Heat transfer: In a triple effect evaporator, the heat transfer for an evaporator is calculated as q = UA (TI - TF) where TI is the initial temperature, TF is the final temperature; U and A are constants. Given that heat transfer for the first evaporator : q(1) = UA (TI - TB); second evaporator : q(2) = UA (TB - TC); third evaporator : q(3) = UA (TC - TF) where q(x) is the heat transfer function, TB is the temperature of second inlet and TC is the temperature of third inlet, prove that the overall heat transfer Q = q(1) q(2) q(3) = UA (TI - TF).
HEAT TRANSFER - ANSWER 5.1 : Q = q(1) + q(2) + q(3) = [ UA (TI - TB) ] + [ UA (TB - TC) ] + [ UA (TC - TF) ] = UATI + (UATB - UATB) + (UATC - UATC) - UATF = UATI - UATF = UA (TI - TF) (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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BIOPROCESS ENGINEERING - EXAMPLE 14.3 : The kinetic behavior of an enzyme could be described using Michalis - Menten equation : Vo = Vmax [S] / (Km + [S]). Derive this equation from [ES] = [E]total [S] / (Km + [S]), Vmax = Kcat [E]total, Vo = Kcat [ES].
ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.21 : The cost of building a biochemical processing plant is increasing due to inflation. Let I = inflation rate, R = nominal interest rate, r = real interest rate. According to Fisher Equation, (1 + r)(1 + I) = (1 + R). According to General Inflation Equation, R = r + I. (a) By assuming that both r and I are fairly small, prove by mathematical calculations that complicated Fisher Equation could be simplified into the General Inflation Equation. (b) By using 2 first order Taylor expansions in the linear approximation, namely 1 / (1 + x) ≈ (1 - x), (1 + x)(1 + y) ≈ 1 + x + y, show by mathematical calculations that (1 + r) = (1 + R) / (1 + I) could be approximated by r ≈ R - I.
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PROCESS CONTROL - EXAMPLE 6.3 : The differential equation is 3 dy / dt + 2y = 1 with y(0) = 1. (a) The Laplace transformation, L for given terms are : L (dy / dt) = sY(s) - y(0), L(y) = Y(s), L(1) = 1 / s. Use such transformation to find Y(s). (b) The initial value theorem states that : When t approaches 0 for a function of y(t), it is equal to a function of sY(s) when s approaches infinity. Use the initial value theorem as a check to the answer found in part (a).
I am writing the chemical engineering subject test for Reliance India Limited. Could someone who has taken the test please list some of the questions from that paper?
CHEMICAL MATERIAL BALANCE - EXAMPLE 2.3 : A 1.5 weight % aqueous salt solution is concentrated to 4 weight % in a single-effect evaporator. The feed rate to the evaporator is F = 7500 kg / h and the feed is at 85 degree Celsius. The evaporator operates at 1 bar. By assuming that only pure solvent of water exists in the form of vapor from the feed, calculate the flow rate of such vapor V.
NATURAL GAS ENGINEERING - QUESTION 26.1 : (a) In natural gas pipe sizing, the length of the pipe from the gas source metre to the farthest appliances is 60 feet. The maximum capacities for typical metallic pipes of 60 feet in length are : 66 cubic feet per hour for pipe size of 0.5 inches; 138 cubic feet per hour for pipe size of 0.75 inches; 260 cubic feet per hour for pipe size of 1 inch. By using the longest run method : (i) Find the best pipe size needed for the capacity of 75 cubic feet per hour. (ii) Estimate the suitable range of capacities for the pipe size of 1 inch. (b) The maximum capacities for typical metallic pipes of 50 feet in length are : 73 cubic feet per hour for pipe size of 0.5 inches; 151 cubic feet per hour for pipe size of 0.75 inches; 285 cubic feet per hour for pipe size of 1 inch. By using the branch method find the best pipe size needed for the capacity of 75 cubic feet per hour when the length of the pipe from the gas source metre to the appliance is 52 feet.
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