTS: The foregoing reasoning assumes that the thermostat measures the true room air temperature and is shielded from radiation exchange with the outside walls.

8 PROBLEM 1.7(g) KNOWN: Fireplace cavity is separated from room air by two glass plates, open at both ends. FIND: Relevant heat transfer processes. The relevant heat transfer processes associated with the double-glazed, glass fire screen are: q rad,1 q rad, Radiation from flames and cavity wall, portions of which are absorbed and transmitted by the two panes, Emission from inner surface of inner pane to cavity, q rad,3 q rad,4 Net radiation exchange between outer surface of inner pane and inner surface of outer pane, Net radiation exchange between outer surface of outer pane and walls of room, q conv,1 Convection between cavity gases and inner pane, q conv q conv,3 Convection across air space between panes, Convection from outer surface to room air, q cond,1 Conduction across inner pane, and q cond, Conduction across outer pane. COMMENTS: (1) Much of the luminous portion of the flame radiation is transmitted to the room interior. () All convection processes are buoyancy driven (free convection).

9 PROBLEM.3 KNOWN: A spherical shell with prescribed geometry and surface temperatures. FIND: Sketch temperature distribution and explain shape of the curve. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction in radial (spherical coordinates) direction, (3) No internal generation, (4) Constant properties. ANALYSIS: Fourier s law, Eq..1, for this one-dimensional, radial (spherical coordinate) system has the form ( π ) dt dt qr = k Ar = k 4 r dr dr where A r is the surface area of a sphere. For steady-state conditions, an energy balance on the system yields E in = Eout, since E g = E st =0. Hence, () qin = qout = qr qr r. That is, q r is a constant, independent of the radial coordinate. Since the thermal conductivity is constant, it follows that dt r = Constant. dr This relation requires that the product of the radial temperature gradient, dt/dr, and the radius squared, r, remains constant throughout the shell. Hence, the temperature distribution appears as shown in the sketch. COMMENTS: Note that, for the above conditions, qr qr() r ; that is, q r is everywhere constant. How does q r vary as a function of radius?

10 PROBLEM.1 KNOWN: Diameter D, thickness L and initial temperature T i of pan. Heat rate from stove to bottom of pan. Convection coefficient h and variation of water temperature T (t) during Stage 1. Temperature T L of pan surface in contact with water during Stage. FIND: Form of heat equation and boundary conditions associated with the two stages. ASSUMPTIONS: (1) One-dimensional conduction in pan bottom, () Heat transfer from stove is uniformly distributed over surface of pan in contact with the stove, (3) Constant properties. ANALYSIS: Stage 1 T 1 T Heat Equation: = x α t Boundary Conditions: T x,0 Initial Condition: ( ) i T q k = q o o = x x= 0 D /4 ( π ) T k = h T( L,t) T ( t) x x= L = T Stage Heat Equation: Boundary Conditions: d T = 0 dx dt k = q o dx x = 0 ( ) = TL T L COMMENTS: Stage 1 is a transient process for which T (t) must be determined separately. As a first approximation, it could be estimated by neglecting changes in thermal energy storage by the pan bottom and assuming that all of the heat transferred from the stove acted to increase thermal energy storage within the water. Hence, with q Mc p d T /dt, where M and c p are the mass and specific heat of the water in the pan, T (t) (q/mc p ) t.

11 PROBLEM.46 KNOWN: Plane wall, initially at a uniform temperature, is suddenly exposed to convective heating. FIND: (a) Differential equation and initial and boundary conditions which may be used to find the temperature distribution, T(x,t); (b) Sketch T(x,t) for these conditions: initial (t 0), steady-state, t, and two intermediate times; (c) Sketch heat fluxes as a function of time for surface locations; (d) Expression for total energy transferred to wall per unit volume (J/m 3 ). ASSUMPTIONS: (1) One-dimensional conduction, () Constant properties, (3) No internal heat generation. ANALYSIS: (a) For one-dimensional conduction with constant properties, the heat equation has the form, T x = 1 T α t and the conditions are: % & K ' K Initial, t 0 T x,0 = T uniform : 1 6 i Boundaries: x = 0 T / x) = 0 adiabatic 0 x = L k T / x) = h T L,t T convection L 1 6 (b) The temperature distributions are shown on the sketch. Note that the gradient at x = 0 is always zero, since this boundary is adiabatic. Note also that the gradient at x = L decreases with time. (c) The heat flux, q x,t = L. x1 6, as a function of time, is shown on the sketch for the surfaces x = 0 and x Continued..

12 PROBLEM.46 (Cont.) For the surface at x = 0, q x 0, t = 0 maximum q x1l,06= h T1L,06 T 1 6 since it is adiabatic. At x = L and t = 0, q x1l,06 is a where T(L,0) = T i. The gradient, and hence the flux, decrease with time. (d) The total energy transferred to the wall may be expressed as I in 0 conv s in s E = q A dt I E = ha T T L,t dt Dividing both sides by A s L, the energy transferred per unit volume is E V I 1 6 h = T T L,t dt J / m L 0 in 3 COMMENTS: Note that the heat flux at x = L is into the wall and is hence in the negative x direction.

13 PROBLEM 3. KNOWN: Temperatures and convection coefficients associated with air at the inner and outer surfaces of a rear window. FIND: (a) Inner and outer window surface temperatures, T s,i and T s,o, and (b) T s,i and T s,o as a function of the outside air temperature T,o and for selected values of outer convection coefficient, h o. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction, (3) Negligible radiation effects, (4) Constant properties. PROPERTIES: Table A-3, Glass (300 K): k = 1.4 W/m K. ANALYSIS: (a) The heat flux may be obtained from Eqs and 3.1, ( ) $ $ T 40 C 10 C,i T,o q = = 1 L m h o k hi 65 W m K 1.4 W m K 30 W m K 50 C q = = 968W m m K W ( ) Hence, with q h ( T T ) $ = i,i,o, the inner surface temperature is q 968W m Ts,i = T,i h = 40 C = 7.7 C i 30 W m K $ $ < q = ho Ts,o T,o find Similarly for the outer surface temperature with ( ) q 968W m Ts,o = T,o h = 10 C = 4.9 C o 65 W m K $ $ < (b) Using the same analysis, T s,i and T s,o have been computed and plotted as a function of the outside air temperature, T,o, for outer convection coefficients of h o =, 65, and 100 W/m K. As expected, T s,i and T s,o are linear with changes in the outside air temperature. The difference between T s,i and T s,o increases with increasing convection coefficient, since the heat flux through the window likewise increases. This difference is larger at lower outside air temperatures for the same reason. Note that with h o = W/m K, T s,i - T s,o, is too small to show on the plot. Continued..

14 PROBLEM 3. (Cont.) Surface temperatures, Tsi or Tso (C) Outside air temperature, Tinfo (C) Tsi; ho = 100 W/m^.K Tso; ho = 100 W/m^.K Tsi; ho = 65 W/m^.K Tso; ho = 65 W/m^.K Tsi or Tso; ho = W/m^.K COMMENTS: (1) The largest resistance is that associated with convection at the inner surface. The values of T s,i and T s,o could be increased by increasing the value of h i. () The IHT Thermal Resistance Network Model was used to create a model of the window and generate the above plot. The Workspace is shown below. // Thermal Resistance Network Model: // The Network: // Heat rates into node j,qij, through thermal resistance Rij q1 = (T - T1) / R1 q3 = (T3 - T) / R3 q43 = (T4 - T3) / R43 // Nodal energy balances q1 + q1 = 0 q - q1 + q3 = 0 q3 - q3 + q43 = 0 q4 - q43 = 0 /* Assigned variables list: deselect the qi, Rij and Ti which are unknowns; set qi = 0 for embedded nodal points at which there is no external source of heat. */ T1 = Tinfo // Outside air temperature, C //q1 = // Heat rate, W T = Tso // Outer surface temperature, C q = 0 // Heat rate, W; node, no external heat source T3 = Tsi // Inner surface temperature, C q3 = 0 // Heat rate, W; node, no external heat source T4 = Tinfi // Inside air temperature, C //q4 = // Heat rate, W // Thermal Resistances: R1 = 1 / ( ho * As ) R3 = L / ( k * As ) R43 = 1 / ( hi * As ) // Convection thermal resistance, K/W; outer surface // Conduction thermal resistance, K/W; glass // Convection thermal resistance, K/W; inner surface // Other Assigned Variables: Tinfo = -10 // Outside air temperature, C ho = 65 // Convection coefficient, W/m^.K; outer surface L = // Thickness, m; glass k = 1.4 // Thermal conductivity, W/m.K; glass Tinfi = 40 // Inside air temperature, C hi = 30 // Convection coefficient, W/m^.K; inner surface As = 1 // Cross-sectional area, m^; unit area

15 PROBLEM 3.37 KNOWN: Inner and outer radii of a tube wall which is heated electrically at its outer surface and is exposed to a fluid of prescribed h and T. Thermal contact resistance between heater and tube wall and wall inner surface temperature. FIND: Heater power per unit length required to maintain a heater temperature of 5 C. ASSUMPTIONS: (1) Steady-state conditions, () One-dimensional conduction, (3) Constant properties, (4) Negligible temperature drop across heater. ANALYSIS: The thermal circuit has the form Applying an energy balance to a control surface about the heater, q = q a + q b To Ti To T q = + ln( r o / ri) ( 1/hπ Do ) + R t,c π k $ $ ( 5-5) C 5 ( 10) C q= + ln ( 75mm/5mm ) m K 1/ 0.01 ( 100 W/m K π 0.15m) + π 10 W/m K W ( ) q = W/m q =377 W/m. < COMMENTS: The conduction, contact and convection resistances are , 0.01 and 0.01 m K/W, respectively,