Question

formula 5 points. water (density = 997.1 kg/m³ = 62.25 lb_m/ft³) is flowing through a 0.78 - inch diameter circular pipe. given that the water flows through the pipe 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, $d=\frac{0.78}{12}=0.065$ ft.

Step2: Calculate cross - sectional area of the pipe

The cross - sectional area of a circle is $A=\pi(\frac{d}{2})^2$. Substituting $d = 0.065$ ft, we get $A=\pi(\frac{0.065}{2})^2=\pi\times(0.0325)^2\approx 0.00332$ ft².

Step3: Calculate volumetric flow rate

The volumetric flow rate $Q = A\times v$, where $v = 1.33$ ft/s. So $Q=0.00332\times1.33 = 0.0044156$ ft³/s.

Step4: Convert time from seconds to hours

Since 1 hour = 3600 s, the volumetric flow rate in ft³/hour is $Q_{h}=0.0044156\times3600 = 15.9$ ft³/hour.

Step5: Calculate mass flow rate in slugs/hour

The density of water $
ho = 62.25$ lb/ft³. First, convert density to slugs/ft³. Since 1 slug = 32.2 lb, $
ho=\frac{62.25}{32.2}\approx1.933$ slugs/ft³. The mass flow rate $\dot{m}=
ho\times Q_{h}$. Substituting $
ho = 1.933$ slugs/ft³ and $Q_{h}=15.9$ ft³/hour, we get $\dot{m}=1.933\times15.9\approx30.7$ slugs/hour.

Answer:

30.7 slugs/hour