Question

Consider the one-dimensional, incompressible flow through the circular channel shown. The velocity is given by U=U₀+U₁sinωt, where U₀=20 m/s, U₁=2 m/s, and ω=0.3rad/s. The channel dimensions are L=1 m, R₁=0.2 m, and R₂=0.1 m. Determine the particle acceleration at the channel exit. Plot the results as a function of time over a complete cycle. On the same plot, show the acceleration at the channel exit if the channel is a constant area, rather than convergent, and explain the difference between the curves.

Answer 1

The acceleration of a fluid particle at the exit of a changing cross-section channel can be found by differentiating the velocity function with respect to time and plotted over a cycle to visualize time-dependent changes. For a constant area channel, the plot of acceleration would differ due to fluid dynamics principles such as the continuity equation and Bernoulli's principle.

Step-by-step explanation:

The student's question relates to fluid dynamics, specifically concerning the acceleration of a fluid particle as it exits a channel. Given the velocity of the fluid U as a function of time U=U₀+U₁sinωt, with specified constants, we can determine the fluid particle acceleration at the exit by differentiating the velocity function with respect to time. This gives the instantaneous acceleration a(t) = dU/dt = U₁ωcosωt.

We'd plot a(t) = 2ωcosωt, where U₁=2 m/s, and ω=0.3rad/s, over a complete cycle (0 to approximately 21 seconds) to show how acceleration varies with time. For a constant area channel, the plot would still be sinusoidal but with a difference in amplitude or frequency related to the change in cross-sectional area. Because the cross-sectional area of the convergent channel changes, the fluid acceleration would vary differently compared to the constant area channel due to the continuity equation and Bernoulli's principle.