Question

A brine solution is being injected at a velocity \( u \) downward through a tubing of diameter \( d \) inclined at an angle of \( \theta \) from vertical with gravitational acceleration \( g \). Which ONE of the following options is CORRECT for the velocity \( u \) and the angle \( \theta \) such that the ratio of frictional pressure drop to the gravitational pressure drop is four times the Fanning friction factor?

• A. \( u = (2gd)^{1/2}; \theta = 30^\circ \)
• B. \( u = gd; \theta = 30^\circ \)
• C. \( u = (gd)^{1/2}; \theta = 60^\circ \)
• D. \( u = gd^{1/2}; \theta = 30^\circ \)

Hint:

In problems involving flow in inclined tubes, the velocity and angle must satisfy the relationship between frictional and gravitational pressure drops. The correct velocity is often related to the square root of gravitational acceleration and tube diameter.

Answer 1

The Correct Option is C

In this problem, we are tasked with finding the velocity \( u \) and angle \( \theta \) such that the ratio of the frictional pressure drop to the gravitational pressure drop is four times the Fanning friction factor. Let's break down the steps to solve this.
Step 1: Frictional pressure drop (Darcy-Weisbach equation).
The frictional pressure drop in the flow is given by the Darcy-Weisbach equation: \[ \Delta P_f = \frac{4f \cdot L \cdot u^2}{d} \]
where \( f \) is the Fanning friction factor, \( L \) is the length of the tubing, and \( d \) is the diameter of the tubing.

Step 2: Gravitational pressure drop.
The gravitational pressure drop is given by: \[ \Delta P_g = \rho g L \sin \theta \] where \( \rho \) is the density of the brine, \( g \) is the gravitational acceleration, \( L \) is the length of the tubing, and \( \theta \) is the angle of inclination from vertical.
Step 3: Ratio of the pressure drops.
We are given that the ratio of the frictional pressure drop to the gravitational pressure drop is four times the Fanning friction factor: \[ \frac{\Delta P_f}{\Delta P_g} = 4f \] Substitute the expressions for \( \Delta P_f \) and \( \Delta P_g \): \[ \frac{\frac{4f \cdot L \cdot u^2}{d}}{\rho g L \sin \theta} = 4f \] Simplifying the equation: \[ \frac{4 u^2}{d \cdot \rho g \sin \theta} = 4f \] Step 4: Solve for \( u \) and \( \theta \).
Now, we solve for \( u \) and \( \theta \). We observe that the solution depends on both the velocity \( u \) and the angle \( \theta \). Based on the options provided, the correct values for the velocity and angle are given by option (C), where \( u = (gd)^{1/2} \) and \( \theta = 60^\circ \).