Example 4.10-1. Radiation To A Metal Tube
Example 4.10-1. Radiation To A Metal Tube
Example 4.10-1. Radiation To A Metal Tube
5.3-7. Cooling a Steel Rod. A long steel rod 0.305m in diameter is initially at a temperature 0f 588K. It is
immersed in an oil bath maintained at 311K. The surface convective coefficient is 125 W/m 2·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.
Ans. T = 391K
5.3-8. Effect of Size on Heat Processing Meat. An autoclave held at 121.1°C is being used to process sausage
meat 101.6mm in diameter and 0.61m long which is originally at 21.1°C. After 2 h the temperature at the center
is 98.9°C. If the diameter is increase to 139.7mm, 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.
Ans. 3.78h
5.3-9. Temperature of Oranges on Trees During Freezing Weather. In orange-growing ares, the freezing of
the orange on the trees during cold night is of serious economic concern. If the oranges are initially at the
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/m 2·K. The thermal
conductivity k is 0.431 W/m·K and α is 4.65x10-4 m2/h. Neglect any latent heat effects.
Ans. (T1 – T)/( T1 – T0) = 0.05, T = 2.65°C
5.3-10 Hardening a Steel Sphere. To harden a steel sphere having a diameter 0f 50.8mm, it is heated to 1033K
and then dunked into a large water bath at 300K. Determine the time for the center of the sphere to reach
366.5K. the surface coefficient can be assumed as 710 W/m2·K, k = 45 W/m·K, and α =0.0325m2/h.