ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.19 : In the purchase of a unit of engineering office, a loan has been made to a bank with the following details : Term N = 30 years; interest rate R = 8.07 % / year; security : primary residence; present value pv = $450000; salary = $75000 / year or $56000 / year after tax. (a) Let the discounted present value PV = [ 1 - 1 / (1 + r) ^ n ] / r for arrears, where r = interest rate of discount, n = number of payment, ^ = symbol for power. If the loan repayment was made monthly : (i) calculate the value of r where r = R / k and R is in decimal value; (ii) find the value of n where n = kN; (iii) estimate the value of k where k = number of repayment per year; (iv) calculate the value of PV based on the formula of discounted present value. (b) Calculate the monthly repayment of the loan, MR based on the following formula : pv = PV x MR. (c) Find the percentage of salary remains after paying the loan every month.

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ENGINEERING PHYSICS - EXAMPLE 30.5 : (a) Let | A > = (Aa Ab Ac), | B > = (Ba Bb Bc), | C > = (Ca Cb Cc). Find | A > + | C > - | B > in term of Aa, Ab, Ac, Ba, Bb, Bc, Ca, Cb and Cc. (b) Let d | E > = d (Ea Eb Ec) = (d Ea d Eb d Ec). If | E > = (6 7 8), find the value of 10 | E >.

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What is the method involved in analyzing powders for composition?

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What is a barometric condenser?

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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).

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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..?

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.30 : A biochemical engineering sales contract adopts Cost Recovery Method in its company accounting. Gross profit is realised only when cash collections exceed the total cost of goods sold. Let Y = Original Cost Recovered, Z = Gross Profit Realised. Its property was sold at A = $500k with cost of B = $300k. Its engineering accountant receives cash of C = $240k in first year, D = $180k in second year and E in third year. (a) Find the values of Y and Z in : (i) first year; (ii) second year; (iii) third year. (b) What is the exact value of E if all cash has been fully received from sales?

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Question 8 - A local utility burns coal having the following composition on a dry basis : Carbon (C) 83.05 %, hydrogen (H) 4.45 %, oxygen (O) 3.36 %, nitrogen (N) 1.08 %, sulfur (S) 0.7 % and ash 7.36 %. Calculate the ash free composition of the coal with reference to C, H, O, N and S.

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What is the difference between cfm (cubic feet per minute) and scfm (standard cubic feet per minute)?

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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.34 : A biochemical engineering company has stock P = 20,000 shares of M % preferred stock of Q = $100 par and N number of shares of $20 par common stock. In year 1 : $10,000 is distributed as dividend. In year 2 : $25,000 is distributed as dividend. Let (Total Dividend) = (Dividend for Preferred Stock) + (Dividend for Common Stock). (a) In year 1, calculate the amount distributed as dividend for preferred stock, when no dividend was distributed for common stock. (b) In year 2, the amount distributed for common stock was $5000. Find the values of M and N, when the dividend per share for common stock is $0.05.

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what is factories act?tell about factories act?

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POLYMER ENGINEERING - QUESTION 24.2 : Let C% be the fractional crystallinity, Rs = density of sample, Ra = density of amorphous form and Rc = density of crystalline form. In a polymer, these unknowns could be related by the equation C% = (Rc / Rs) (Rs - Ra) / (Rc - Ra). (a) Find the equation of Rc as a function of C%, Rs and Ra. (b) Two samples of a polymer, C and D exist. For sample C, C% = 0.513 when Rs = 2.215 unit. For sample D, C% = 0.742 when Rs = 2.144 unit. Both samples C and D have the same values of Ra and Rc. Find the values of Ra and Rc in 6 decimal places.

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