Question

Metal Shield = 0.8 4-02 4. (Total 8pts) Hot gases flows into a long pipe of 0.05-m diameter and at a certain location the flow has a temperature of T.-700K and heat transfer coefficient of h. For safety reasons, the pipe is to be covered with a thin concentric metal shield at Tai - 337K with a 0.1-m diameter. The emissivity of the pipe wall surface is Epi-. The inside and outside surface emissivities of the shield are a = 0.2 and & = 0.8, respectively. The pipe with shield cover is placed in an ambient surrounding at temperature Tyu-300K. We wish to determine the heat rate per unit length extracted from the hot gas and the temperature of pipe surface. -5.67x 10" wm-K Hot Gas (1) The heat transfer modes involved in this problem are (A) In series (B) In parallel (C) Combined series and parallel (D) None of above (2) The heat rate per unit length (W/m) extracted from the hot gas is most nearly? (A) 8.542 (B) 17.08 (C) 34.17 (D) 68.33 (3) The temperature of pipe surface (K) is more nearly (A) 378.8 (B) 410.0 (C) 435.4 (D) 475.9 (4) If the temperature of pipe surface is 690K, the heat transfer coefficient between hot gas and pipe wall (W/m²K) is most nearly (A) 43.52 (B) 21.76 (C) 10.88 (D) 5.44 (5) My favorite class at Sac State has been? A. Heat Transfer B. Heat Transfer C. Heat Transfer D. Heat Transfer

Answer 1

1. As the heat goes in a straight line from the gas to the surroundings without there being any diversions, the heat transfer is in series

2. 68.33 W/m . As the system is maintaining its temperature, the net heat gained by the metal shield is the same as the net heat lost by it. Hence we find the net heat transfer from the metal pipe.

3. 378.8 . The heat transferred by the pipe is equated to the heat calculated above and the surface temperature of the pipe is calculated

4. 43.52 .The heat is transferred by convection between the hot gas and the pipe. Using convective heat transfer equations we find the value of "h."

Ambient the Page No Date Metal shield @337K 80=0.8 B=0.2 I da olm. do Joosm O 8=1 , took o towards atmosphue Net neat by metal lost plate- que & r. (Tm Tay Ann qu= 0.88 5.67x 108 (3374-300*) x Txo.1 - mak²x W/M²K² qm= 68.37159 W/m. Now pipe must bansmit equal - energy 68.37159 = 5.67*108 CT - 3377 x 780051 T= 378.7326k. There is connective between gas heat transfer & pipe. i.e q= h. A.AT 68.37159 = h. Txo.o5x1x 10.- h = 43.5267.