Two parallel pumppipe assemblies shown in Fig 810 deliver wa

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.

Solution

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