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Условие Отговори Решение

IZhO 2015, Теория — Задача 2. Vessel with water

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Всички материали от IZhO 2015

XI International Zhautykov Olympiad -Theoretical Competition- · 13 януари 2015 г. · 7 т.

Внимание: Бележка към темата

The separate answer sheets with figures 1.1, 1.2 and 1.3 of Problem 1 are not included in the archived problems file.

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The time available for the theoretical competition is 4 hours. There are three Problems.
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Условие

In the cylindrical vessel with the cross section area $S = 0.500\ m^2$ and height $h = 0.500\ m$ a tube of length $L = 2.00\ m$ with opened ends is inserted vertically through the hermetically sealed lid. The lower end of the tube is a bit above the bottom of the vessel. Water with the density $\rho = 1000\ kg/m^3$ is poured into the vessel as shown in the figure on the right. The cross section area of the tube is much smaller than the cross section area of the vessel and the vessel wall material conducts heat very well. Assume that the atmospheric pressure is $p_0 = 1.01\cdot 10^5\ Pa$ , the ambient temperature is $T_0 = 293\ K$ and the acceleration of gravity is $g = 9.80\ m/s^2$ .

Cylindrical vessel of height h with a sealed lid standing on a base; a vertical tube of length L passes through the lid down to just above the bottom; the vessel is partly filled with water (grey) to height x.

1. Find the height of the water level in the vessel $x = x_0$ at the time moment when the tube is completely filled with water. Express your answer in terms of $p_0, \rho, g, h, L$ , and find its numerical value. [2 т.]

Now, assume that the walls of the vessel and the tube are coated with a material which does not conduct heat at all. The air inside the vessel is then heated fast enough such that the water does not have enough time to warm up.

2. Find air pressure inside the vessel $p(x)$ as a function of $x$ . Express your answer in terms of $p_0, \rho, g, L, x$ . [0,5 т.]

3. Find air temperature inside the vessel $T(x)$ as a function of $x$ . Express your answer in terms of $p_0, \rho, g, L, x$ . [1 т.]

4. Find the temperature $T_m$ to which the air in the vessel must be heated in order to displace all the water from the vessel. Express your answer in terms of $p_0, \rho, g, L, T_0$ and find its numerical value. [1 т.]

5. Find the amount of heat $Q$ , which must be given to the air in the vessel in order to displace all the water from the vessel. Express your answer in terms of $p_0, \rho, g, h, L, S$ , and find its numerical value. [2,5 т.]

Отговори

Решение

While pouring the water into the vessel the air is compressed and its pressure increases. At the moment when the tube is completely filled with water the air pressure inside the vessel is equal to

p = p_0 + \rho g(L - x_0). \qquad (1)

Since the vessel wall highly conducts heat, the temperature of the air inside the vessel does not change, so the equations of state are

p_0Sh = vRT_0, \qquad (2)

pS(h - x_0) = vRT_0, \qquad (3)

where $S$ is the cross section area of the vessel, $v$ is number of moles of the air inside the vessel.

From Eq. (1) - (3) the following quadratic equation is obtained:

\rho g x_0^2 - [p_0 + \rho g(L + h)]x_0 + \rho g h L = 0, \qquad (4)

which has the obvious solution:

x_0 = \frac{1}{2}\left[\frac{p_0}{\rho g} + L + h \pm \sqrt{\left(\frac{p_0}{\rho g} + L + h\right)^2 - 4hL}\right]. \qquad (5)

Among two possible solutions (5), we should choose the one with the less value since it should be $x_0 = h$ at $p_0 = 0$ or $x_0 = 0$ at $L = 0$ , that is,

x_0 = \frac{1}{2}\left[\frac{p_0}{\rho g} + L + h - \sqrt{\left(\frac{p_0}{\rho g} + L + h\right)^2 - 4hL}\right]. \qquad (6)

Substituting the numerical values gives

x_0 = 7.86\cdot 10^{-2}\ m. \qquad (7)

The water is at equilibrium and, thus, the air pressure inside the vessel is found as a function of $x$ as follows

p(x) = p_0 + \rho g(L - x). \qquad (8)

Equation of state of an ideal gas for an arbitrary $x$ is given by

p(x)S(h - x) = vRT(x), \qquad (9)

which, together with Eq.(1), yields

T(x) = T_0\left(1 - \frac{x}{h}\right)\left(1 + \frac{\rho g(L - x)}{p_0}\right). \qquad (10)

The temperature, at which the air displaces water out of the vessel, is determined by the condition $x = 0$ , which, in accordance with (10), leads to

T_m = T_0\left(1 + \frac{\rho g L}{p_0}\right), \qquad (11)

and the corresponding numerical value is evaluated as

T_m = 350\ K. \qquad (12)

The change of the internal energy of the air is obtained as

\Delta U = \frac{5}{2}vR(T - T_0) = \frac{5}{2}\rho g L S h, \qquad (13)

and the work done by the air to displace the water, is calculated as

\begin{aligned} A = \int_0^{x_0} p(x)Sdx = \frac{1}{2}p_0SL\left(1 + \frac{p_0}{2\rho g L} + \frac{\rho g L}{2p_0}\left[1 + \frac{2h}{L} - \frac{h^2}{L^2}\right]\right) - \\ - \frac{1}{4}p_0SL\left(1 + \frac{\rho g(L - h)}{p_0}\right)\sqrt{\left(1 + \frac{h}{L} + \frac{p_0}{\rho g L}\right)^2 - \frac{4h}{L}}. \end{aligned} \qquad (14)

According to the first law of thermodynamics, the heat given to the air is found as

Q = \Delta U + A, \qquad (15)

which, together with Eqs. (13) and (14), yields

\begin{aligned} Q = \frac{1}{2}p_0SL\left(1 + \frac{p_0}{2\rho g L} + \frac{\rho g L}{2p_0}\left[1 + \frac{12h}{L} - \frac{h^2}{L^2}\right]\right) - \\ - \frac{1}{4}p_0SL\left(1 + \frac{\rho g(L - h)}{p_0}\right)\sqrt{\left(1 + \frac{h}{L} + \frac{p_0}{\rho g L}\right)^2 - \frac{4h}{L}}. \end{aligned} \qquad (16)

Substituting the numerical values gives

Q = 17.0\ kJ. \qquad (17)

Marking scheme

№	Content	баллы
1	Formula (1) $p = p_0 + \rho g(L - x_0)$	0,25	2,0
	Formula (2) $p_0Sh = vRT_0$	0,25
	Formula (3) $pS(h - x_0) = vRT_0$	0,25
	Formula (4) $\rho g x_0^2 - [p_0 + \rho g(L + h)]x_0 + \rho g h L = 0$	0,25
	Formula (5) $x_0 = \frac{1}{2}\left[\frac{p_0}{\rho g} + L + h \pm \sqrt{\left(\frac{p_0}{\rho g} + L + h\right)^2 - 4hL}\right]$	0,25
	Formula (6) $x_0 = \frac{1}{2}\left[\frac{p_0}{\rho g} + L + h - \sqrt{\left(\frac{p_0}{\rho g} + L + h\right)^2 - 4hL}\right]$	0,5
	Formula (7) $x_0 = 7{,}86\cdot 10^{-2}\ m$	0,25
2	Formula (8) $p(x) = p_0 + \rho g(L - x)$	0,5	0,5
3	Formula (9) $p(x)S(h - x) = vRT(x)$	0,5	1,0
	Formula (10) $T(x) = T_0\left(1 - \frac{x}{h}\right)\left(1 + \frac{\rho g(L - x)}{p_0}\right)$	0,5
4	Formula (11) $T_m = T_0\left(1 + \frac{\rho g L}{p_0}\right)$	0,5	1,0
	Formula (12) $T_m = 350\ K$	0,5
5	Formula (13) $\Delta U = \frac{5}{2}vR(T - T_0) = \frac{5}{2}\rho g L S h$	0,5	2,5
	Formula (14) $A = \frac{1}{2}p_0SL\left(1 + \frac{p_0}{2\rho g L} + \frac{\rho g L}{2p_0}\left[1 + \frac{2h}{L} - \frac{h^2}{L^2}\right]\right) - \frac{1}{4}p_0SL\left(1 + \frac{\rho g(L - h)}{p_0}\right)\sqrt{\left(1 + \frac{h}{L} + \frac{p_0}{\rho g L}\right)^2 - \frac{4h}{L}}.$	0,5
	Formula (15) $Q = \Delta U + A$	0,5
	Formula (16) $Q = \frac{1}{2}p_0SL\left(1 + \frac{p_0}{2\rho g L} + \frac{\rho g L}{2p_0}\left[1 + \frac{12h}{L} - \frac{h^2}{L^2}\right]\right) - \frac{1}{4}p_0SL\left(1 + \frac{\rho g(L - h)}{p_0}\right)\sqrt{\left(1 + \frac{h}{L} + \frac{p_0}{\rho g L}\right)^2 - \frac{4h}{L}}.$	0,5
	Formula (17) $Q = 17{,}0\ kJ$	0,5
Total			7,0

Оригинал в Архива: IZhO-2015-Theory_eng.pdf · официални решения: IZhO-2015-Theory_eng_sol.pdf