How Long Does It Take for a Water Droplet to Move Radially on a Turbine?

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SUMMARY

The discussion focuses on calculating the time it takes for a water droplet to move radially on a turbine, specifically from a radius of r0 = 0.08 m to r = 4.13 m, with the turbine rotating at a constant angular velocity of w = 402 rad/s. The relevant equation derived is V = w√(r² - r0²) = dr/dt, which leads to the integral ∫dt = ∫(dr/(w√(r² - r0²))). The solution provided suggests that the time can be expressed as t = (1/w) * ln(r + √(r² - r0²)). However, confusion arises regarding the placement of the square root term in the equation, indicating a need for clarity in the integration process.

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Homework Statement


A water droplet condenses at r0=0.80 m. The turbine is rotating at a fixed speed of N RPM.

You can treat the water droplet as a point mass of m=0.002 kg. Starting from rest at its formation point, it starts moving outward on the radial vane as shown. Its mass stays constant during this motion (no new condensation nor evaporation).
Calculate the time (in seconds) it takes for the droplet to move from the point of formation (ro) to the radius r.
r0= 0.08m
r= 4.13m
w= 402rad/s

Homework Equations



V = w√(r^2-r0^2) = dr/dt

The Attempt at a Solution



∫dt = ∫(dr/w√(r^2-r0^2))

dt is from 0->t and dr is ro->r

I know that this is the equation for determining the time taken but I do now how to to integrate that function. I have been told it is t= 1/w*ln(r+√r^2-r0^2). But when i attempted to calculate the time it was always wrong.
 
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Why don't you show your calculations?
 
GaryTravis said:
∫dt = ∫(dr/w√(r^2-r0^2))
Is the square root term in the numerator or denominator? If the latter, then you need another set of parentheses to show it.
 

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