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.
ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.19 : Value of k has to be found first : 12 months / year, then k = 12. (a) (i) r = R / k = 0.0807 / 12 = 0.006725. (ii) n = kN = 12 x 30 = 360. (iii) k = 12. (iv) PV = [ 1 - 1 / (1 + r) ^ n ] / r = [ 1 - 1 / (1 + 0.006725) ^ 360 ] / 0.006725 = 135.382. (b) MR = pv / PV = $450000 / 135.382 = $3323.928. (c) Annual repayment = MR x k = $3323.928 x 12 = $39887.136. Percentage of salary remains after paying loan = 100 [ 1 - (annual repayment) / (annual salary after tax) ] = 100 (1 - 39887.136 / 56000) = 28.773 %. 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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solution =2oolitre ,solute =Polyelectrolyte,Solution=water ,we have to prepare 1% solution.how much solute And Solvent?
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CHEMICAL MATERIAL BALANCE – EXAMPLE 2.6 : According to Raoult's law for ideal liquid, x (PSAT) = yP where x is mole fraction of component in liquid, y is mole fraction of component in vapor, P is overall pressure and PSAT is saturation pressure. A liquid with 60 mole % component 1 and 40 mole % component 2 is flashed to 1210 kPa. The saturation pressure for component 1 is ln (PSAT) = 15 - 3010 / (T + 250) and for component 2 is ln (PSAT) = 14 - 2700 / (T + 205) where PSAT is in kPa and T is in degree Celsius. By assuming the liquid is ideal, calculate (a) the fraction of the effluent that is liquid; (b) the compositions of the liquid and vapor phases. The outlet T is 150 degree Celsius.
QUANTUM COMPUTING - EXAMPLE 32.6 : (a) Let H | 0 > = 0.707 ( | 0 > + | 1 > ), H | 1 > = 0.707 ( | 0 > - | 1 > ). Find the values for H | 0 > + H | 1 > and H | 0 > - H | 1 >. (b) In quantum computing, a qubyte is a quantum byte, or 8 quantum bits, a sequence processed as a unit. A qubit is a quantum bit. According to Alexander Holevo in his theorem, n qubits cannot carry more than n classical bits of information. What is the maximum amount of classical bits of information that can be carried by 1 qubyte.
PROCESS DESIGN - EXAMPLE 21.2 : The names of the flow streams could be represented by : H1 for first hot stream, H2 for second hot stream, C1 for first cold stream, C2 for second cold stream. Data of supply temperature Ts in degree Celsius : 150 for H1, 170 for H2, 30 for C1, 30 for C2. Data of target temperature Tt in degree Celsius : 50 for H1, 169 for H2, 150 for C1, 40 for C2. Data of heat capacity Cp in kW / degree Celsius : 3 for H1, 360 for H2, 3 for C1, 30 for C2. (a) Find the enthalpy changes, dH for all streams of flow H1, H2, C1 and C2 in the unit of kW. Take note of the formula dH = (Cp) (Tt - Ts). (b) Match the hot streams H1 and H2 with the suitable cold streams C1 and C2 to achieve the maximum energy efficiency.
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