Mixing Ethanol Blends

Transcription

1 Example 1 Mixing Ethanol Blends Flex fuel vehicles are designed to operate on a gasoline-ethanol mixture. This gasoline-ethanol mixture is called E85 and is 85% ethanol and 15% gasoline. Flex fuel vehicles perform almost identically to gasoline powered vehicles. Since E85 is less expensive than gasoline, it would appear that that flex fuel vehicles would be the wave of the future. However, the energy content of E85 is less than that of gasoline so flex fuel vehicles get worse mileage than gasoline powered vehicles. In general, the higher the percentage of ethanol in the fuel, the worse the mileage is. A recent study determined that several vehicles designed to run on gasoline are able to run well on gasoline-ethanol blends. In tests using E10 (10% ethanol), E20 (20% ethanol) and E0 (0% ethanol), the cost per mile traveled by vehicles using the three blends was lowest for vehicles using E20 or E0. Because of these results, gasoline stations around the country are beginning to sell several blends of ethanol to cash in on this result. A gasoline station in Rapid City, South Dakota carries three blends, E10, E20,

2 and E85. The cost per gallon of each of these fuels is shown in the table below. Ethanol Blend Cost per Gallon E10 $2.75 E20 $2.68 E85 $2.14 A customer wants to purchase 10 gallons of E20. The customer could simply purchase the 10 gallons at $2.68 per gallon. Or the customer could blend their own E20 using the available E10 and E85. How many gallons of E10 and E85 would the customer need to blend in their 10 gallon tank to effectively have E20? Would it be cheaper to blend E10 and E85 or to simply pump E20?

3 Solution In this problem, we need to compare the cost of pumping 10 gallons of straight E20 to the cost of a specific combination of E10 and E85 that will yield 20% ethanol. It is straightforward to calculate cost of each type of fuel, cost cost per gallon number of gallons For instance, 10 gallons of E20 would cost $ gallon cost gallons $26.80 The key question is to determine if a blend of E10 and E85 would be cheaper. How much of each type of fuel would be needed to fill the tank and result in a mixture that is 20% ethanol? We ll start with two variables, one to represent the amount E10 in gallons and another to represent the amount of E85 in gallons:

4 E10: amount of 10% ethanol fuel in gallons E85: amount of 85% ethanol fuel in gallons Writing out this description is key to our strategy. Not only does it help you to focus on what the unknowns are, but it also includes the units, gallons. The problem includes copious amounts of information. However two pieces of information relate to the variables. First of all, the customer needs to fill a 10 gallon tank with a combination of two fuels. Since the variables represent the individual amount of fuels, the sum of the variables corresponds to the total amount of fuel, E10 E total amount of fuel

5 The second important piece of information is that the mixture of fuel must contain 20% ethanol. For a 10 gallon tank, the amount of ethanol is 20% of gallons 2 gallons. This ethanol will come from the 10 gallons or E10 which is 10% ethanol and the E85 which is 85% ethanol. By multiplying the percent times the gallons of fuel, we can determine how much ethanol is in the fuel. For instance 8 gallons of E85 and 2 gallons of E10 would have amount of ethanol in 2 gallons of E10:.10 2 gallons + amount of ethanol in 8 gallons of E85:.85 8 gallons 7 gallons

6 A mixture of 10 gallons of fuel with 7 gallons of ethanol would be a 70% mixture. This is clearly way too much ethanol. However, we could construct a table with differing amounts of fuel to help us determine the solution:

7 Amount of E10 (gallons) Amount of E85 (gallons) Total amount of Ethanol (gallons) ? 6? 6?? As the amount of E10 is increased and the amount of E85 is decreased, the total amount of ethanol in the mixture drops. Based on the table, the amount of E10 must be between 8 and 9 gallons and the amount of E85 must be between 1 and 2 gallons. It is unlikely that we could find the exact amounts of each type of fuel by expanding the table.

8 However, we can replace the amounts of fuel with variables that represent these amounts to give the equation.10 E10.85 E85 2 total amount of ethanol in blend This gives us two equations in two variables that we can solve by the substitution method or the elimination method. For this example, we ll solve the system E10 E E10.85 E85 2 Solve the first equation for E10 to yield E10 10 E85. Substitute this expression into the second equation to obtain E85.85 E85 2

9 Solve the equation.1010 E85.85 E85 2 for E85: 1.10 E85.85 E E E85 1 Remove parentheses Combine like terms Subtract 1 from both sides E Divide both sides by.75 Now that we know that we ll need 4 gallon of E85, we can use the first equation to find the amount of E10: E Now let s find the cost of this blend, $ 26 $ 4 gallon gallon cost 2.75 gallons 2.14 gallons $26.69

10 Compared to pumping 10 gallons of E20 (for a total of $26.80), pumping a blend of E10 and E85 (from the same pump!) saves about 11 cents. How does this match the graph of each equation? Figure 1 By solving each equation for E10, we can graph E10 = 10 E85 (blue) and E10 = E85 (red) to see that the point of intersection is at the correct ordered pair.

11 Before moving on, we need to make sure the solution makes sense. In the table we created earlier, we noted that the amount of E10 had to between 8 and 9 gallons. Since , our solution is consistent. The amount of E85 is 4 1. which is between 1 and 2 as expected. Example 2 Mixing Ethanol Blends In Example 1, we created a system of equations to describe a mix of E10 and E85 ethanol, E10 E E E85 2 where E10 is the amount of 10% ethanol pumped in gallons and E85 is the amount of 85% ethanol pumped in gallons. The first equation describes the total amount in the mixture, 10 gallons. The second equation describes the total amount of ethanol in the mixture, 20% of 10 gallons or 2 gallons. Solve this system of equations by finding the reduced row echelon form for the augmented matrix.

12 Solution The augmented matrix for this system is where the first column corresponds to E10 and the second column corresponds to E85. To convert this matrix to reduced row echelon form, we must create pivots in the first and second columns and put zeros above and below the pivots. There is already a 1 in the first row and column, so we need to use row operations to put a 0 below it: 0.10 R1 : R2 : R1R becomes R 2

13 We can place a pivot in the second row and column by multiplying the row by the reciprocal of 0.75, R2 becomes R A 0 is placed above the pivot in the second column by multiplying the second row by -1, adding it to the first row, and putting the result in the first row: 1 R : 0 1 R : Based on this reduced row echelon form, E10 and E85. This means that mixing 26 gallons of 10% ethanol with 4 gallons of 85% ethanol will yield 10 gallons of 20% ethanol R R becomes R

14 Example Mixing Three Types of Ethanol In some Midwestern states, three different blends of ethanol are available. E10 contains 10% ethanol, E0 contains 0% ethanol, and E85 contains 85% ethanol. How much of each of the blends must be mixed to make 10 gallons of 20% ethanol? If the price per gallon for each type is given in the table below, what is the least that the 10 gallon blend would cost? Blend Cost per Gallon E10 $2.75 E0 $2.61 E85 $2.14

15 Solution This problem adds another type of ethanol that should contribute to the total number of gallons in the mixture and the total amount of ethanol in the mixture. Each of the two equations in 1 needs another term corresponding to E0: Total amount of mixture: E10 E0 E85 10 Total amount of ethanol: 0.10E10 0.0E0 0.85E where E10 is the number of gallons of 10% ethanol, E0 is the number of gallons of 0% ethanol, and E85 is the number of gallons of 85% ethanol. The augmented matrix for this system is

16 As in the previous example, the pivot in the first column is already in place so we simply need to use row operations to put a 0 below it: 0.10 R1 : R2 : R1 R becomes R 2

17 To put a pivot in the second column, multiply the second row by the reciprocal of 0.20: R2 becomes R To put a 0 above the pivot, 1 R : R1 : R R becomes R Writing this matrix as an a system of equations leads to E E85 5 E0.75E85 5

18 E85 appears in both equations so we ll solve for the other variables in terms of E85, E E85 E0 5.75E85 This system has many solutions, but we must be careful. These variables represent gallons of different ethanol blends. Because of this, the variables cannot be negative. As long we pick reasonable values for E85, we can insure that all the variables are nonnegative. For instance, if we set E85 0 we get E E This solution makes sense since mixing equal amounts of E10 and E0 should yield a mixture with a percent ethanol midway between 10% and 0%.

19 Notice that as we increase the amount of E85, the amount of E10 increases (the E85 term is added in the E10 equation) and the amount of E0 decreases (the E85 term is subtracted in the E0 equation). Eventually the amount of E0 will equal 0 as E85 is increased. This occurs when E85.75E85 5 E85 or Reasonable values for E85 are from 0 to 4 including 0 and 4 since this leads to non-negative values for E10 and E0 as well. We can calculate the total cost of any combination using cost per gallon of each type of ethanol. For instance, 5 gallons of E10 and 5 gallons of E0 would cost total cost

20 Here is a table of some of the possible solutions based on E E85 E0 5.75E85 and their corresponding total costs: E85 (gallons) E10 (gallons) E0 (gallons) Approximate Total Cost (dollars)

21 As the amount of E85 is increased, the total cost drops. Each increase of 0.25 gallons of E85 leads to a drop in the total cost of about 0.02 dollars. The lowest cost appears to come from mixing 4 gallons of E85 and 26 gallons of E10. Using any more E85 would require us to use a negative amount of E0. Even though this would yield a lower cost, it is not a reasonable value for this problem.