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water (density = 997.1 kg/m^3 = 62.25 lb_m/ft^3) is flowing through a c…

Question

water (density = 997.1 kg/m^3 = 62.25 lb_m/ft^3) is flowing through a circular pipe. given that the water flows through a pipe with a mass flowrate of 1.4 lb - mass/sec and the velocity in the pipe is known to be 1.2 ft/sec, what is the diameter of the pipe in inches? report your answer to two decimal places. answer

Explanation:

Step1: Recall mass - flow rate formula

The mass - flow rate formula is $\dot{m}=
ho vA$, where $\dot{m}$ is the mass - flow rate, $
ho$ is the density, $v$ is the velocity, and $A$ is the cross - sectional area of the pipe. The cross - sectional area of a circular pipe is $A = \frac{\pi}{4}d^{2}$, where $d$ is the diameter of the pipe. So, $\dot{m}=
ho v\frac{\pi}{4}d^{2}$.

Step2: Rearrange the formula for diameter

We can re - arrange the formula $\dot{m}=
ho v\frac{\pi}{4}d^{2}$ to solve for $d$. First, multiply both sides by $\frac{4}{
ho v\pi}$: $d^{2}=\frac{4\dot{m}}{
ho v\pi}$. Then, take the square - root of both sides: $d=\sqrt{\frac{4\dot{m}}{
ho v\pi}}$.

Step3: Identify the given values

We are given that $\dot{m} = 1.4\ lb - mass/sec$, $
ho=62.25\ lb_{m}/ft^{3}$, and $v = 1.2\ ft/sec$.

Step4: Substitute the values into the formula

$d=\sqrt{\frac{4\times1.4}{62.25\times1.2\times\pi}}$. First, calculate the denominator: $62.25\times1.2\times\pi=62.25\times3.769911 = 234.65$. Then, calculate the numerator: $4\times1.4 = 5.6$. So, $d=\sqrt{\frac{5.6}{234.65}}=\sqrt{0.02386}$.

Step5: Calculate the diameter in feet and convert to inches

$d=\sqrt{0.02386}\approx0.1544\ ft$. Since 1 foot = 12 inches, $d = 0.1544\times12=1.8528\ inches\approx1.85\ inches$.

Answer:

$1.85$ inches