Calculate Acceleration at A, B and C in Fluid Flow Through Venturi Cone

[Varidius]

Fluid is flowing through a venturi-like cone, 0.1m long, horizontally along a streamline. At the start, it is traveling at 6m/s and after 0.1m it is traveling at 18m/s. Velocity is also stated to be a linear function of distance along the streamline. The question asks to determine the acceleration at at the point where it is 6m/s (A), at the point where it is 18m/s (B) and at a distance halfway between (C).

Since the problem says that velocity is linear with distance, I feel safe in saying that nothing too weird happens in between and basic physics apply (no crazy decelerations).

I'm thinking of using basic physics equations (ie. that use starting velocity, resultant velocity, displacement and acceleration) and solving for acceleration for each leg (A-C, C-B), but that'll only give me two acceleration values.

How do I solve this?

 

[Delta2]

Velocity linear with distance means that velocity $$v=v_A+cs$$ where $$v_A=6m/s$$. We know that for $$s=0.1, v_B=18$$ hence we can find
$$c=\frac{v_B-v_A}{s}=120$$

Use the fact that velocity is linear with distance to prove that acceleration is linear with velocity more specifically that $$a=cv=120(6+120s)$$.

 

[Varidius]

Sorry, stupid question, but what's c?

 

[Delta2]

It is just the constant of the linear relationship that velocity has with distance. It is also called the slope or gradient.

Problem states that velocity is a linear function of distance, which means that there are constants b and c such that $$v=b+cs$$. If we plot this equation on a diagram with velocity v on vertical axis and distance s on horizontal axis then what we get is a line, hence the word linear.

Check http://en.wikipedia.org/wiki/Linear_equation

 

[DeeNos]

Dear scientist,

Thank you for your question. In order to calculate the acceleration at points A, B, and C in fluid flow through a venturi cone, we can use the basic equation for acceleration, a = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time taken for the change in velocity to occur.

At point A, the initial velocity is 6m/s and the final velocity is also 6m/s, since the fluid is not changing velocity at this point. Therefore, the acceleration at point A is 0m/s^2.

At point B, the initial velocity is 6m/s and the final velocity is 18m/s, and the distance traveled is 0.1m. Using the equation a = (18m/s - 6m/s)/(0.1m), we get an acceleration of 120m/s^2.

At point C, the initial velocity is 6m/s and the final velocity is 12m/s, and the distance traveled is 0.05m (halfway between A and B). Using the same equation, we get an acceleration of 120m/s^2.

It is important to note that these calculations assume a constant acceleration throughout the distance traveled. If there are any changes in the flow or external factors, the acceleration may vary.

I hope this helps in your calculations. Please let me know if you have any further questions.

Best,