ENVIRONMENTAL ENGINEERING - QUESTION 22.2 : Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) + Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).
ENVIRONMENTAL ENGINEERING - ANSWER 22.2 : (a) BOD = (8.8 - 5.9) (300 / 30) = 29 mg / L. (b) BOD = (8.8 - 4.2) (600 / 100) = 27.6 mg / L. (c) BOD = (29 + 27.6) / 2 = 28.3 mg / L. (d) NBOD = 45 - 28.3 = 16.7 mg / L. 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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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.3 : (a) In the M / M / 1 queue that happens with randomness, let State 0 = the queue and server are empty, State 1 = the server is in use and the queue is empty, State 2 = the server is in use and 1 is in the queue, State 3 = the server is in use and 2 in the queue. Let P (0) = probability of State 0, P (1) = probability of State 1, P (2) = probability of State 2, P (3) = probability of State 3 and so on. If c = constant, P (1) = c P (0), P (2) = c [ c P (0) ], P (3) = c { c [ c P (0) ] }, write an equation that involves P (N), P (N + 1) and c. (b) Let L = market price of risk, r = riskless rate, m = expected return, s = volatility. Given that L = (m - r) / s related to oil prices, expected return = 12 %, s = 20 %, riskless rate = 8 %, calculate the market price of risk.
hi.i m student of chemical engg. and studying in 5 th sem.how can i improve my technical knowledge..can u sugest me a link like book or any website..?
QUANTUM COMPUTING - EXAMPLE 32.8 : In quantum computing, a quantum state is given by S = a | 00 > + b | 01 > + g | 10 > + d | 11 >. (a) Find S in term of | 0 > and | 1 > etc. (b) The probability of getting x is P(x). For S = 0.5 | 00 > + 0.5 | 01 > + 0.5 | 10 > + 0.5 | 11 >, find P(0) and P(1). Hint : P(00) + P(01) = P(0) = a x a + b x b, P(10) + P(11) = P(1) = g x g + d x d.
I HAVE DOMAIN KNOWLEDGE IN PROCESS MANUFACTURINGAND I GOT AN APPARTUNITY AS A BUSINESS ANALYST AND I DNT HAVE ANY KNOWLEDGE ON ORACLE APPS/GIVE ME ANY VALUBLE SUGGESTION?
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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.37 : A biochemical engineering professional applies 5-step Du Pont formula in the accounting. Let Net Income = A, Earning Before Tax (EBT) = B, Earning Before Interest, Tax (EBIT) = C, Net Sales = D, Total Assets = E, Shareholders Equity = F. Let Tax Burden = A / B = 0.75, Interest Burden = B / C = 1.05, EBIT Margin = C / D = 0.27, Asset Turnover = D / E = 0.66, Return on Equity = A / F = 0.37. (a) Calculate the value of Leverage = E / F. (b) If AB + BC + CD + DE + EF = $$ 141932, find the values of A, B, C, D, E and F.
ENGINEERING MECHANIC - EXAMPLE 15.3 : A biochemical trolley of mass 15 kg is towing a trailer of mass 5 kg along a straight horizontal pathway. The trailer and the trolley are connected by a light inextensible tow-bar. The engine of the trolley exerts a driving force of magnitude 100 N. The trailer and the trolley experience resistances of magnitude 10 N and 30 N respectively. (a) Form 2 equations with unknowns T and a, that represents the equilibrium for the 2 systems of the trolley and trailer. (b) Solve the simultaneous equations from the 2 equations that are obtained in part (a) of this question. T is the tension of the tow-bar and a is the acceleration.
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Question 33 – By using Excel program either on laptop or desktop PC, solve the differential equation dy / dx = -2y + x + 4 with h = 0.005, initial values : x = 0, y = 1. The 4th order Runge-Kutta method provides : y(N + 1) = y(N) + (1/6) (k1 + 2k2 +2k3 + k4), k1 = h [ -2y(N) + x(N) + 4 ], k2 = h { -2 [ y(N) + k1 / 2 ] + x(N) + h / 2 + 4 }, k3 = h { -2 [ y(N) + k2 / 2 ] + x(N) + h / 2 + 4 }, k4 = h { -2 [ y(N) + k3 ] + x(N) + h + 4 }. What is the value of y at x = 0.5?
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