--%>

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 : Haloalkanes define primary secondary

    define primary secondary and tertiary alkyl halides with examples

  • Q : Quastion of finding vapour pressure

    Vapour pressure of CCl425Degree C at is 143mm of Hg0.5gm of a non-volatile solute (mol. wt. = 65) is dissolved in 100ml CCl4 .Find the vapour pressure of the solution (Density of CCl4 = = 1.58g /cm2): (a)141.43mm (b)

  • Q : M ive me answer of this question. When

    ive me answer of this question. When mercuric iodide is added to the aqueous solution of potassium iodide, the: (a) Freezing point is raised (b) Freezing point is lowered (c) Freezing point does not change (d) Boiling point does not change

  • Q : Rotational energy and entropy due to

    The entropy due to the rotational motion of the molecules of a gas can be calculated. Linear molecules: as was pointed out, any rotating molecule has a set of allowed rotational energies. For a linear molecule the

  • Q : Ionic radius of chloride ion The edge

    The edge length of the unit cell of Nacl crystal lattice is 552 pm. If ionic radius of sodium ion is 95. What is the ionic radius of chloride ion:(a) 190 pm  (b) 368 pm  (c) 181 pm  (d) 276 pm     <

  • Q : Influence of temperature Can someone

    Can someone please help me in getting through this problem. With increase of temperature, which of the following changes: (i) Molality (ii) Weight fraction of solute (iii) Fraction of solute present in water (iv) Mole fraction.

  • Q : Determining concentration in ppm A 500

    A 500 gm tooth paste sample has 0.2g fluoride concentration. Determine the concentration of F in terms of ppm level: (a) 250 (b) 200 (c) 400 (d) 1000Answer: (c) F-ions in ppm = (0.2/500) x 106 = 400

  • Q : Direction of dipole moment expected

    Illustrate the direction of the dipole moment expected for hydrogen bromide?

  • Q : Liquid surfaces The surface between a

    The surface between a liquid and a vapour distinguishes these fluids. The surface tension of liquids can be looked upon as that the property which draws a liquid together and forms a liquid vapour interface, therefore, distinguishing liquids from gases.<

  • Q : Problem on moles of solution The number

    The number of moles of a solute in its solution is 20 and total no. of moles are 80. The mole fraction of solute wil be: (a) 2.5 (b) 0.25 (c) 1 (d) 0.75