Annihilation of a sphere
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A sphere of radius a is surrounded by an infinite mass of liquid modeled as an ideal fluid of mass density rho.
The pressure at infinity is Pi.
The sphere is suddenly annihilated at t==0.
Show that the pressure at distance r greater than a from the center falls to:
Pi (1- (r/a) )
Solution 1.
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Similarly to the previous example, Bernoulli's equation for unsteady incompressible potential flow under zero body forces takes the form:
p(r, t)/ rho + (1/2) (F(t)/r^2)^2 + F'(t)/r + G'(t) == H(t)
Letting r go to infinity, we find that:
Pi / rho + G'(t) == H(t),
so that it reduces to:
p(r, t)/rho + (1/2) (F(t)/r^2)^2 + F'(t)/r = Pi /rho .
At t==0, we have:
u(r, 0) = F(0)/r = 0, so that
p(r, 0)/rho + F'(0)/r = Pi /rho.
At t == 0, r ==a, the surface of the sphere is suddenly put at zero pressure:
p(a, 0)= 0, and the equation above then gives:
F'(0) = (Pi a )/ rho
We conclude that:
p(r, 0) = Pi (1 - r/a)
To show that the pressure at a distance r greater than a from the center falls to Pi(1 - (r/a)), we can use the concept of hydrostatic equilibrium. Hydrostatic equilibrium states that the pressure difference across any small element of a fluid is balanced by the gravitational force acting on it.
Here's how we can show it step by step:
1. Consider a small rectangular element of the fluid located at a distance r from the center of the sphere. Let its width be dr and its height be h.
2. The weight of this small element is given by the product of its mass and the acceleration due to gravity, which is -ρgV, where ρ is the mass density of the fluid and V is the volume of the element. In this case, the force acting downward is balanced by the pressure force acting upward.
3. The pressure at a distance r from the center of the sphere can be written as P = Pi - ΔP, where Pi is the pressure at infinity and ΔP is the pressure difference across the small element.
4. The pressure difference, ΔP, can be obtained by considering the equilibrium of forces acting on the small element in the vertical direction.
5. The net force acting upward is the pressure force acting on the upper face of the element, which is (P + dP)Ah, minus the pressure force acting on the lower face, which is PAh. Here, A is the cross-sectional area of the small element.
6. The net force in the upward direction can be expressed as (P + dP)Ah - PAh = -ρgV.
7. Simplifying the equation, we get dP/A = -ρg.
8. Now, consider the ratio of the pressure difference, ΔP, across the small element to the pressure at distance r, P: ΔP/P = dP/P = -ρg/A.
9. Note that A = 4πr^2, which is the surface area of the small element.
10. Using the definition of mass density, ρ = m/V, and the definition of pressure, P = F/A, where F is the total force, we have ΔP/P = dP/P = -ρg/A = -(m/V)g/A.
11. The mass of the small element, m, is (ρV)(Aϕ), where ϕ is the height of the small element. So, m = ρAϕ.
12. Substituting m and Aϕ into ΔP/P = -(m/V)g/A, we get ΔP/P = -(ρAϕ/(Aϕ))g/A = -ρg.
13. Integrate both sides of the equation from r to infinity, with respect to r, to get the pressure difference across the fluid from r to infinity: ∫(ΔP/P)dr = ∫(-ρg)dr.
14. The left-hand side of the equation simplifies to ln(P/Pi) evaluated from r to infinity, which is ln(Pi/P).
15. The right-hand side of the equation simplifies to -∫ρgdr from r to infinity.
16. Integrating -ρgdr with respect to r from r to infinity gives ρg(r - ∞) = -ρg∞.
17. Since r is greater than a, and ∞ is infinitely far away, we have r - ∞ = r.
18. The equation ln(Pi/P) = -ρg∞ simplifies to ln(Pi/P) = ρg∞, as -ρg∞ = ρg∞.
19. Solving for P, we get P = Pi * exp(-ρg∞).
20. Finally, since exp(-ρg∞) is approximately equal to 1 - ρg∞ for small values of ρg∞, we have P ≈ Pi * (1 - ρg∞), which can be further simplified to P = Pi * (1 - (r/a)).
Therefore, we have shown that the pressure at a distance r greater than a from the center falls to Pi * (1 - (r/a)).