Question
Cnnsider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir. The purpose of the venturi is to create a vacuum in the reservoir when the venturi is placed in an airstream. (The rncunun is defined as the pressure difference helow the outside ambient pressure.) The venturi has a throat-10-inlet area ratio of 0.85 . Calculate the maximum vacuum obainable in the reservoir when the venturi is placed in an airstream of $90 \mathrm{~m} / \mathrm{s}$ at standard sea level conditions.
Step 1
Step 1: Calculate the velocity of the air at the throat of the venturi using the equation $v = \frac{Q}{A}$, where $Q$ is the volumetric flow rate and $A$ is the cross-sectional area of the throat. Show more…
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Consider a venturi with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir. The purpose of the venturi is to create a vacuum in the reservoir when the venturi is placed in an airstream. (The vacuum is defined as the pressure difference below the outside ambient pressure.) The venturi has a throat-to-inlet area ratio of 0.85. Calculate the maximum vacuum obtainable in the reservoir when the venturi is placed in an airstream of 90 m/s at standard sea level conditions.
Venturi Tube Consider the venturi tube of Problem 55 and Fig. $15-48$ without the manometer. Let $A$ equal $5 a .$ Suppose that the pressure $P_{1}$ at $A$ is $2.0 \mathrm{~atm} .$ Compute the values of $(\mathrm{a})\left|\vec{V}_{A}\right|$ at $A$ and (b) $\left|\vec{v}_{a}\right|$ at $a$ that would make the pressure $P_{2}$ at $a$ equal to zero. (c) Compute the corresponding volume flow rate if the diameter at $A$ is $5.0 \mathrm{~cm}$. The phenomenon that occurs at $a$ when $P_{2}$ falls to nearly zero is known as cavitation. The water vaporizes into small bubbles.
A vertical venturi-meter carries a liquid of relative density $0.8$ and has inlet and throat diameters of $150 \mathrm{~mm}$ and $75 \mathrm{~mm}$ respectively. The pressure connection at the throat is $150 \mathrm{~mm}$ above that at the inlet. If the actual rate of flow is $40 \mathrm{~L} \cdot \mathrm{s}^{-1}$ and the coefficient of discharge is $0.96$, calculate (a) the pressure difference between inlet and throat, and $(b)$ the difference of levels in a vertical U-tube mercury manometer connected between these points, the tubes above the mercury being full of the liquid. (Relative density of mercury = 13.56.)
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