2-14E The power consumed by the resistance wire of an iron is given. The heat generation and the heat flux are to be determined. Assumptions Heat is generated uniformly in the resistance wire. q = 1000 W Analysis A 1000 W iron will convert electrical energy into heat in the wire at a rate of 1000 W. Therefore, the rate of heat generation in a D = 0.08 in resistance wire is simply equal to the power rating of a resistance heater. Then the rate of heat generation in the wire per unit volume is L = 15 in determined by dividing the total rate of heat generation by the volume of the wire to be G G 1000 W  3.412 Btu/h  7 3 g = = =  = 7.820 × 10 Btu/h ⋅ ft  2 2 V wire (πD / 4) L [π(0.08 / 12 ft) / 4](15 / 12 ft)  1W  Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determined by dividing the total rate of heat generation by the surface area of the wire to be G G 1000 W  3.412 Btu/h  5 2 = = q =   = 1.303 × 10 Btu/h ⋅ ft Awire πDL π (0.08 / 12 ft)(15 / 12 ft)  1W  Discussion Note that heat generation is expressed per unit volume in Btu/h⋅ft3 whereas heat flux is expressed per unit surface area in Btu/h⋅ft2. 2-26 For a medium in which the heat conduction equation is given by 1 ∂  2 ∂T  1 ∂T = r r 2 ∂r  ∂r  α ∂ t (a) Heat transfer is transient, (b) it is one-dimensional, (c) there is no heat generation, and (d) the thermal conductivity is constant. 2-42 A long pipe of inner radius r1 , outer radius r2 , and thermal conductivity k is considered. The outer surface of the pipe is subjected to convection to a medium at T∞ with a heat transfer coefficient of h. Assuming steady one-dimensional conduction in the radial direction, the convection boundary condition on the outer surface of the pipe can be expressed as h T∞ dT (r2 ) −k = h[T (r2 ) − T∞ ] dr r1 r2 2-56 A large plane wall is subjected to specified temperature on the left surface and convection on the right surface. The mathematical formulation, the variation of temperature, and the rate of heat transfer are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Properties The thermal conductivity is given to be k = 2.3 W/m⋅°C. Analysis (a) Taking the direction normal to the surface of the wall to be the x direction with x = 0 at the left surface, the mathematical formulation of this problem can be expressed as d 2T =0 dx 2 and T (0) = T1 = 80° C T1=80°C A=20 m2 k L=0.4 m T∞ =15°C h=24 W/m2.°C x dT ( L) = h[T ( L) − T∞ ] dx (b) Integrating the differential equation twice with respect to x yields dT = C1 dx −k T ( x ) = C1x + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give x = 0: T (0) = C1 × 0 + C2 x = L: − kC1 = h[(C1 L + C2 ) − T∞ ] → → C2 = T1 C1 = − h(C2 − T∞ ) k + hL → C1 = − h(T1 − T∞ ) k + hL Substituting C1 and C2 into the general solution, the variation of temperature is determined to be T ( x) = − =− h(T1 − T∞ ) x + T1 k + hL (24 W / m 2 ⋅° C)(80 − 15)° C (2.3 W / m⋅° C) + (24 W / m 2 ⋅° C)(0.4 m) = 80 − 1311 .x x + 80° C (c) The rate of heat conduction through the wall is h(T1 − T∞ ) dT Q wall = −kA = −kAC1 = kA dx k + hL (24 W/m 2 ⋅ °C)(80 − 15)°C = (2.3 W/m ⋅ °C)(20 m 2 ) (2.3 W/m ⋅ °C) + (24 W/m 2 ⋅ °C)(0.4 m) = 6030 W Note that under steady conditions the rate of heat conduction through a plain wall is constant. 2-62E A steam pipe is subjected to convection on the inner surface and to specified temperature on the outer surface. The mathematical formulation, the variation of temperature in the pipe, and the rate of heat loss are to be determined for steady one-dimensional heat transfer. Assumptions 1 Heat conduction is steady and one-dimensional since the pipe is long relative to its thickness, and there is thermal symmetry about the center line. 2 Thermal conductivity is constant. 3 There is no heat generation in the pipe. Properties The thermal conductivity is given to be k = 7.2 Btu/h⋅ft⋅°F. Analysis (a) Noting that heat transfer is one-dimensional in the radial r direction, the mathematical formulation of this problem can be expressed as d  dT  =0 r dr  dr  and −k dT (r1 ) = h[T∞ − T (r1 )] dr T (r2 ) = T2 = 160° F T =160°F Steam 250°F h=1.25 (b) Integrating the differential equation once with respect to r gives r dT = C1 dr L = 15 ft Dividing both sides of the equation above by r to bring it to a readily integrable form and then integrating, dT C1 = dr r T (r ) = C1 ln r + C2 where C1 and C2 are arbitrary constants. Applying the boundary conditions give C1 = h[T∞ − (C1 ln r1 + C2 )] r1 r = r1: −k r = r2: T (r2 ) = C1 ln r2 + C2 = T2 Solving for C1 and C2 simultaneously gives C1 = T2 − T∞ r k ln 2 + r1 hr1 and C2 = T2 − C1 ln r2 = T2 − T2 − T∞ ln r2 r k ln 2 + r1 hr1 Substituting C1 and C2 into the general solution, the variation of temperature is determined to be = T (= r ) C1 ln r + T2 − C1 ln= r2 C1 (ln r − ln r2 ) + T 2 T2 − T∞ r ln + T2 r k r2 ln 2 + r1 hr1 (160 − 250)°F r r = + 160°F = −2.59 ln + 160°F ln 2.4 7.2 Btu/h ⋅ ft ⋅°F 2.4 in 2.4 in + ln 2 (1.25 Btu/h ⋅ ft 2 ⋅°F)(2 /12 ft) (c) The rate of heat conduction through the pipe is dT C T −T −kA = −k (2π rL) 1 = −2π Lk 2 ∞ Q = r k dr r ln 2 + r1 hr1 = −2π (15 ft)(7.2 Btu/h ⋅ ft ⋅°F) (160 − 250)°F = 1758 Btu/h 2.4 7.2 Btu/h ⋅ ft ⋅°F ln + 2 (1.25 Btu/h ⋅ ft 2 ⋅°F)(2 /12 ft) 2-86 A long resistance heater wire is subjected to convection at its outer surface. The surface temperature of the wire is to be determined using the applicable relations directly and by solving the applicable differential equation. Assumptions 1 Heat transfer is steady since there is no indication of any change with time. 2 Heat transfer is onedimensional since there is thermal symmetry about the center line and no change in the axial direction. 3 Thermal conductivity is constant. 4 Heat generation in the wire is uniform. Properties The thermal conductivity is given to be k = 15.1 W/m⋅°C. Analysis (a) The heat generation per unit volume of the wire is Q gen Q gen 2000 W g = . = = = 1061 × 108 W / m 3 2 Vwire πro L π (0.001 m) 2 (6 m) The surface temperature of the wire is then (Eq. 2-68) T∞ h T∞ h k g 0 ro r Ts = T∞ + o gr × 108 W / m 3 )(0.001 m) (1061 . = 409° C = 30° C + 2h 2(140 W / m 2 . ° C) (b) The mathematical formulation of this problem can be expressed as 1 d  dT  g + = 0 r r dr  dr  k dT (r0 ) = h[T (r0 ) − T∞ ] (convection at the outer surface) dr dT (0) = 0 (thermal symmetry about the centerline) dr Multiplying both sides of the differential equation by r and integrating gives g dT g r 2 d  dT  (a) =− + C1 =− r → r r dr k 2 dr  dr  k Applying the boundary condition at the center line, dT (0) g B.C. at r = 0: 0× =− × 0 + C1 → C1 = 0 dr 2k Dividing both sides of Eq. (a) by r to bring it to a readily integrable form and integrating, g 2 dT g → (b) T (r ) = − r + C2 =− r 4k dr 2k Applying the boundary condition at r = r0 , and − k gr0 gr g 2  g 2  = h − r0 + C 2 − T∞  → C 2 = T∞ + 0 + r0 2k 2h 4k  4k  Substituting this C2 relation into Eq. (b) and rearranging give  gr g 2 T (r ) = T∞ + (r0 − r 2 ) + 0 4k 2h which is the temperature distribution in the wire as a function of r. Then the temperature of the wire at the surface (r = r0 ) is determined by substituting the known quantities to be B. C. at r = r0 : T (r0 ) = T∞ + −k   gr gr × 108 W / m 3 )(0.001 m) g 2 (1061 . = 409° C (r0 − r02 ) + 0 = T∞ + o = 30° C + 4k 2h 2h 2(140 W / m 2 . ° C) Note that both approaches give the same result.