IPhO 2010 · Theory problem 2

Chimney physics

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:

\[ \frac{1}{2}\rho v^2(z)+\rho gz+p(z)=const, \]

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

Figure 1

Fig 1. Sketch of a chimney of height \(h\) with a furnace at temperature \(T_{Smoke}\).

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 \(B\), \(A\), \(T_{Air}\), \(g=9.81\,\mathrm{m/s^2}\), \(\Delta T=T_{Smoke}-T_{Air}\). Important: in all subsequent tasks assume that this minimal height is the height of the chimney. (3.5 points)

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 \(100\,\mathrm{m}\). How high is the other chimney? (0.5 points)

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)

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.

Figure 2

Fig 2. Sketch of a solar power plant.

Task 2

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

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

Task 3

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

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

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

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)

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

Official English text of the problem.

Answers

Take \(z=0\) at the chimney entrance and \(z=h\) at its outlet. Below, \(\rho_h\) denotes the density of the smoke or heated air.

  • 1(a) For \(\Delta T>0\), \[ \boxed{h_{\min}=\frac{B^2T_{Air}}{2gA^2\Delta T}}. \]

  • 1(b) \[ \boxed{h_{\mathrm{warm}}\simeq145\,\mathrm{m}}. \]

  • 1(c) \[ \boxed{v(z)=\frac BA\qquad(0<z<h)}. \] The velocity rises from nearly zero in the furnace to \(B/A\) in the entrance region, then remains constant throughout the chimney.

  • 1(d) \[ \boxed{p(z)=p_{Air}(h)+\rho_hg(h-z)}. \] The pressure decreases linearly with height, with slope \(-\rho_hg\), reaching the external atmospheric pressure at the outlet.

  • 2(a) \[ \boxed{\eta=\frac{gh}{cT_{Air}}}. \]

  • 2(b) The graph of \(\eta\) against \(h\) is a straight line through the origin, with slope \(g/(cT_{Air})\).

  • 3(a) \[ \boxed{\eta=0.00641\simeq0.64\%}. \]

  • 3(b) The ideal available power is \[ \boxed{P\simeq4.49\times10^4\,\mathrm{W}\simeq45\,\mathrm{kW}}. \]

  • 3(c) For an effective illumination time \(\tau\), \[ \boxed{E=P\tau}. \] For \(12\,\mathrm{h}\) at the stated irradiance, \(E\simeq0.54\,\mathrm{MWh}=1.9\,\mathrm{GJ}\). If the stated irradiance is a full-day average, the \(24\,\mathrm{h}\) result is \(1.08\,\mathrm{MWh}=3.88\,\mathrm{GJ}\).

  • 4(a) \[ \boxed{\Delta T= \left[\frac{(GS)^2T_{Air}}{2gh\,\rho_h^2A^2c^2}\right]^{1/3} \simeq9.1\,\mathrm{K}}. \]

  • 4(b) \[ \boxed{\dot m=\frac{GS}{c\Delta T} =\left[\frac{2gh(\rho_hA)^2GS}{cT_{Air}}\right]^{1/3} \simeq7.7\times10^2\,\mathrm{kg\,s^{-1}}}. \]

Solution

Write \(\rho_a=\rho_{Air}\), \(\rho_h=\rho_{Smoke}\), and \(p_0=p_{Air}(0)\). For the solar chimney, \(\rho_h\) instead denotes the heated-air density. Absolute temperatures in all formulas are in kelvin.

1(a)

The chimney operates by natural draft: the external air is denser than the hot gas, so its pressure falls more rapidly with height. At the limiting operating condition, the nearly stationary gas in the furnace is at the atmospheric pressure \(p_0\). This is the boundary condition for discharging the gas without requiring furnace overpressure.

By the hydrostatic law outside the chimney,

\[ p_{Air}(h)=p_0-\rho_a gh. \]

At the open outlet, the gas pressure equals the local atmospheric pressure. Bernoulli’s equation between the furnace, where the velocity is negligible, and the outlet gives

\[ p_0=p_{Air}(h)+\rho_hgh+\frac12\rho_hv^2. \]

Consequently,

\[ \frac12\rho_hv^2=(\rho_a-\rho_h)gh. \]

The smoke and air have the same density at equal temperature and pressure. Applying the ideal-gas law at the common base pressure therefore gives

\[ \frac{\rho_a}{\rho_h} =\frac{T_{Smoke}}{T_{Air}} =1+\frac{\Delta T}{T_{Air}}. \]

Thus

\[ \boxed{v^2=2gh\frac{\Delta T}{T_{Air}}}. \]

Because the density change between the furnace and chimney is neglected, the volumetric discharge capacity is \(Av\). Releasing all the gas requires

\[ Av\geq B. \]

At the minimum height, equality holds:

\[ \left(\frac BA\right)^2 =2gh_{\min}\frac{\Delta T}{T_{Air}}. \]

Hence

\[ \boxed{h_{\min}=\frac{B^2T_{Air}}{2gA^2\Delta T}}. \]

This result requires \(\Delta T>0\). For \(B>0\) and \(\Delta T=0\), no finite height produces the required natural draft under these assumptions.

1(b)

Both chimneys have the same \(B\), \(A\), and furnace temperature, so

\[ h\propto\frac{T_{Air}}{T_{Smoke}-T_{Air}}. \]

The relevant temperatures and temperature differences are

\[ T_c=243.15\,\mathrm{K},\qquad T_w=303.15\,\mathrm{K}, \]

\[ \Delta T_c=430\,\mathrm{K},\qquad \Delta T_w=370\,\mathrm{K}. \]

Therefore,

\[ h_w =h_c\frac{T_w}{T_c}\frac{\Delta T_c}{\Delta T_w} =(100\,\mathrm{m}) \frac{303.15}{243.15}\frac{430}{370}. \]

This gives

\[ \boxed{h_w=144.9\,\mathrm{m}\simeq145\,\mathrm{m}}. \]

The warmer surroundings provide a smaller relative density difference, requiring a taller chimney.

1(c)

Conservation of mass in steady flow gives

\[ \dot m=\rho_hAv(z). \]

Since both \(\rho_h\) and \(A\) are constant throughout the chimney, \(v(z)\) must also be constant. With the stipulated minimum height, the volume flow is exactly \(B\), so

\[ \boxed{v(z)=\frac BA =\sqrt{\frac{2gh\Delta T}{T_{Air}}}}. \]

For the requested sketch, put \(z\) on the horizontal axis and \(v\) on the vertical axis. Mark the chimney entrance at \(z=0\). Show a rapid rise from the negligible furnace velocity to \(B/A\) in the entrance region, followed by a horizontal line at \(v=B/A\) from the entrance to \(z=h\). If the entrance region is idealized as having negligible extent, its acceleration can be represented by a step at \(z=0\).

1(d)

Within the chimney, Bernoulli’s equation is

\[ \frac12\rho_hv^2+\rho_hgz+p(z)=\text{constant}. \]

Since \(v\) is constant,

\[ \frac{dp}{dz}=-\rho_hg. \]

Using the outlet boundary condition \(p(h)=p_{Air}(h)\),

\[ \boxed{p(z)=p_{Air}(h)+\rho_hg(h-z)}. \]

Equivalently,

\[ p(z)=p_0-\rho_agh+\rho_hg(h-z). \]

The exterior pressure at height \(z\) is \(p_{Air}(z)=p_0-\rho_agz\). Subtracting,

\[ \boxed{p(z)-p_{Air}(z)=-(\rho_a-\rho_h)g(h-z)}. \]

Thus the gas pressure is below the surrounding atmospheric pressure everywhere inside the chimney below its outlet, and the two pressures become equal at the outlet.

Immediately after the entrance,

\[ p(0^+) =p_0-(\rho_a-\rho_h)gh =p_0-\frac12\rho_hv^2. \]

This entrance pressure reduction accompanies the acceleration from the furnace into the chimney. Thereafter, the pressure decreases linearly with height.

2(a)

Consider a mass \(m\) of air heated by \(\Delta T\) in the collector. The absorbed heat is

\[ Q=mc\Delta T. \]

Its kinetic energy in the chimney is

\[ K=\frac12mv^2. \]

According to the definition in the question,

\[ \eta=\frac KQ=\frac{v^2}{2c\Delta T}. \]

Applying the result of 1(a),

\[ v^2=\frac{2gh\Delta T}{T_{Air}}, \]

gives

\[ \boxed{\eta=\frac{gh}{cT_{Air}}}. \]

Within the stated model, the efficiency is independent of the temperature rise, collector area, and chimney cross-section. It measures the kinetic power available relative to the absorbed solar power.

2(b)

At fixed \(T_{Air}\) and \(c\),

\[ \eta(h)=\frac{g}{cT_{Air}}h. \]

The requested diagram has \(h\) on the horizontal axis and \(\eta\) on the vertical axis. It is a straight line through \((0,0)\), with positive slope

\[ \boxed{\frac{d\eta}{dh}=\frac{g}{cT_{Air}}}. \]

Doubling the height doubles the efficiency. This linear relation applies only within the approximations of the problem; it should not be extrapolated to heights where changes of density become significant.

3(a)

Substituting the prototype data,

\[ \eta =\frac{(9.81\,\mathrm{m\,s^{-2}})(195\,\mathrm{m})} {(1012\,\mathrm{J\,kg^{-1}K^{-1}})(295\,\mathrm{K})} =0.0064077. \]

Therefore,

\[ \boxed{\eta\simeq0.64\%}. \]

3(b)

The collector radius is \(244/2=122\,\mathrm{m}\), so its area is

\[ S=\pi(122\,\mathrm{m})^2 =4.67595\times10^4\,\mathrm{m^2}. \]

All incident solar power is assumed to heat the air:

\[ P_{\mathrm{solar}}=GS =(150\,\mathrm{W\,m^{-2}}) (4.67595\times10^4\,\mathrm{m^2}) =7.01392\times10^6\,\mathrm{W}. \]

The available kinetic power is

\[ P=\eta GS =(0.0064077)(7.01392\times10^6\,\mathrm{W}), \]

hence

\[ \boxed{P=4.4943\times10^4\,\mathrm{W}\simeq45\,\mathrm{kW}}. \]

This is the ideal power available for conversion under the problem’s definition; no additional turbine or generator efficiency is specified.

3(c)

For an effective illumination time \(\tau\) at the stated irradiance,

\[ \boxed{E=P\tau=\frac{ghGS}{cT_{Air}}\tau}. \]

Thus each hour at the stated conditions supplies approximately \(44.9\,\mathrm{kWh}\).

If a typical sunny day is estimated as \(12\,\mathrm{h}\) at the stated irradiance,

\[ E=(44.943\,\mathrm{kW})(12\,\mathrm{h}) =539.3\,\mathrm{kWh}, \]

or

\[ \boxed{E\simeq0.54\,\mathrm{MWh} \simeq1.9\times10^9\,\mathrm{J}}. \]

If \(150\,\mathrm{W\,m^{-2}}\) is instead interpreted as an average over the entire \(24\,\mathrm{h}\) day,

\[ \boxed{E=(44.943\,\mathrm{kW})(24\,\mathrm{h}) \simeq1.08\,\mathrm{MWh} =3.88\times10^9\,\mathrm{J}}. \]

The statement does not specify the effective illumination time or averaging convention.

4(a)

In steady operation, the solar heating power equals the rate at which heat is carried away by the air:

\[ GS=\dot m\,c\Delta T. \]

Mass conservation gives

\[ \dot m=\rho_hAv. \]

Combining these with the chimney speed,

\[ v=\sqrt{\frac{2gh\Delta T}{T_{Air}}}, \]

we obtain

\[ GS =\rho_hAc\Delta T \sqrt{\frac{2gh\Delta T}{T_{Air}}} =\rho_hAc\sqrt{\frac{2gh}{T_{Air}}}\, (\Delta T)^{3/2}. \]

Solving for \(\Delta T\),

\[ \boxed{\Delta T =\left[ \frac{(GS)^2T_{Air}} {2gh\,\rho_h^2A^2c^2} \right]^{1/3}}. \]

For the prototype,

\[ A=\pi(5\,\mathrm{m})^2=78.5398\,\mathrm{m^2}. \]

Using the given hot-air density,

\[ \Delta T= \left[ \frac{(7.01392\times10^6)^2(295)} {2(9.81)(195)(0.9)^2(78.5398)^2(1012)^2} \right]^{1/3}\mathrm{K}. \]

Therefore,

\[ \boxed{\Delta T=9.05\,\mathrm{K}\simeq9.1\,\mathrm{K}}. \]

The air enters the chimney at approximately \(304\,\mathrm{K}\).

4(b)

From the heating balance,

\[ \dot m=\frac{GS}{c\Delta T}. \]

Numerically,

\[ \dot m =\frac{7.01392\times10^6\,\mathrm{W}} {(1012\,\mathrm{J\,kg^{-1}K^{-1}})(9.0503\,\mathrm{K})} =765.8\,\mathrm{kg\,s^{-1}}. \]

Thus

\[ \boxed{\dot m\simeq7.7\times10^2\,\mathrm{kg\,s^{-1}}}. \]

Eliminating \(\Delta T\) also gives the general expression

\[ \boxed{\dot m= \left[ \frac{2gh(\rho_hA)^2GS}{cT_{Air}} \right]^{1/3}}. \]

As a numerical check, the corresponding speed is

\[ v=\sqrt{\frac{2(9.81)(195)(9.0503)}{295}} =10.83\,\mathrm{m\,s^{-1}}, \]

which gives \(\rho_hAv=765.8\,\mathrm{kg\,s^{-1}}\) and kinetic power \(\tfrac12\dot m v^2=44.94\,\mathrm{kW}\), consistent with 3(b).

The solution was checked against the official solution.