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Chapter 4: Problem 123

Incompressible liquid of negligible viscosity is pumped, at total volume flow rate \(Q,\) through two small holes into the narrow gap between closely spaced parallel disks as shown. The liquid flowing away from the holes has only radial motion. Assume uniform flow across any vertical section and discharge to atmospheric pressure at \(r=R\) Obtain an expression for the pressure variation and plot as a function of radius. Hint: Apply conservation of mass and the momentum equation to a differential control volume of thickness \(d r\) located at radius \(r\)

Short Answer

Expert verified
The pressure distribution for the incompressible liquid between the disks is given by \(p = -Q^2 / (2 \pi^2 h^2) (1/r - 1/R)\). The pressure decreases as the radius increases from the center.

Step by step solution

01

Apply Conservation of Mass

For an incompressible fluid, the volume flow rate is constant. Therefore: \(Q = 2\pi r h u\), where \(h\) is the gap between the discs and \(u\) is the velocity of the fluid at a radius \(r\). You can rearrange this equation to solve for velocity \(u\): \(u = Q / (2\pi r h)\). This establishes a relationship between the fluid velocity and the radius from its source.
02

Apply Momentum Equation

Next, balance the forces on a cylindrical fluid element of radius \(r\) and thickness \(dr\), situated at a distance \(r\) from the center. Since the flow is radial and there's no body force, the only forces are the pressure force and the resistance from the fluid element at \(r+dr\). This gives: \((p+p(dp/dr)dr)(2\pi r h) - p(2\pi r h) = (Q^2/r^2 - Q^2/(r+dr)^2)\((\pi h)\). Considering the limit for \(dr\) tending to zero and rearranging the equation results in the following differential equation for pressure: \(-dp/dr = Q^2 / (2 r^2 \pi^2 h^2)\) or \(dp/dr = -Q^2 / (2 \pi^2 r^2 h^2)\). This equation describes how the pressure changes as a function of radius.
03

Solve the Pressure Equation

Integrate the pressure differential equation from Step 2 over the limits from \(r\) to \(R\) (where the fluid discharges to atmospheric pressure). Integrating \(\int dp = - Q^2 / (2 \pi^2 h^2) \int (dr/r^2)\) from \(r\) to \(R\) gives: \(p - p_r = -Q^2 / (2 \pi^2 h^2) (1/r - 1/R)\). Here, \(p_r\) is the pressure at radius \(R\), which discharges to atmospheric pressure, hence, \(p_r = 0\). Therefore the pressure distribution is: \(p = -Q^2 / (2 \pi^2 h^2) (1/r - 1/R)\)
04

Plot Pressure Distribution

The pressure distribution function established in step 3 can then be plotted as a function of radius.

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