Question

water (density = 997.1 kg/m³ = 62.25 lbm/ft³) is flowing through a 0.78 inch diameter at a velocity of 1.33 ft/s. what is the mass flowrate of the water in slugs/hour? report your answer to one decimal place.

Explanation:

Step1: Convert diameter to feet

The diameter $d = 0.78$ inches. Since 1 foot = 12 inches, $d=0.78/12 = 0.065$ ft.

Step2: Calculate cross - sectional area

The cross - sectional area of the pipe $A=\pi(\frac{d}{2})^2$. Substituting $d = 0.065$ ft, we get $A=\pi(\frac{0.065}{2})^2=\pi\times(0.0325)^2\approx 0.00332$ $ft^{2}$.

Step3: Calculate volume flow rate

The volume flow rate $Q = A\times v$, where $v = 1.33$ ft/s. So $Q=0.00332\times1.33 = 0.0044156$ $ft^{3}/s$.

Step4: Convert density to slugs/ft³

We know that 1 slug = 32.174 lbₘ. The density $
ho = 62.25$ lbₘ/ft³. So $
ho=\frac{62.25}{32.174}\approx1.935$ slugs/ft³.

Step5: Calculate mass flow rate

The mass flow rate $\dot{m}=
ho\times Q$. First, convert the volume flow rate to $ft^{3}/h$: $Q = 0.0044156\times3600=15.9$ $ft^{3}/h$. Then $\dot{m}=1.935\times15.9\approx30.8$ slugs/h.

Answer:

30.8 slugs/h