Temperature Response in Cooling a Wire. A small copper wire with a diameter of 0.792 mm and initially at 366.5 K is suddenly immersed in a liquid held constant at 311 K. The convection coefficient h = 85.2 W/m² · K. The physical properties can be assumed constant and are k = 374 W/m · K, c_p = 0.389 kJ/kg · K, and ρ = 8890 kg/m³.

Quenching Lead Shot in a Bath. Lead shot having an average diameter of 5.1 mm is at an initial temperature of 204.4°C. To quench the shot it is added to a quenching oil bath held at 32.2°C and falls to the bottom. The time of fall is 15 s. Assuming an average convection coefficient of h = 199 W/m² · K, what will be the temperature of the shot after the fall? For lead, ρ = 11 370 kg/m³ and c_p = 0.138 kJ/kg · K.

Unsteady-State Heating of a Stirred Tank. A vessel is filled with 0.0283 m³ of water initially at 288.8 K. The vessel, which is well stirred, is suddenly immersed in a steam bath held at 377.6 K. The overall heat-transfer coefficient U between the steam and water is 1136 W/m² · K and the area is 0.372 m². Neglecting the heat capacity of the walls and agitator, calculate the time in hours to heat the water to 338.7 K. [Hint: Since the water is well stirred, its temperature is uniform. Show that Eq. (5.2-3) holds by starting with Eq. (5.2-1).]

Temperature in a Refractory Lining. A combustion chamber has a 2-in.-thick refractory lining to protect the outer shell. To predict the thermal stresses at start-up, the temperature 0.2 in. below the surface is needed 1 min after start-up. The following data are available. The initial temperature T₀ = 100°F, the hot gas temperature T₁ = 3000°F, h = 40 btu/h · ft² · °F, k = 0.6 btu/h · ft · °F, and α = 0.020 ft²/h. Calculate the temperature at a 0.2-in. depth and a 0.6-in. depth. Use Fig. 5.3-3 and justify its use by seeing if the lining acts as a semi-infinite solid during this 1-min period.

Freezing Temperature in the Soil. The average temperature of the soil to a considerable depth is approximately 277.6 K (40°F) during a winter day. If the outside air temperature suddenly drops to 255.4 K (0°F) and stays there, how long will it take for a pipe 3.05 m (10 ft) below the surface to reach 273.2 K (32°F)? The convective coefficient is h = 8.52 W/m² · K (1.5 btu/h · ft² · °F). The soil physical properties can be taken as 5.16 × 10⁻⁷ m²/s (0.02 ft²/h) for the thermal diffusivity and 1.384 W/m · K (0.8 btu/h · ft · °F) for the thermal conductivity. (Note: The solution is trial and error, since the unknown time appears twice in the graph for a semi-infinite solid.)

Cooling a Slab of Aluminum. A large piece of aluminum that can be considered a semi-infinite solid initially has a uniform temperature of 505.4 K. The surface is suddenly exposed to an environment at 338.8 K with a surface convection coefficient of 455 W/m² · K. Calculate the time in hours for the temperature to reach 388.8 K at a depth of 25.4 mm. The average physical properties are α = 0.340 m²/h and k = 208 W/m · K.

Transient Heating of a Concrete Wall. A wall made of concrete 0.305 m thick is insulated on the rear side. The wall at a uniform temperature of 10°C (283.2 K) is exposed on the front side to a gas at 843°C (1116.2 K). The convection coefficient is 28.4 W/m² · K, the thermal diffusivity is 1.74 × 10⁻³ m²/h, and the thermal conductivity is 0.935 W/m · K.

Cooking a Slab of Meat. A slab of meat 25.4 mm thick originally at a uniform temperature of 10°C is to be cooked from both sides until the center reaches 121°C in an oven at 177°C. The convection coefficient can be assumed constant at 25.6 W/m² · K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m · K and the thermal diffusivity 5.85 × 10⁻⁴ m²/h. Use the Heisler chart.

Unsteady-State Conduction in a Brick Wall. A flat brick wall 1.0 ft thick is the lining on one side of a furnace. If the wall is at a uniform temperature of 100°F and one side is suddenly exposed to a gas at 1100°F, calculate the time for the furnace wall at a point 0.5 ft from the surface to reach 500°F. The rear side of the wall is insulated. The convection coefficient is 2.6 btu/h · ft² · °F and the physical properties of the brick are k = 0.65 btu/h · ft · °F and α = 0.02 ft²/h.

Cooling a Steel Rod. A long steel rod 0.305 m in diameter is initially at a temperature of 588 K. It is immersed in an oil bath maintained at 311 K. The surface convective coefficient is 125 W/m² · K. Calculate the temperature at the center of the rod after 1 h. The average physical properties of the steel are k = 38 W/m · K and α = 0.0381 m²/h.

Effect of Size on Heat Processing Meat. An autoclave held at 121.1°C is being used to process sausage meat 101.6 mm in diameter and 0.61 m long which is originally at 21.1°C. After 2 h the temperature at the center is 98.9°C. If the diameter is increased to 139.7 mm, how long will it take for the center to reach 98.9°C? The heat-transfer coefficient to the surface is h = 1100 W/m² · K, which is very large, so the surface resistance can be considered negligible. (Show this.) Neglect the heat transfer from the ends of the cylinder. The thermal conductivity k = 0.485 W/m · K.

Temperature of Oranges on Trees During Freezing Weather. In orange-growing areas, the freezing of the oranges on the trees during cold nights is of serious economic concern. If the oranges are initially at a temperature of 21.1°C, calculate the center temperature of the orange if exposed to air at −3.9°C for 6 h. The oranges are 102 mm in diameter and the convective coefficient is estimated as 11.4 W/m² · K. The thermal conductivity k is 0.431 W/m · K and α is 4.65 × 10⁻⁴ m²/h. Neglect any latent heat effects.

Hardening a Steel Sphere. To harden a steel sphere having a diameter of 50.8 mm, it is heated to 1033 K and then dunked into a large water bath at 300 K. Determine the time for the center of the sphere to reach 366.5 K. The surface coefficient can be assumed as 710 W/m² · K, k = 45 W/m · K, and α = 0.0325 m²/h.

Unsteady-State Conduction in a Short Cylinder. An aluminum cylinder is initially heated to a uniform temperature of 204.4°C. Then it is plunged into a large bath held at 93.3°C, where h = 568 W/m² · K. The cylinder has a diameter of 50.8 mm and is 101.6 mm long. Calculate the center temperature after 60 s. The physical properties are α = 9.44 × 10⁻⁵ m²/s and k = 207.7 W/m · K.

Conduction in Three Dimensions in a Rectangular Block. A rectangular steel block 0.305 m by 0.457 m by 0.61 m is initially at 315.6°C. It is suddenly immersed in an environment at 93.3°C. Determine the temperature at the center of the block after 1 h. The surface convection coefficient is 34 W/m² · K. The physical properties are k = 38 W/m · K and α = 0.0379 m²/h.

Schmidt Numerical Method for Unsteady-State Conduction. A material in the form of an infinite plate 0.762 m thick is at an initial uniform temperature of 366.53 K. The rear face of the plate is insulated. The front face is suddenly exposed to a temperature of 533.2 K. The convective resistance at this face can be assumed as zero. Calculate the temperature profile after 0.875 h using the Schmidt numerical method with M = 2 and slabs 0.1524 m thick. The thermal diffusivity is 0.0929 m²/h.

Unsteady-State Conduction with Nonuniform Initial Temperature Profile. Use the same conditions as in Problem 5.4-1 but with the following change. The initial temperature profile is not uniform but is 366.53 K at the front face and 422.1 K at the rear face with a linear variation between the two faces.

Unsteady-State Conduction Using the Digital Computer. Repeat Problem 5.4-2 but use the computer and a spreadsheet. Use slabs 0.03048 m thick and M = 2.0. Calculate the temperature profile after 0.875 h.

Chilling Meat Using Numerical Methods. A slab of beef 45.7 mm thick and initially at a uniform temperature of 283 K is being chilled by a surface contact cooler at 274.7 K on the front face. The rear face of the meat is insulated. Assume that the convection resistance at the front surface is zero. Using five slices and M = 2, calculate the temperature profile after 0.54 h. The thermal diffusivity is 4.64 × 10⁻⁴ m²/h.

Cooling Beef with Convective Resistance. A large slab of beef is 45.7 mm thick and is at an initial uniform temperature of 37.78°C. It is being chilled at the front surface in a chilled air blast at −1.11°C with a convective heat-transfer coefficient of h = 38.0 W/m² · K. The rear face of the meat is insulated. The thermal conductivity of the beef is k = 0.498 W/m · K and α = 4.64 × 10⁻⁴ m²/h. Using a numerical method with five slices and M = 4.0, calculate the temperature profile after 0.27 h. [Hint: Since there is a convective resistance, the value of N must be calculated. Also, Eq. (5.4-7) should be used.]

Cooling Beef Using the Digital Computer. Repeat Problem 5.4-5 using the digital computer. Use 20 slices and M = 4.0. Use a spreadsheet calculation.

Convection and Unsteady-State Conduction. For the conditions of Example 5.4-3, continue the calculations for a total of 12 time increments. Plot the temperature profile.

Alternative Convective Boundary Condition for Numerical Method. Repeat Example 5.4-3 but instead use the alternative boundary condition, Eq. (5.4-11). Also, use M = 4. Calculate the profile for the full 12 time increments.

Numerical Method for Semi-infinite Solid and Convection. A semi-infinite solid initially at a uniform temperature of 200°C is cooled at its surface by convection. The cooling fluid at a constant temperature of 100°C has a convective coefficient of h = 250 W/m² · K. The physical properties of the solid are k = 20 W/m · K and α = 4 × 10⁻⁵ m²/s. Using a numerical method with Δx = 0.040 m and M = 4.0, calculate the temperature profile after 50 s total time.

Chilling Slab of Beef. Repeat Example 5.5-1, where the slab of beef is cooled to 10°C at the center, but use air of 0°C at a lower value of h = 22.7 W/m² · K.

Chilling Fish Fillets. Codfish fillets originally at 10°C are packed to a thickness of 102 mm. Ice is packed on both sides of the fillets and wet-strength paper separates the ice and fillets. The surface temperature of the fish can be assumed as essentially 0°C. Calculate the time for the center of the fillets to reach 2.22°C and the temperature at this time at a distance of 25.4 mm from the surface. Also, plot temperature versus position for the slab. The physical properties are (B1) k = 0.571 W/m · K, ρ = 1052 kg/m³, and c_p = 4.02 kJ/kg · K.

Average Temperature in Chilling Fish. Fish fillets having the same physical properties given in Problem 5.5-2 are originally at 10°C. They are packed to a thickness of 102 mm with ice on each side. Assuming that the surface temperature of the fillets is 0°C, calculate the time for the average temperature to reach 1.39°C. (Note: This is a case where the surface resistance is zero. Can Fig. 5.3-13 be used for this case?)

Time to Freeze a Slab of Meat. Repeat Example 5.5-2 using the same conditions except that a plate or contact freezer is used where the surface coefficient can be assumed as h = 142 W/m² · K.

Freezing a Cylinder of Meat. A package of meat containing 75% moisture and in the form of a long cylinder 5 in. in diameter is to be frozen in an air-blast freezer at −25°F. The meat is initially at the freezing temperature of 27°F. The heat-transfer coefficient is h = 3.5 btu/h · ft² · °F. The physical properties are ρ = 64 lb_m/ft³ for the unfrozen meat and k = 0.60 btu/h · ft · °F for the frozen meat. Calculate the freezing time.

Heat Generation Using Equation of Energy Change. A plane wall with uniform internal heat generation of W/m³ is insulated at four surfaces, with heat conduction only in the x direction. The wall has a thickness of 2L m. The temperature at one wall at x = +L and at the other wall at x = −L is held constant at T_w K. Using the differential equation of energy change, Eq. (5.6-18), derive the equation for the final temperature profile.

Heat Transfer in a Solid Using Equation of Energy Change. A solid of thickness L is at a uniform temperature of T₀ K. Suddenly the front surface temperature of the solid at z = 0 m is raised to T₁ at t = 0 and held there, and at z = L at the rear to T₂ and held constant. Heat transfer occurs only in the z direction. For constant physical properties and using the differential equation of energy change, do as follows:

Radial Temperature Profile Using the Equation of Energy Change. Radial heat transfer is occurring by conduction through a long, hollow cylinder of length L with the ends insulated.

Heat Conduction in a Sphere. Radial energy flow is occurring in a hollow sphere with an inside radius of r_i and an outside radius of r_o. At steady state the inside surface temperature is constant at T_i and constant at T_o on the outside surface.

Variable Heat Generation and Equation of Energy Change. A plane wall is insulated so that conduction occurs only in the x direction. The boundary conditions which apply at steady state are T = T₀ at x = 0 and T = T_L at x = L. Internal heat generation per unit volume is occurring and varies as , where and ß are constants. Solve the general differential equation of energy change for the temperature profile.

Thermal and Hydrodynamic Boundary Layer Thicknesses. Air at 294.3 K and 101.3 kPa with a free stream velocity of 12.2 m/s is flowing parallel to a smooth, flat plate held at a surface temperature of 383 K. Do the following:

Boundary-Layer Thicknesses and Heat Transfer. Air at 37.8°C and 1 atm abs flows at a velocity of 3.05 m/s parallel to a flat plate held at 93.3°C. The plate is 1 m wide. Calculate the following at a position 0.61 m from the leading edge: