what is difference between open loop system & close loop system by overview.
Answer / viswa
To put it simply, for example in a closed loop system the measured value is sent back to the comparator to compare the value with the set point where as in open loop it is not.
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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.17 : In the engineering calculations of interest rate caused by inflation, General Inflation Effect and Fisher Effect may be considered. Let I = inflation rate, R = nominal interest rate, r = real interest rate. According to Fisher Effect, (1 + R) = (1 + r) (1 + I). According to General Inflation Effect, r = R - I. (a) If I = 0.1 for all effects, both the values of R and r in the Fisher Effect are the same as R and r in the General Inflation Effect, find the values of R and r. (b) If R has the same value caused by both General Inflation Effect and Fisher Effect, find the possible values of R, r and I in term of R etc.
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?
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).
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