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Chapter 4: Problem 111

A horizontal axisymmetric jet of air with 0.5 in. diameter strikes a stationary vertical disk of 8 in. diameter. The jet speed is 225 ft/s at the nozzle exit. A manometer is connected to the center of the disk. Calculate (a) the deflection, \(h,\) if the manometer liquid has \(S G=1.75\) and (b) the force exerted by the jet on the disk.

Short Answer

Expert verified
The deflection h in the manometer and the force F exerted by the jet on the disk can be calculated using the established equations. Specific details and calculations might depend on constant values such as the density of air, the acceleration due to gravity and the density of water.

Step by step solution

01

Calculate the pressure at the center of the disc

Firstly, assign the velocity (V) of 225 ft/s, diameter (D) of the jet as 0.5 in, and the specific gravity (SG) of the manometer liquid as 1.75. Using Bernoulli's equation, derive the pressure at the center of the disc by equating it to the pressure in the jet. Given that atmospheric pressure (p0) = 0, the equation becomes, \(P = \frac{1}{2} \times \rho \times V^2\) where \(\rho\) is the density of air.
02

Determine the deflection

The deflection (h) in the manometer, is given by the manometer equation, which states that \(P = \rho_{liq} \times g \times h\), where \(\rho_{liq} = SG \times \rho_{water}\). By rearranging this equation, the deflection can be solved as, \(h = \frac{P}{\rho_{liq} \times g}\)
03

Calculate the force exerted by the jet

The force (F) exerted by a fluid jet on a flat plate is given by \(F = \rho \times q \times V\), where q is the flow rate of the jet, which can be calculated using the area (A) and velocity of the jet, yielding \(q = A \times V\). Substituting this back into the force equation, \(F = \rho \times A \times V^2\).

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