Question

1.33 m³ of fluid flows out of a pipe in 24.5 s. the fluid leaves the pipe at 3.55 m/s. what is the area of the pipe? ? m²

Explanation:

Step1: Recall the formula for volumetric flow rate

The volumetric flow rate \( Q \) is given by the volume \( V \) divided by time \( t \), so \( Q=\frac{V}{t} \). Also, the volumetric flow rate can be expressed as \( Q = A \times v \), where \( A \) is the cross - sectional area of the pipe and \( v \) is the velocity of the fluid.
First, calculate the volumetric flow rate \( Q \). We know that \( V = 1.33\ m^{3} \) and \( t=24.5\ s \). So \( Q=\frac{V}{t}=\frac{1.33}{24.5}\ m^{3}/s \).

Step2: Solve for the area \( A \)

Since \( Q = A\times v \), we can re - arrange the formula to \( A=\frac{Q}{v} \). We already know that \( Q=\frac{1.33}{24.5}\ m^{3}/s \) and \( v = 3.55\ m/s \). Substitute these values into the formula for \( A \):
\( A=\frac{\frac{1.33}{24.5}}{3.55}=\frac{1.33}{24.5\times3.55} \)
First, calculate \( 24.5\times3.55 = 24.5\times(3 + 0.55)=24.5\times3+24.5\times0.55 = 73.5+13.475 = 86.975 \)
Then, \( A=\frac{1.33}{86.975}\approx0.0153\ m^{2} \)

Answer:

\( 0.0153\) (or more precisely, after calculating \(\frac{1.33}{24.5\times3.55}=\frac{1.33}{86.975}\approx0.0153\))