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IPhO 2010, Теория — Задача 2. Chimney physics

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41st International Physics Olympiad, Croatia – Theoretical competition, July 19th, 2010 · 19 юли 2010 г.

Условие

Introduction

Gaseous products of burning are released into the atmosphere of temperature $T_{Air}$ through a high chimney of cross-section $A$ and height $h$ (see Fig. 1). The solid matter is burned in the furnace which is at temperature $T_{Smoke}$ . The volume of gases produced per unit time in the furnace is $B$ .

Assume that:

the velocity of the gases in the furnace is negligibly small
the density of the gases (smoke) does not differ from that of the air at the same temperature and pressure; while in furnace, the gases can be treated as ideal
the pressure of the air changes with height in accordance with the hydrostatic law; the change of the density of the air with height is negligible
the flow of gases fulfills the Bernoulli equation which states that the following quantity is conserved in all points of the flow: $\frac{1}{2}\rho v^2(z) + \rho g z + p(z) = const,$
where $\rho$ is the density of the gas, $v(z)$ is its velocity, $p(z)$ is pressure, and $z$ is the height
the change of the density of the gas is negligible throughout the chimney

Sketch of a vertical chimney of cross-section A and height h standing on a horizontal furnace. Inside the chimney upward arrows and the label ρ_Smoke; outside, the labels ρ_Air and T_Air. In the furnace: smoke clouds, a flame over coal, a circled B, and the labels T_Smoke and ρ_Smoke. — Fig 1. Sketch of a chimney of height h with a furnace at temperature T_Smoke.

Task 1

Task 1 a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas into the atmosphere? Express your result in terms of $B$ , $A$ , $T_{\mathrm{Air}}$ , $g=9.81\mathrm{m/s^2}$ , $\Delta T=T_{Smoke}-T_{Air}$ . Important: in all subsequent tasks assume that this minimal height is the height of the chimney. (3.5 points)

Task 1 b) Assume that two chimneys are built to serve exactly the same purpose. Their cross sections are identical, but are designed to work in different parts of the world: one in cold regions, designed to work at an average atmospheric temperature of -30 $^{\circ}\mathrm{C}$ and the other in warm regions, designed to work at an average atmospheric temperature of 30 $^{\circ}\mathrm{C}$ . The temperature of the furnace is 400 $^{\circ}\mathrm{C}$ . It was calculated that the height of the chimney designed to work in cold regions is 100 m. How high is the other chimney? (0.5 points)

Task 1 c) How does the velocity of the gases vary along the height of the chimney? Make a sketch/diagram assuming that the chimney cross-section does not change along the height. Indicate the point where the gases enter the chimney. (0.6 points)

Task 1 d) How does the pressure of the gases vary along the height of the chimney? (0.5 points)

Solar power plant

The flow of gases in a chimney can be used to construct a particular kind of solar power plant (solar chimney). The idea is illustrated in Fig. 2. The Sun heats the air underneath the collector of area $S$ with an open periphery to allow the undisturbed inflow of air (see Fig. 2). As the heated air rises through the chimney (thin solid arrows), new cold air enters the collector from its surrounding (thick dotted arrows) enabling a continuous flow of air through the power plant. The flow of air through the chimney powers a turbine, resulting in the production of electrical energy. The energy of solar radiation per unit time per unit of horizontal area of the collector is $G$ . Assume that all that energy can be used to heat the air in the collector (the mass heat capacity of the air is $c$ , and one can neglect its dependence on the air temperature). We define the efficiency of the solar chimney as the ratio of the kinetic energy of the gas flow and the solar energy absorbed in heating of the air prior to its entry into the chimney.

Sketch of a solar chimney power plant: a tall dark chimney of height h (double arrow labelled h) rising from the centre of a flat collector labelled 'collector of area S'; yellow wavy arrows labelled 'solar radiation, G' fall on the collector; thick blue dotted arrows show cold air entering the collector periphery and thin red arrows show heated air flowing to and up the chimney and out of its top; hills and clouds in the background. — Fig 2. Sketch of a solar power plant.

Task 2

Task 2 a) What is the efficiency of the solar chimney power plant? (2.0 points)

Task 2 b) Make a diagram showing how the efficiency of the chimney changes with its height. (0.4 points)

Manzanares prototype

The prototype chimney built in Manzanares, Spain, had a height of 195 m, and a radius 5 m. The collector is circular with diameter of 244 m. The specific heat of the air under typical operational conditions of the prototype solar chimney is 1012 J/kg K, the density of the hot air is about 0.9 kg/m $^3$ , and the typical temperature of the atmosphere $T_{Air}$ = 295 K. In Manzanares, the solar power per unit of horizontal surface is typically 150 W/m $^2$ during a sunny day.

Task 3

Task 3 a) What is the efficiency of the prototype power plant? Write down the numerical estimate. (0.3 points)

Task 3 b) How much power could be produced in the prototype power plant? (0.4 points)

Task 3 c) How much energy could the power plant produce during a typical sunny day? (0.3 points)

Task 4

Task 4 a) How large is the rise in the air temperature as it enters the chimney (warm air) from the surrounding (cold air)? Write the general formula and evaluate it for the prototype chimney. (1.0 points)

Task 4 b) What is the mass flow rate of air through the system? (0.5 points)

Отговори

Решение

Solution - Chimney physics

This problem was inspired and posed by using the following two references:

W.W. Christie, Chimney design and theory, D. Van Nostrand Company, New York, 1902.
J. Schlaich, R. Bergermann, W. Schiel, G. Weinrebe, Design of Commercial Solar Updraft Tower Systems — Utilization of Solar Induced Convective Flows for Power Generation, Journal of Solar Energy Engineering 127, 117 (2005).

Solution of Task 1

a) What is the minimal height of the chimney needed in order that the chimney functions efficiently, so that it can release all of the produced gas in the atmosphere?

Let $p(z)$ denote the pressure of air at height $z$ ; then, according to one of the assumptions $p(z) = p(0) - \rho_{Air} gz$ , where $p(0)$ is the atmospheric pressure at zero altitude.

Throughout the chimney the Bernoulli law applies, that is, we can write

\frac{1}{2}\rho_{Smoke} v(z)^2 + \rho_{Smoke} gz + p_{Smoke}(z) = const., \qquad (1)

where $p_{Smoke}(z)$ is the pressure of smoke at height $z$ , $\rho_{Smoke}$ is its density, and $v(z)$ denotes the velocity of smoke; here we have used the assumption that the density of smoke does not vary throughout the chimney. Now we apply this equation at two points, (i) in the furnace, that is at point $z = -\varepsilon$ , where $\varepsilon$ is a negligibly small positive number, and (ii) at the top of the chimney where $z = h$ to obtain:

\frac{1}{2}\rho_{Smoke} v(h)^2 + \rho_{Smoke} gh + p_{Smoke}(h) \approx p_{Smoke}(-\varepsilon) \qquad (2)

On the right hand side we have used the assumption that the velocity of gases in the furnace is negligible (and also $-\rho_{Smoke} g\varepsilon \approx 0$ ).

We are interested in the minimal height at which the chimney will operate. The pressure of smoke at the top of the chimney has to be equal or larger than the pressure of air at altitude $h$ ; for minimal height of the chimney we have $p_{Smoke}(h) \approx p(h)$ . In the furnace we can use $p_{Smoke}(-\varepsilon) \approx p(0)$ . The Bernoulli law applied in the furnace and at the top of the chimney [Eq. (2)] now reads

\frac{1}{2}\rho_{Smoke} v(h)^2 + \rho_{Smoke} gh + p(h) \approx p(0). \qquad (3)

From this we get

v(h) = \sqrt{2gh\left(\frac{\rho_{Air}}{\rho_{Smoke}} - 1\right)}. \qquad (4)

The chimney will be efficient if all of its products are released in the atmosphere, i.e.,

v(h) \ge \frac{B}{A}, \qquad (5)

from which we have

h \ge \frac{B^2}{A^2}\frac{1}{2g}\frac{1}{\frac{\rho_{Air}}{\rho_{Smoke}} - 1}. \qquad (6)

We can treat the smoke in the furnace as an ideal gas (which is at atmospheric pressure $p(0)$ and temperature $T_{Smoke}$ ). If the air was at the same temperature and pressure it would have the same density according to our assumptions. We can use this to relate the ratio $\rho_{Air}/\rho_{Smoke}$ to $T_{Smoke}/T_{Air}$ that is,

\frac{\rho_{Air}}{\rho_{Smoke}} = \frac{T_{Smoke}}{T_{Air}}, \qquad (7)

and finally

h \ge \frac{B^2}{A^2}\frac{1}{2g}\frac{T_{Air}}{T_{Smoke} - T_{Air}} = \frac{B^2}{A^2}\frac{1}{2g}\frac{T_{Air}}{\Delta T}. \qquad (8)

For minimal height of the chimney we use the equality sign.

b) How high should the chimney in warm regions be?

\frac{h(30)}{h(-30)} = \frac{\dfrac{T(30)}{T_{Smoke} - T(30)}}{\dfrac{T(-30)}{T_{Smoke} - T(-30)}}; h(30) = 145m. \qquad (9)

c) How does the velocity of the gases vary along the height of the chimney?

The velocity is constant,

v = \sqrt{2gh\left(\frac{\rho_{Air}}{\rho_{Smoke}} - 1\right)} = \sqrt{2gh\left(\frac{T_{Smoke}}{T_{Air}} - 1\right)} = \sqrt{2gh\frac{\Delta T}{T_{Air}}}. \qquad (10)

This can be seen from the equation of continuity $Av = const.$ ( $\rho_{Smoke}$ is constant). It has a sudden jump from approximately zero velocity to this constant value when the gases enter the chimney from the furnace. In fact, since the chimney operates at minimal height this constant is equal to $B$ , that is $v = B/A$ .

d) At some height $z$ , from the Bernoulli equation one gets

p_{smoke}(z) = p(0) - (\rho_{Air} - \rho_{Smoke})gh - \rho_{Smoke} gz. \qquad (11)

Thus the pressure of smoke suddenly changes as it enters the chimney from the furnace and acquires velocity.

Solution of Task 2

a) The kinetic energy of the hot air released in a time interval $\Delta t$ is

E_{kin} = \frac{1}{2}(Av\Delta t\rho_{Hot})v^2 = Av\Delta t\rho_{Hot} gh\frac{\Delta T}{T_{Atm}}, \qquad (12)

Where the index “Hot” refer to the hot air heated by the Sun. If we denote the mass of the air that exits the chimney in unit time with $w = Av\rho_{Hot}$ , then the power which corresponds to kinetic energy above is

P_{kin} = wgh\frac{\Delta T}{T_{Air}}. \qquad (13)

This is the maximal power that can be obtained from the kinetic energy of the gas flow.

The Sun power used to heat the air is

P_{Sun} = GS = wc\Delta T. \qquad (14)

The efficiency is evidently

\eta = \frac{P_{kin}}{P_{Sun}} = \frac{gh}{cT_{Atm}}. \qquad (15)

b) The change is apparently linear.

Solution of Task 3

a) The efficiency is

\eta = \frac{gh}{cT_{Atm}} = 0.0064 = 0.64\%. \qquad (16)

b) The power is

P = GS\eta = G(D/2)^2\pi\eta = 45\,\mathrm{kW}. \qquad (17)

c) If there are 8 sunny hours per day we get 360kWh.

Solution of Task 4

The result can be obtained by expressing the mass flow of air $w$ as

w = Av\rho_{Hot} = A\sqrt{2gh\frac{\Delta T}{T_{Air}}}\rho_{Hot} \qquad (18)

w = \frac{GS}{c\Delta T} \qquad (19)

which yields

\Delta T = \left(\frac{G^2S^2T_{Atm}}{A^2c^2\rho_{Hot}^2 2gh}\right)^{1/3} \approx 9.1\,\mathrm{K}. \qquad (20)

From this we get

w = 760\,\mathrm{kg/s}. \qquad (21)

Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf