Fourth Midterm—Take Home part

Paper companies usually operate hydroelectric generating stations on bodies of water, preferably rivers. Water is piped from a dam to the power station, the rates at which the water flows through the different pipes varying depending on external conditions.

Let us design a power station with three different hydroelectric turbines, each with a known power function that gives the amount of electric power generated, as a function of the water flow arriving at the turbine. The incoming water can be apportioned in different volumes to each turbine, so the goal is to determine how to distribute water among the turbines to give the maximum total energy production for any rate of flow.

Let us denote Q_j the flow through turbine j in cubic feet per second; K_j, the power generated by turbine j in kilowatts; Q_T, the total flow through the station in cubic feet per second. Using experimental evidence, the following models are determined for the power output of each turbine:

\begin{array}{rcl} K_1 &=& \big( -18.89 + 0.1277 Q_1 - 4.08 \cdot 10^{-5} Q_1^2 \big) \big( 170 - 1.6 \cdot 10^{-6} Q_T^2 \big) \\  K_2 &=& \big( -24.51 + 0.1358 Q_2 - 4.69 \cdot 10^{-5} Q_2^2 \big) \big( 170 - 1.6 \cdot 10^{-6} Q_T^2 \big) \\ K_3 &=& \big( -27.02 +0.138 Q_3 - 3.84 \cdot 10^{-5} Q_3^2 \big) \big( 170 - 1.6 \cdot 10^{-6} Q_T^2 \big). \end{array}

These are the allowable flows of operation:

250 \leq Q_1 \leq 1110, \quad 250 \leq Q_2 \leq 1110, \quad 250 \leq Q_3 \leq 1225.

Problem 1. If all three turbines are being used, we wish to determine the flow Q_j to each turbine that will give the maximum total energy production. Our limitations are that the flows must sum to the total incoming flow and the given domain restrictions must be observed: Maximize the total energy production K_1 + K_2 + K_3 subject to the constraint Q_1 + Q_2 + Q_3 = Q_T and the domain restrictions on each Q_j.

Problem 2. For which values of Q_T is your result valid?

Problem 3. Is it possible in some situations that more power could be produced by using only one turbine? Make a graph of the three power functions, and use it to help decide if an incoming flow of 1000 ft3/s should be distributed to all three turbines, or routed to just one. If you determine that only one turbine should be used, which one would it be?

Problem 4. For some flow levels it would be advantageous to use two turbines. If the incoming flow is 1500 ft3/2, which two turbines would you recommend using? For this flow, is using two turbines more efficient than using all three?

Problem 5. If the incoming flow is 3400 ft3/s, what would you recommend to the company?

Solution

As we discussed in class, the best way to approach the solution of the first problem is with Lagrange multipliers. Take first the function to be maximized:

\begin{array}{rcl} f(Q_1,Q_2,Q_3,Q_T) &=& \big( -60.42 + 0.1277 Q_1 - 4.08 \cdot 10^{-5} Q_1^2 + 0.1358 Q_2 - 4.69 \cdot 10^{-5} Q_2^2 \\ && \hspace{0.25cm}+0.138 Q_3 - 3.84 \cdot 10^{-5} Q_3^2 \big) \big( 170 - 1.6 \cdot 10^{-6} Q_T^2 \big) \end{array}

and then build your constraint:

g(Q_1,Q_2,Q_3,Q_T)=Q_1+Q_2+Q_3-Q_T.

We need to solve the system

\big\{ \nabla f = \lambda \nabla g, \quad g=0 \big\},

which gives us five equations:

(1) \big( 0.1277-8.16\cdot 10^{-5}Q_1 \big)  \big( 170 - 1.6 \cdot 10^{-6} Q_T^2 \big) = \lambda
(2) \big( 0.1358-9.38\cdot 10^{-5}Q_2 \big)  \big( 170 - 1.6 \cdot 10^{-6} Q_T^2 \big) = \lambda
(3) \big( 0.138-7.68\cdot 10^{-5}Q_3 \big)  \big( 170 - 1.6 \cdot 10^{-6} Q_T^2 \big) = \lambda
(4) 3.2\cdot10^{-6}Q_T \big( -60.42 + 0.1277 Q_1 - 4.08 \cdot 10^{-5} Q_1^2 + 0.1358 Q_2 - 4.69 \cdot 10^{-5} Q_2^2 +0.138 Q_3 - 3.84 \cdot 10^{-5} Q_3^2 \big) = \lambda
(5) Q_1+Q_2+Q_3=Q_T

Equations (1) and (2) allow us to write Q_2 in terms of Q_1 (rounding here and there to two decimal places):

(6) Q_2= 0.87 Q_1 + 83.35.

Equations (1) and (3) allow us to write Q_3 in terms of Q_1:

(7) Q_3= 1.06 Q_1 + 134.11.

Equations (5), (6) and (7) allow us to write Q_T in terms of Q_1:

(8) Q_T = 2.93 Q_1+220.47

Rewriting equations (1) and (4) in terms of Q_1 alone, setting them equal to each other and solving for Q_1 gives us all the values of Q_1 that maximize/minimize the energy function. There are only three solutions: two complex and one real.

Q_1=-2300, Q_1=1300-0.25i, Q_1=3240+0.125i

A quick check tells us that the solution we obtain with the only real value, Q_1=-2300, can only offer a minimum of the energy function. Besides, the value is obviously outside of the allowed interval. That leaves us with a function that has no absolute maximum!

Note how at no point have we used the constraints on the Q’s: we do so now, since this guarantees us the existence of an actual absolute maximum in the closed set formed with the constraints. A quick check is enough to conclude that the maximum of our energy function will happen precisely when all the variables attain their maximum possible values; that is: Q_1=1110, Q_2=1110, and Q_3=1225 (all of them in cubic feet per second) with a total flow of Q_T=3445 cubic feet per second. The power in this case is 33987.50 kilowatts.

On with the second part. If we decide to use only one turbine and a total flow of 1000 cubic feet per second, then we have only three possibilities:

  • Q_1=Q_T=1000, Q_2=Q_3=0: In this case, f=K_1=11452.884 kilowatts.
  • Q_2=Q_T=1000, Q_1=Q_3=0: In this case, f=K_2=10843.276 kilowatts.
  • Q_3=Q_T=1000, Q_1=Q_2=0: In this case, f=K_3=12222.472 kilowatts.

If only one turbine is to be used, the one offering more power is the third. Let us now compare with the maximum power using the three turbines. For this task, we have another optimization by means of Lagrange multipliers, but now the functions change to simpler expressions:

\begin{array}{rcl} f(Q_1,Q_2,Q_3) &=& 168.4 \big( -60.42 + 0.1277 Q_1 - 4.08 \cdot 10^{-5} Q_1^2 + 0.1358 Q_2  \\ && \hspace{1.25cm}- 4.69 \cdot 10^{-5} Q_2^2+0.138 Q_3 - 3.84 \cdot 10^{-5} Q_3^2 \big)\\ g(Q_1,Q_2,Q_3)&=&Q_1+Q_2+Q_3 \end{array}

Note the absence of Q_T. Now our Lagrange multipliers looks like this:

(9) 168.4 \big( 0.1277-8.16\cdot 10^{-5}Q_1 \big)  = \lambda
(10) 168.4 \big( 0.1358-9.38\cdot 10^{-5}Q_2 \big) = \lambda
(11) 168.4 \big( 0.138-7.68\cdot 10^{-5}Q_3 \big) = \lambda
(12) Q_1+Q_2+Q_3=1000

The solution of this one is much simpler, and yields Q_1=265.83, Q_2=317.61, Q_3=416.56, all of them in cubic feet per second. The maximum attained is then 8397.39 kilowatts, clearly smaller than what we can obtain with any single turbine.

Problem 4 is more complicated, since we need to perform four different optimizations with Lagrange multipliers! In all four cases, Q_T=1500:

  • Q_3=0: We have then f=K_1+K_2 and g(Q_1,Q_2)=Q_1+Q_2.
  • Q_2=0: We have then f=K_1+K_3 and g(Q_1,Q_3)=Q_1+Q_3.
  • Q_1=0: We have then f=K_2+K_3 and g(Q_2,Q_3)=Q_2+Q_3.
  • f=K_1+K2+K_3 and g(Q_1,Q_2,Q_3)=Q_1+Q_2+Q_3 as before, but with constraint g=1500.

Note that in this case it does not make sense to restrict the flow to a single turbine, because none of them can withstand by themselves that large a quantity of cubic feet per second.

Problem 5, it is a bit easier, since the constraint Q_T=3400 clearly indicates that the three turbines must be used simultaneously (note how one or two turbines together could never reach this flow). It is just a matter of finding the maximum with the same technique.

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  1. November 28, 2012 at 8:29 pm

    is -3.84×10^-6 supposed to be changed to -3.84×10^-5 in k3? It was x10^-6 on the handout, but this page has x10^-5

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