A microgrid power station utilizes two types of energy storage units: Lithium-ion batterie… — GMAT

A microgrid power station utilizes two types of energy storage units: Lithium-ion batteries and Flow batteries. A primary grid configuration consisting of $4$ Lithium-ion batteries and $3$ Flow batteries provides a total continuous storage capacity of $140\text{ MWh}$ . An alternative configuration consisting of $2$ Lithium-ion batteries and $5$ Flow batteries provides a total continuous storage capacity of $126\text{ MWh}$ . Assuming that each battery of a given type contributes a constant storage capacity, what is the individual storage capacity, in MWh, of one Lithium-ion battery and one Flow battery, respectively?

$23\text{ MWh}$ for a Lithium-ion battery and $16\text{ MWh}$ for a Flow batteryAnswer
B
$16\text{ MWh}$ for a Lithium-ion battery and $23\text{ MWh}$ for a Flow battery
C
$20\text{ MWh}$ for a Lithium-ion battery and $18\text{ MWh}$ for a Flow battery
D
$25\text{ MWh}$ for a Lithium-ion battery and $14\text{ MWh}$ for a Flow battery
E
$21\text{ MWh}$ for a Lithium-ion battery and $19\text{ MWh}$ for a Flow battery

Answer

The storage capacity is

23\text{ MWh}

for a Lithium-ion battery and

16\text{ MWh}

for a Flow battery.

The system of equations

4L + 3F = 140

and

2L + 5F = 126

uniquely determines

L = 23\text{ MWh}

and

F = 16\text{ MWh}

. The option specifying

23\text{ MWh}

for Lithium-ion and

16\text{ MWh}

for Flow correctly matches the values of

L

and

F

in their respective order.

Step-by-Step Solution

Set up a system of two linear equations in two variables.

Let

L

be the capacity of one Lithium-ion battery and

F

be the capacity of one Flow battery:
Equation 1:

4L + 3F = 140

Equation 2:

2L + 5F = 126

Translate the physical battery configurations and total capacities into algebraic equations.

Eliminate variable

L

by multiplying Equation 2 by

2

Equation 2 scaled:

2(2L + 5F) = 2(126) \Rightarrow 4L + 10F = 252

Align the coefficients of

L

to subtract Equation 1 from the scaled Equation 2.

Subtract Equation 1 from the scaled equation to solve for

F

(4L + 10F) - (4L + 3F) = 252 - 140 \Rightarrow 7F = 112 \Rightarrow F = 16

Eliminating

L

isolate the single-variable linear equation for

F

Substitute

F = 16

back into Equation 2 to solve for

L

2L + 5(16) = 126 \Rightarrow 2L + 80 = 126 \Rightarrow 2L = 46 \Rightarrow L = 23

Determine the value of

L

once

F

is known.

Key Concept

Simultaneous Linear Equations (System of 2 Equations with 2 Variables)

Estimated Time:2m 0s

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