Surface of the Cylinder

Surface of the Cylinder The lumped system analysis is more likely to be applicable for the body allowed to cool in the air since the Biot number is proportional to the convection heat transfer coefficient, which is larger in water than it is in air because of the larger thermal conductivity of water. Therefore, the Biot number is more likely to be less than 0.1 for the case of the solid cooled in the airThe temperature of a gas stream is to be measured by a thermocouple. The time it takes to register 99 percent of the initial ∆ T is to be determined. Assumptions 1 The junction is spherical in shape with a diameter of D = 0.0012 m. 2 The thermal properties of the junction are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 Radiation effects are negligible. 5 The Biot number is Bi < 0.1 so that the lumped system analysis is applicable (this assumption will be verified). Properties hTe properties of the junction are given to be k = 35 W / m.° C, ρ = 8500 kg / m3, and C p = 320 J / kg.° C . Analysis The characteristic length of the junction and the Biot number are Lc = Bi = V Asurface = πD 3 / 6 D 0.0012 m = = = 0.0002 m 6 6 πD 2 hLc (65 W / m 2 . ° C)(0.0002 m) = = 0.00037 < 01 . k (35 W / m. ° C) Since Bi < 0.1 , the lumped system analysis is applicable. Then the time period for the thermocouple to read 99% of the initial temperature difference is determined from Gas h, T∞ T (t ) − T∞ = 0.01 Ti − T∞ b= Junction D T(t) hA h 65 W / m 2 . ° C = 01195 s-1 = = . ρC pV ρC p Lc (8500 kg / m3 )(320 J / kg. ° C)(0.0002 m) -1 T (t ) − T∞ = e −bt  → 0.01 = e − ( 0.1195 s ) t  → t = 38.5 s Ti − T∞ A long cylindrical shaft at 400°C is allowed to cool slowly. The center temperature and the heat transfer per unit length of the cylinder are to be determined. Assumptions 1 Heat conduction in the shaft is one-dimensional since it is long and it has thermal symmetry about the center line. 2 The thermal properties of the shaft are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of stainless steel 304 at room temperature are given to be k = 14.9 W/m.°C, ρ = 7900 kg/m3, Cp = 477 J/kg.°C, α = 3.95×10-6 m2/s Analysis First the Biot number is calculated to be Bi = hro (60 W/m 2 .°C)(0.175 m ) = = 0.705 k (14.9 W/m.°C) The constants λ 1 and A1 corresponding to this Biot number are, from Table 4-1, λ 1 = 10935 . and A1 = 11558 . The Fourier number is Air T = 150°C Steel shaft Ti = 400°C τ= αt = 2 L (3.95 × 10 −6 m 2 /s)(20 × 60 s) (0.175 m) 2 = 0.1548 which is very close to the value of 0.2. Therefore, the one-term approximate solution (or the transient temperature charts) can still be used, with the understanding that the error involved will be a little more than 2 percent. Then the temperature at the center of the shaft becomes θ 0,cyl = 2 2 T0 − T∞ = A1 e −λ1 τ = (1.1558)e −(1.0935) (0.1548) = 0.9605 Ti − T∞ T0 − 150 = 0.9605  → T0 = 390 °C 400 − 150 The maximum heat can be transferred from the cylinder per meter of its length is m = ρV = ρπro 2 L = (7900 kg/m 3 )[π(0.175 m) 2 (1 m)] = 760.1 kg Q max = mC p [T∞ − Ti ] = (760.1 kg )(0.477 kJ/kg.°C)(400 − 150)°C = 90,638 kJ Once the constant J1 = 0.4689 is determined from Table 4-2 corresponding to the constant λ 1 =1.0935, the actual heat transfer becomes  Q  Q  max   T − T∞  = 1 − 2 0  T −T  ∞  cyl  i  J 1 (λ 1 )  390 − 150  0.4689  = 0.177 = 1 − 2   λ  400 − 150  1.0935 1  Q = 0.177(90,638 kJ ) = 16,015 kJ Steaks are cooled by passing them through a refrigeration room. The time of cooling is to be determined. Assumptions 1 Heat conduction in the steaks is one-dimensional since the steaks are large relative to their thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steaks are constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of steaks are given to be k = 0.45 W/m.°C and α = 0.91×10-7 m2/s Analysis The Biot number is Bi = hL (9 W/m 2 .°C)(0.01 m) = = 0.200 (0.45 W/m.°C) k The constants λ 1 and A1 corresponding to this Biot number are, from Table 4-1, λ 1 = 0.4328 and A1 = 10311 . The Fourier number is 2 T ( L , t ) − T∞ = A1e −λ1 τ cos(λ1 L / L ) Ti − T∞ 2 2 − (−11) = (1.0311)e −(0.4328) τ cos(0.4328)  → τ = 5.601 > 0.2 25 − (−11) Steaks 25°C Refrigerated air 11°C Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the length of time for the steaks to be kept in the refrigerator is determined to be t= (5.601)(0.01 m) 2 τL2 = = 6155 s = 102.6 min α (0.91× 10 −7 m 2 /s) The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a particular location is to be determined. Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and thus the soil can be considered to be a semi-infinite medium with a specified surface temperature. 2 The thermal properties of the soil are constant. Properties The thermal properties of the soil are given to be k = 0.35 W/m.°C and α = 0.15×10-6 m2/s. Analysis The length of time the snow pack stays on the ground is t = (60 days)(24 hr / days)(3600 s / hr) = 5.184 × 106 s The surface is kept at -18°C at all times. The depth at which freezing at 0°C occurs can be determined from the analytical solution, Ts =Soil Ti = 8°C  x  T ( x, t ) − Ti  = erfc  Ts − Ti  αt    x 0−8   = erfc 6 2 6 − −8−8  2 (0.15 × 10 m /s)(5.184 × 10 s)     x   → 0.444 = erfc   1.7636  Water pipe x = 0.5297  → x = 0.934 m 1.7636 Then from Table 4-3 we get Discussion The solution could also be determined using the chart, but it would be subject to reading error. A thick wood slab is exposed to hot gases for a period of 5 minutes. It is to be determined whether the wood will ignite. Assumptions 1 The wood slab is treated as a semi-infinite medium subjected to convection at the exposed surface. 2 The thermal properties of the wood slab are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. Properties The thermal properties of the wood are k = 0.17 W/m.°C and α = 1.28×10-7 m2/s. Analysis The one-dimensional transient temperature distribution in the wood can be determined from  x T ( x, t ) − Ti = erfc T∞ − Ti  2 αt    hx h 2αt     erfc x + h αt   − exp +   k  2 αt k  k 2      where 2 -7 h αt (35 W/m .°C) (1.28 × 10 m / s)(5 × 60 s) = = 1.276 0.17 W/m.°C k 2 h αt k 2 2  h αt   = 1.276 2 = 1.628 =  k    Wood Slab Ti = 2 Hot gases T∞ = L=0.3 Noting that x = 0 at the surface and using Table 4-3 for erfc values, 0 x T ( x, t ) − 25 = erfc(0) − exp(0 + 1.628)erfc(0 + 1.276 ) 550 − 25 = 1 − (5.0937)(0.0727) = 0.630 Solving for T(x, t) gives T ( x, t ) = 356°C which is less than the ignition temperature of 450°C. Therefore, the wood will not ignite. A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface as well as the total heat transfer from the cylinder for 15 min of cooling are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of brass are given to be ρ = 8530 kg / m3 , C p = 0.389 kJ/kg ⋅ °C , k = 110 W/m ⋅°C , and α = 3.39 × 10 −5 m 2 / s . Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 4 cm and a plane wall of thickness 2L = 15 cm. We measure x from the midplane. (a) The Biot number is calculated for the plane wall to be Bi = hL (40 W/m 2 .°C)(0.075 m) = = 0.02727 k (110 W/m.°C) D0 = 8 cm The constants λ 1 and A1 corresponding to this Biot Air number are, from Table 4-1, T∞ = λ1 = 0.164 and A1 = 1.0050 τ= L2 (3.39 × 10 −5 m 2 / s)(15 min × 60 s / min) = r L = 15 Brass cylinder The Fourier number is αt z (0.075 m) 2 Ti = 150°C = 5.424 > 0.2 Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from θ o, wall = 2 2 T0 − T∞ = A1 e −λ1 τ = (1.0050)e −(0.164) (5.424) = 0.869 Ti − T∞ We repeat the same calculations for the long cylinder, Bi = hr0 (40 W/m 2 .°C)(0.04 m) = = 0.01455 k (110 W/m.°C) λ 1 = 01704 . and A1 = 10038 . τ= αt ro 2 θ o ,cyl = = (3.39 × 10 −5 m 2 / s)(15 × 60 s) (0.04 m) 2 = 19.069 > 0.2 2 2 T0 − T∞ = A1e − λ1 τ = (10038 . )e − ( 0.1704 ) (19.069 ) = 0.577 Ti − T∞ Then the center temperature of the short cylinder becomes  T (0,0, t ) − T∞  = θ o, wall × θ o,cyl = 0.869 × 0.577 = 0.501   short  Ti − T∞  cylinder T (0,0, t ) − 20 = 0.501  → T (0,0, t ) = 85.1°C 150 − 20 (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to determine the dimensionless temperature at the surface of the wall. θ ( L, t ) wall = 2 2 T ( x , t ) − T∞ = A1 e −λ1 τ cos(λ1 L / L) = (1.0050)e −(0.164) (5.424) cos(0.164) = 0.857 Ti − T∞ Then the center temperature of the top surface of the cylinder becomes  T ( L,0, t ) − T∞  = θ ( L, t ) wall × θ o,cyl = 0.857 × 0.577 = 0.494   short  Ti − T∞  cylinder T ( L,0, t ) − 20 = 0.494  → T ( L,0, t ) = 84.2°C 150 − 20 (c) We first need to determine the maximum heat can be transferred from the cylinder [ ] m = ρV = ρπro 2 L = (8530 kg/m 3 ) π(0.04 m) 2. (0.15 m) = 6.43 kg Q max = mC p (Ti − T∞ ) = (6.43 kg )(0.389 kJ/kg.°C)(150 − 20)°C = 325 kJ Then we determine the dimensionless heat transfer ratios for both geometries as  Q  Q  max  sin(λ 1 ) sin(0.164)  = 1 − θ o, wall = 1 − (0.869) = 0.135  0.164 λ 1  wall  Q  Q  max  J (λ ) 0.0846  = 1 − 2θ o,cyl 1 1 = 1 − 2(0.577) = 0.427  0.1704 λ1  cyl The heat transfer ratio for the short cylinder is  Q  Q  max   Q    short =  Q  cylinder  max   Q    plane +  Q  wall  max     Q   long 1 −  Q  cylinder   max       plane  = 0.135 + (0.427)(1 − 0.135) = 0.504  wall   Then the total heat transfer from the short cylinder during the first 15 minutes of cooling becomes Q = 0.503Q max = (0.504)(325 kJ) = 164 kJ