Practice Question: Bernoulli’s Energy Theorem

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Solution

Solution: C) \(HL = 12.12 ft\)

Given the discharge of the pipe that carrying oil is:

\[Q_1= Q_2=5.30 ft³/sec\]

Velocity can be related to the flow rate Q and area A by the flow rate equation:

\[v_1=v_2= {Q\over A}\]

\[v_1 = v_2 ={{Q} \over{π\over4} (d)²}\]

Velocity of the Pipe is calculated as:

\[v_1={5.30 ft³/sec \over {π\over 4} (0.49ft)²}=28.11 ft/sec\]
\[v_2={5.30 ft³/sec \over {π\over 4} (1.48ft)²}=3.08 ft/sec\]

Using Bernoulli’s Energy Equation, Taking 1 as datum:

\[E_1 ={{v_1²\over2g}+ {p_1\over \gamma}+z_1}\]

\[E_1 ={{v_1²\over2g}+ {p_1\over \gamma}+z_1}= {(28.11 ft/s)²\over(2(32.2 ft/s²))} + {(0.09lb/ft²)\over((62.4 lb/ft³))} +0\]\[E_1 = 12.27 ft\]

\[E_2 ={{v_2²\over2g}+ {p_2\over \gamma}+z_2}\]\[E_2 ={{v_2²\over2g}+ {p_2\over \gamma}+z_2}= - {(3.08 ft/s)²\over(2(32.2 ft/s²))} + {(0.09lb/ft²)\over((62.4 lb/ft³))} +0.49\]\[E_2 = 0.15 ft\]

Since  , the flow is from 1 to 2

Head Loss:\[HL= E_1-E_2=12.27 ft-0.15 ft\]

 \[HL =12.12 ft\]

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