--%>
Assignment Help - homework help

Problem on convection coefficient

An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel casting involves insertion of thermocouples in the casting at distances of 10 mm and 20 mm from the surface.  When the experiment was performed in a well insulated air duct so that the temperature of the walls surrounding the steel surface was the same as the air temperature, the thermocouples measured the temperatures of 50°C and 40°C in the steel, respectively, when the air and the surrounding wall temperatures were 200oC.

a) If the steel has a thermal conductivity of 15 W/m2K and surface emissivity of 0.95, what is the convection coefficient between the steel and hot air?

b) Determine the contributions of convective and radiation heat fluxes to steel casting?

In the recurring experiments the duct insulation was eliminated and even though the hot air temperature was similar as before, i.e., 200oC, the surrounding walls temperature dropped to 20oC. As a result the steady state heat flux to the steel casting dropped by 13.3% compared to the first experiment.

c) Supposing similar convection coefficient as the value coputed in part a), what would be the temperatures measured by the thermocouples?

d) Assume that the steel casting were well insulated from each and evry side except for the surface from which the distance to thermocouples is measured. Assuming the same convection coefficient as the value calculated in part a), what would be the temperatures of the thermocouples when:

i) The experiment is performed in the well insulated duct at T∞ = Tsurr = 200oC?
ii) The experiment is executed in non-insulated duct at T∞ = 100oC and Tsurr = 20oC?

   Related Questions in Chemistry

  • Q : Explain various chemicals associated

    During processing of food, several chemicals are added to it to augment its shelf life and to make it more attractive as well. Main types of food addi

    		•
  • Q : Molality of glucose Help me to go

    Help me to go through this problem. Molecular weight of glucose is 180. A solution of glucose which contains 18 gms per litre is : (a) 2 molal (b) 1 molal (c) 0.1 molal (d)18 molal

    		•
  • Q : Problem based on molality of glucose

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

    		•
  • Q : Pressure Phase Diagrams The occurrence

    The occurrence of different phases of a one component system can be shown on a pressure temperature. The phases present in a one line system at various temperatures can be conveniently presented on a P- versus-T diagram. An example is pro

    		•
  • Q : Vant Hoff factor The Van't Hoff factor

    The Van't Hoff factor of the compound K3Fe(CN)6 is: (a) 1  (b) 2  (c) 3  (d) 4  Answer: (d) K3[Fe(CN)6] → 3K+

    		•
  • Q : Amines why o-toluidine is a weaker base

    why o-toluidine is a weaker base than aniline?

    		•
  • Q : Molar mass Select the right answer of

    Select the right answer of the question. Which is heaviest: (a)25 gm of mercury (b)2 moles of water (c)2 moles of carbon dioxide (d)4 gm atoms of oxygen

    		•
  • Q : Molar concentration Choose the right

    Choose the right answer from following. Molar concentration (M) of any solution : a) No. of moles of solute/Volume of solution in litre (b) No. of gram equivalent of solute / volume of solution in litre (c) No. of moles os solute/ Mass of solvent in kg  (

    		•
  • Q : Question related to colligative

    The colligative properties of a solution depend on: (a) Nature of solute particles present in it (b) Nature of solvent used (c) Number of solute particles present in it (d) Number of moles of solvent only

    		•
  • Q : Problem on molarity-normality-molality

    Can someone please help me in getting through this problem. The solution ofAl2(SO4)3 d = 1.253gm/m comprise 22% salt by weight. The molarity, normality and molality of the solution is: (1) 0.805 M, 4.83 N, 0.825 M (2)

    		•