Question

# A commercial steel annulus 40 ft long, with a = 1 in and b = ½ in, connects two reservoirs that differ in surface height by 20 ft. Compute the flow rate in ft³/s through the annulus if the fluid is water at 20°C.

Solution

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The flow rate can be easily solved using Bernoulli's Energy Equation. Solve first the hydraulic diameter which will be used as a diameter in solving other values. Hydraulic diameter has a formula $D_h=\dfrac{4A}{\mathcal{P}}$ where $A$ is the cross sectional area of the outlet and $\mathcal{P}$ is the wetted perimeter of the outlet.

The area for annulus is the area of the larger circle minus the area of the smaller one $A=\pi{a^2}-\pi{b^2}$. The wetted perimeter is the perimeter of the smaller and larger circles $\mathcal{P}=2\pi a+2 \pi b$. Solve now for the hydraulic diameter.

\begin{align*} D_h&=\dfrac{4A}{\mathcal{P}}\\ &=\dfrac{4(\pi{a^2}-\pi{b^2})}{2\pi a+2 \pi b}\\ &=\dfrac{4(\pi(1^2)-\pi(0.5^2))}{2\pi(1)+2 \pi (0.5)}\\ &=1~\text{in}=\frac{1}{12}~\text{ft} \end{align*}

Solve the friction factor which is needed for the head loss in the Bernoulli's Energy Equation.

The friction factor can be solved by solving first the Reynold's number and the ratio $\dfrac{\epsilon}{D_h}$ where $\epsilon$ is the roughness coefficient from Table 6.1 and $D_h$ is the hydraulic diameter. $\epsilon$ of a commercial steel is equal to $0.0015~\text{ft}$.

\begin{align*} &\frac{\epsilon}{D_h}=\frac{0.00015}{\frac{1}{12}}\\ &\frac{\epsilon}{D_h}=0.0018 \end{align*}

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