Съдържание

APhO 2006, Теория — Задача 3A. Part A

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Всички материали от APhO 2006

7th Asian Physics Olympiad — Theoretical Question 3 · 2,5 т.

Условие

This problem consists of four not related parts.

A. [2.5 points] The Mariana Abyss in the Pacific Ocean has a depth of $H = 10920\ m$ . Density of salted water at the surface of the ocean is $\rho_0 = 1025\ kg/m^3$ , bulk modulus is $K = 2{,}1\cdot 10^9\ Pa$ , acceleration of gravity is $g=9.81 m/s^2$ . Neglect the change in the temperature and in the acceleration of gravity with the depth, and also neglect the atmospheric pressure.

A1) Find the relation between the density $\rho(x)$ and pressure $P(x)$ at the depth of $x$ .

A2) Find the numerical value of the pressure $P(H)$ at the bottom of the Mariana Abyss. You may use iterative methods to solve this part.

Note: The fluids have very small compressibility. Compressibility coefficient is defined as

\kappa = -\frac{1}{V}\left(\frac{dV}{dP}\right)_{T=const}

Bulk modulus $K$ is the inverse of $\kappa$ : $K=1/\kappa$ .

Отговори

Решение

The change of the pressure is related to the change in the density via

\Delta P = -K\frac{\Delta V}{V} = K\frac{\Delta\rho}{\rho_0}, \qquad (1)

where $\rho_0$ is the density of water at the surface.

\rho = \rho_0 + \Delta\rho = \rho_0\left(1+\frac{\Delta\rho}{\rho_0}\right) = \rho_0\left(1+\frac{\Delta P}{K}\right), \qquad (2)

where $\Delta P \approx P$ (we neglect the atmospheric pressure). Then

(A1)

\rho(x) = \rho_0\left(1+\frac{P(x)}{K}\right). \qquad (3)

The change of the hydrostatic pressure with the depth equals

dP = g\cdot\rho(x)dx, \quad \frac{dP}{dx} = g\rho(x) = g\rho_0 + g\rho_0\frac{P(x)}{K}, \qquad (4)

\frac{dP(x)}{dx} - \frac{g\rho_0}{K}P(x) = g\rho_0. \qquad (5)

The solution of this differential equation with boundary condition $P(0) = 0$ is

P(x) = K\left(\exp\frac{g\rho_0}{K}x - 1\right). \qquad (6)

Since $\frac{g\rho_0}{K}H \ll 1$ , we can use the expansion

\exp z \approx 1 + z + \frac{z^2}{2!} + \ldots, \qquad (7)

thus

P(x) \cong g\rho_0 x + \frac{1}{2K}\left(g\rho_0 x\right)^2 \qquad (8)

The last formula can be simply derived using the method of successive iterations. First, the pressure can be estimated without compressibility taken into account:

P_0(x) = g\rho_0 x. \qquad (9)

Correction to the density in the first approximation can be obtained using $P_0(x)$ :

\rho_1(x) = \rho_o\left(1 + \frac{g\rho_0 x}{K}\right). \qquad (10)

Now, correction to pressure can be obtained using $\rho_1(x)$ :

P_1(H) = \int_0^H \rho_1(x)g\,dx = g\rho_0 x + \frac{1}{2K}\left(g\rho_0 x\right)^2, \qquad (11)

as obtained earlier.

Putting in the numerical values, we get

(A2)

P(H) = (1098\cdot 10^5 + 28{,}7\cdot 10^5)\,\text{Па} \approx 1{,}13\cdot 10^8\ \text{Па}. \qquad (12)

Mark Distribution

No	Total Pt	Partial Pt	Contents
A	2.5	0.5	Equation
		1.5	Differential equation method	1.0	Equation (4) or (5)
				0.5	Equation (6)
			Iterative method	0.5	$P_0(x)$
				0.5	$\rho_1(x)$
				0.5	$P_1(H)$
		0.5	Numerical value of $P(H)$

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf