Heat transfer: In a steady state one dimensional conduction with no heat generation, the differential equation is d / dx (k dT / dx) = 0. Prove that T(x) = ax b, where k, a and b are constants. (b) At x = 0, T = c and at x = L, T = d. Prove that T(x) = (d - c) x / L c for boundary conditions.
HEAT TRANSFER - ANSWER 5.2 : When d / dx (k dT / dx) = 0, d (dT) / [ (dx) (dx) ] = 0. Integrate both sides gives dT / dx = a. Second integration gives T(x) = ax + b for both sides (proven). (b) T(0) = a(0) + b = b = c. T(L) = d = aL + c then a = (d - c) / L. Substitute in T(x) = ax + b gives T(x) = (d - c) x / L + c (proven). The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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ENGINEERING MATERIAL - EXAMPLE 12.1 : In crystal material, hexagonal crystal system could form 4-digit index in certain direction of solid. For [1(-1)0] direction in the hexagonal crystal systems of particular catalyst applied in fume removal of incinerator, what is the four-digit index for this direction? Hint : The transformation equations between the 3-digit [h'k'l'] and the 4-digit [hkil] indices are : h = (1/3) (2h'-k'); i = - (h + k); k = (1/3) (2k'-h'); l = l' A. [(-1)100] B. [1(-1)00] C. [(-1)000] D. [00(-1)(-1)] E. [(-1)0(-1)0]
A concentric, counter-current heat exchanger is used to cool lubricating oil. Water is used as the coolant. The mass flow rate of oil into the heat exchanger is 0.1 kg / s = FO. For oil, the inlet temperature TIO = 100 degree Celsius and the outlet temperature TOO = 55 degree Celsius. For water, the inlet temperature TIW = 35 degree Celsius and the outlet temperature TOW = 42 degree Celsius. What is the mass flow rate of water in kg / s, FW needed to maintain these operating conditions? Constant for heat capacity of oil is CO = 2131 J /(kg K) and for water is CW = 4178 J /(kg K). Use the equation (FO)(CO)(TIO – TOO) = (FW)(CW)(TOW – TIW).
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Question 77 - The molecular weights M in kg / mol of 3 different monomers a, b and c in a polymer are Ma = 14, Mb = 16 and Mc = 18, with their respective quantities in N units having the ratio of Na : Nb : Nc = 2 : 3 : 5. (a) Find the numerical average molecular weight of the polymer by using the formula (Ma Na + Mb Nb + Mc Nc) / (Na + Nb + Nc). (b) Find the weighted average molecular weight of the polymer by using the formula (Ma Na Ma + Mb Nb Mb + Mc Nc Mc) / (Ma Na + Mb Nb + Mc Nc). (c) Calculate the polydispersity Q by using the answer in (b) divided by answer in (a). (d) Find the volumetric average molecular weight of the polymer by using the formula (Ma Na Ma Ma + Mb Nb Mb Mb + Mc Nc Mc Mc) / (Ma Na Ma + Mb Nb Mb + Mc Nc Mc). (e) Estimate the polydispersity Q by using the answer in (d) divided by answer in (b).
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