In thermodynamics and fluid mechanics, the pressure loss gradient along a pipe is often modeled using the Darcy-Weisbach equation expressed as $h_f = f \frac{L}{D} \frac{v^2}{2g}$, where $h_f$ is the head loss due to friction, $f$ is the Darcy friction factor, $L$ is pipe length, $D$ is hydraulic diameter, $v$ is flow velocity, and $g$ is gravitational acceleration. Solve this literal equation explicitly for the flow velocity $v$, assuming all quantities are strictly positive real numbers.

In thermodynamics and fluid mechanics, the pressure loss gradient along a pipe is often modeled using the Darcy-Weisbach equation expressed as $h_f = f \frac{L}{D} \frac{v^2}{2g}$ , where $h_f$ is the head loss due to friction, $f$ is the Darcy friction factor, $L$ is pipe length, $D$ is hydraulic diameter, $v$ is flow velocity, and $g$ is gravitational acceleration. Solve this literal equation explicitly for the flow velocity $v$ , assuming all quantities are strictly positive real numbers.

Final answer

$v = \sqrt{\frac{2g D h_f}{f L}}$

Steps

Combine the terms on the right-hand side

To solve the literal equation for the flow velocity $v$ , isolate the variable using inverse operations.

Write the right-hand side as a single rational expression:

$h_f = \frac{f \cdot L \cdot v^2}{2g \cdot D}$

Clear the denominator

Multiply both sides of the equation by the entire denominator, $2gD$ , to clear the fraction:

$h_f \cdot (2gD) = f L v^2$

$2g D h_f = f L v^2$

Isolate $v^2$

Divide both sides by the coefficient of $v^2$ , which is $fL$ :

$v^2 = \frac{2g D h_f}{f L}$

Solve for $v$ by taking the square root

Take the square root of both sides to isolate $v$ :

$v = \pm\sqrt{\frac{2g D h_f}{f L}}$

Common Pitfall: When taking the square root of both sides of an algebraic equation, both positive and negative roots are mathematically possible ( $\pm$ ). However, because all physical quantities in this context are strictly positive real numbers ( $v > 0$ ), we reject the negative root.

Thus, the explicit solution for $v$ is:

$v = \sqrt{\frac{2g D h_f}{f L}}$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student inverts the placement of the factors in the numerator and denominator when clearing the denominators and isolating v².

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$h_f = \frac{f L v^2}{2g D}$
Step 2
$v^2 = \frac{f L}{2g D h_f}$
Correct
$v^2 = \frac{2g D h_f}{f L}$
To isolate $v^2$ , you must multiply both sides by the denominator $2g D$ and divide by the coefficient $f L$ . Instead, you inverted the resulting fraction, placing $f L$ in the numerator and $2g D h_f$ in the denominator.
Step 3
$v = \sqrt{\frac{f L}{2g D h_f}}$
Correct
$v = \sqrt{\frac{2g D h_f}{f L}}$
You correctly took the positive square root to solve for $v$ , but the result inherits the inverted fraction from the previous step.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly grouped the right-hand side factors into a single rational expression.

You correctly recognized that taking the square root isolates $v$ from $v^2$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Literal Equation Inversion Error

You inverted the numerator and denominator when rearranging the equation to isolate $v^2$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Conflating multiplication and division during multi-step isolation: factors in the denominator of the original fraction move to the numerator of the opposite side via multiplication, whereas factors multiplying the target variable must move to the denominator via division.

REASONING PATH

Your ACEpath teacher’s suggested approach

Inverse Operations for Fractions: To move a term across the equals sign, perform the inverse operation: divide by what is multiplied, and multiply by what is divided.
Multiply both sides by $2g D$ first to obtain $2g D h_f = f L v^2$ , then divide by $f L$ to get $v^2 = \frac{2g D h_f}{f L}$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Multiply both sides of $h_f = \frac{f L v^2}{2g D}$ by the denominator $2g D$ : $2g D h_f = f L v^2$
Divide both sides by the coefficient of $v^2$ , which is $f L$ : $v^2 = \frac{2g D h_f}{f L}$
Take the principal square root of both sides (since $v > 0$ ): $v = \sqrt{\frac{2g D h_f}{f L}}$

Final answer

$v = \sqrt{\frac{2g D h_f}{f L}}$

NEXT STEPS

Final Takeaways from Your Teacher

Write out the intermediate step $2g D h_f = f L v^2$ explicitly before isolating the squared variable to prevent fraction inversion.
Double-check your rearranged formula using dimensional analysis or by substituting simple test values.

OverviewACEpath Teacher's Diagnosis

02 The student correctly isolates v² but forgets to take the square root to solve for v.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
2g D h_f = f L v^2
Step 2
v^2 = $\frac{2g D h_f}{f L}$
Step 3
v = $\frac{2g D h_f}{f L}$
Correct
v = $\sqrt{\frac{2g D h_f}{f L}}$
To solve for $v$ , you must apply the inverse operation of squaring (the square root) to both sides of the equation. Removing the exponent on the left requires taking the square root of the expression on the right.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared the fraction by multiplying both sides of the equation by $2gD$ .

You isolated the squared variable term $v^2$ accurately by dividing both sides by $fL$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted square root when solving for squared variable

You dropped the exponent $2$ on the left side to write $v$ , but you left the right side unchanged instead of taking its square root.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete application of inverse operations when solving higher-degree literal equations: treating the removal of an exponent as simply erasing the power rather than applying the matching radical to both sides.

REASONING PATH

Your ACEpath teacher’s suggested approach

Invert Powers with Roots: For any positive real quantity $x$ , the inverse of $x^2 = A$ is $x = \sqrt{A}$ .
Whenever you change $x^2$ to $x$ , immediately place the entire opposite side under a square root radical.

CORRECTED PATH

ACEpath Teacher's Demonstration

Take the principal square root of both sides: $\sqrt{v^2} = \sqrt{\frac{2g D h_f}{f L}}$
Simplify the radical, applying the condition that $v > 0$ : $v = \sqrt{\frac{2g D h_f}{f L}}$

Final answer

v = $\sqrt{\frac{2g D h_f}{f L}}$

NEXT STEPS

Final Takeaways from Your Teacher

Double-check your final variable against the prompt: if the prompt asks for $v$ , ensure you have not stopped at $v^2$ .
Practice using balanced inverse operations when solving physics and geometry formulas involving squared variables, such as kinetic energy ( $E_k = \frac{1}{2}mv^2$ ) or cylinder volume ( $V = \pi r^2 h$ ).

OverviewACEpath Teacher's Diagnosis

03 The student treats the factor of 2 in the denominator as an exponent or coefficient on v, incorrectly dividing by 2 instead of taking the square root.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
2g D h_f = f L v^2 $\implies$ v^2 = $\frac{2g D h_f}{f L}$
Step 2
v = $\frac{g D h_f}{f L}$
Correct
v = $\sqrt{\frac{2g D h_f}{f L}}$
To undo an exponent of $2$ , you must take the principal square root of the entire expression rather than dividing by $2$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared the compound denominator by multiplying by $2gD$ .

You isolated the squared variable $v^2$ by dividing by the coefficient $fL$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Inverse Operation Confusion (Exponent vs. Coefficient)

Dividing by $2$ to undo squaring instead of applying a square root.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the exponent $2$ in $v^2$ as a multiplicative factor of $2$ , which led to dividing the numerator by $2$ instead of taking the square root of the entire fraction.

REASONING PATH

Your ACEpath teacher’s suggested approach

Inverse of an Exponent: The inverse operation of squaring ( $x^2 = C$ ) is taking the square root ( $x = \sqrt{C}$ for $x > 0$ ), not dividing by $2$ .
Whenever a variable has an exponent $n$ , apply the radical $\sqrt[n]{\dots}$ to both sides rather than dividing by $n$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Take the principal square root of both sides: $v = \sqrt{\frac{2g D h_f}{f L}}$ .

Final answer

v = $\sqrt{\frac{2g D h_f}{f L}}$

NEXT STEPS

Final Takeaways from Your Teacher

Review the difference between isolating a variable with a coefficient ( $2x = k \implies x = \frac{k}{2}$ ) versus a variable with an exponent ( $x^2 = k \implies x = \sqrt{k}$ ).
Practice solving literal formulas containing squared variables, such as kinetic energy ( $E_k = \frac{1}{2}mv^2$ ) and the Pythagorean theorem.

Lock it in

Question

If $y = \frac{a x}{b}$ , what is the correct two-step inverse process to isolate $x$ ?

Answer

$x = \frac{b y}{a}$ . Multiply both sides by denominator $b$ , then divide both sides by coefficient $a$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student inverts the placement of the factors in the numerator and denominator when clearing the denominators and isolating v².

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student correctly isolates v² but forgets to take the square root to solve for v.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student treats the factor of 2 in the denominator as an exponent or coefficient on v, incorrectly dividing by 2 instead of taking the square root.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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