Two parallel pump-pipe assemblies shown in Fig. 8-10 deliver water from a common
ID: 2326885 • Letter: T
Question
Two parallel pump-pipe assemblies shown in Fig. 8-10 deliver water from a common source to a common destination. The total volume flow rate required at the destination is 0.01 m^3/s. The drops in pressure in the two lines are functions of the square of the flow rates, Delta _P1, Pa = 2.1 times 10^10 F_1^2 and Delta _p2, Pa = 3.6 times 10^10 F_2^2 where F_1 and F_2 are the respective flow rates in cubic meters per second. The two pumps have the same efficiency, and the two motors that drive the pumps also have the same efficiency. (a) It is desired to minimize the total power equipment. Set up the objective function and constraints in terms of F_1 and F_2. (b) Solve for the optimal values of the flow rates that result in minimum total water power using the method of Lagrange multipliers. Ans.: F_1 = 0.00567.Explanation / Answer
a)
For pressure drop to be equal in both the pipes,
2.1*1010 F12 = 3.6*1010 F22
F12 = 1.714 F22..............eqn1
Also by mass conservation,
F1 + F2 = 0.01...............eqn2
Power input = Pressure drop*Flow rate / Efficiency
Since efficiency is constant we drop it from calculations.
Power input for both pumps = 2.1*1010 F12 *F1 + 3.6*1010 F22 * F2
= 2.1*1010 F13 + 3.6*1010 F23
Hence, objective function is to minimize 2.1*1010 F13 + 3.6*1010 F23
Constraint is F1 + F2 = 0.01 or F1 + F2 - 0.01 = 0
b)
L(F1, F2, l) = (2.1*1010 F13 + 3.6*1010 F23) + l*(F1 + F2 - 0.01)
delL / del F1 = 6.3*1010 F12 + l = 0
delL / del F2 = 10.8*1010 F22 + l = 0
delL / del l = F1 + F2 - 0.01 = 0
Solving these eqns we get,
F1 = sqrt [-l / (6.3*1010)]
F2 = sqrt [-l / (10.8*1010)]
sqrt [-l / (6.3*1010)] + sqrt [-l / (10.8*1010)] - 0.01 = 0
sqrt (-l) / (2.51*105) + sqrt (-l) / (3.286*105) - 0.01 = 0
Solving it, we get l = -2025164
Thus, F1 = sqrt [2025164 / (6.3*1010)]
F1 = 0.00569
F2 = 0.01 - F1
F2 = 0.00433
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