Chapter 7, The solar chimney and tower Video Solutions, Solar Energy: An Introduction

Problem 1

The draft in a chimney is important to ensure smoke goes up the chimney rather than back into the building. Draft is measured in terms of the pressure drop into a fireplace $\left(P_1-P_2 \equiv \Delta P\right.$, see Fig. 7.1) and a draft of $20 \mathrm{~Pa}$ is considered good for a fireplace in a household residence. The equation for draft is written $$\Delta P=C P_1 H\left[\frac{1}{T_1}-\frac{1}{T_3}\right]$$ where $C$ is a group of variables to make a constant factor. Determine $C$ and calculate $T_3$ if the chimney is $6 \mathrm{~m}$ high and $\Delta P$ is $20 \mathrm{~Pa}$.

James Kiss

Numerade Educator

07:12

Problem 2

It is proposed to make a solar chimney that has a rock wall, rather than a transparent wall as in Fig. 7.2 , which absorbs the solar radiation during the day to release it only at night when required. This is accomplished by closing a damper during the day and when the outside air temperature becomes low enough, at night, it is opened. Assume the rock heats to $40^{\circ} \mathrm{C}$ while the outside air temperature is $20^{\circ} \mathrm{C}$ at night. The rock gradually (linearly) cools to the surroundings temperature over $5 \mathrm{~h}$. Determine the mass flow rate of air over the $5 \mathrm{~h}$ period, in $\mathrm{kg} / \mathrm{s}$, and the $\mathrm{ACH}$ as a function of time. Assume the house has a volume of $400 \mathrm{~m}^3$ and the rock wall is $10 \mathrm{~m}$ high by $2 \mathrm{~m}$ wide by $1 \mathrm{~m}$ thick. Also assume the flow area for the air through the chimney to be $1.5 \mathrm{~m}^2$ and the exit air temperature, the same as the rock. Rock has a heat capacity of $1 \mathrm{~kJ} / \mathrm{kg}-\mathrm{K}$ and a density of $2500 \mathrm{~kg} / \mathrm{m}^3$.

Nick Auwerda

Numerade Educator

07:12

Problem 3

A solar chimney can be used during daylight hours to draw hot air underground, where it is cooled, and then circulated through the house, as shown in the figure below. Using the assumptions given below, determine the temperature of the house, $T_2$. Assume the cool air enters the house at $13{ }^{\circ} \mathrm{C}$ call this $T_1$, enters the chimney at temperature $T_2$ and exits the chimney at temperature $T_3$. Also, assume the house air temperature $T_2$ is the same throughout the entire house (this is sometimes called the well-mixed assumption). Use the First Law of Thermodynamics to determine $T_2$ (see eqn (3.12)) $$\dot{m} C_p\left[T_2-T_1\right]=\dot{Q}_{i n}$$ assuming the house is the system, the solar chimney is thermally isolated from the house and there are no heat losses from the house. Here $\dot{Q}_{i n}$ is the rate of solar energy input into the house. To find this, assume the insolation is $500 \mathrm{~W} / \mathrm{m}^2$ and the (effective) area of the house that the insolation can be transmitted through is $30 \mathrm{~m}^2$ (multiplying these two together yields $\dot{Q}_{i n}$ ). Of course $\dot{m}$ is the mass flow rate of air through the house and $C_p$ is the heat capacity of air. The solar chimney is $4 \mathrm{~m}$ high (this is short since the air exits from the second floor after travelling through first floor), the device area $A_D$ to gather the insolation is $13 \mathrm{~m}^2$ and the chimney exit (flow) area is $2 \mathrm{~m}^2$. Finding the temperature $T_2$ will require you to use the design and operating lines for the solar chimney together with the FLOT discussed above.

Sri Datta Vikas Buchemmavari

Numerade Educator

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Problem 4

Determine if Leonardo da Vinci was right and a spit could be turned by rising air within a chimney, as shown in Fig. 7.4. Assume the temperature rise of the air is $200{ }^{\circ} \mathrm{C}$, the diameter and height of the cylindrical chimney are $1 \mathrm{~m}$ and $10 \mathrm{~m}$, respectively, and the ambient air temperature is $300 \mathrm{~K}$. To do this one must realize that the rate of kinetic energy imparted by flowing air to the fan (turbine) is $\dot{K E}=\frac{1}{2} \dot{m} v^2$, where $\dot{m}$ is the air mass flow rate and $v$ the air velocity. See the discussion which follows Example 7.3. Assume the fan or turbine sweeps a circular area almost equal to the chimney area. The minimum power to turn the spit is $0.25 \mathrm{hp}$.

Victor Salazar

Numerade Educator

02:11

Problem 5

Repeat Exercise 7.4 for a solar chimney. In this case assume the chimney is also circular in crosssection, has the same dimensions as above and the half facing the Sun is transparent and the other half is the absorber which absorbs all the radiation. Use eqn (7.20) to find the rate of kinetic energy generation assuming $P_D(\beta)=500 \mathrm{~W} / \mathrm{m}^2$. Comment on the amount of power generated in light of the size and scale of proposed solar tower facilities.

Dominador Tan

Numerade Educator

03:20

Problem 6

Pretorius and Kröger (Trans. ASME, 128 (2006) 302) have given the following relation for the heat transfer coefficient from the cover of a solar tower, which is more complicated than eqn (7.27) and is written as $h\left(\mathrm{~W} / \mathrm{m}^2-\mathrm{K}\right)=\frac{0.2106+0.0026 v_w\left[\frac{\rho_{\mathrm{air}} T_m}{\mu_{\mathrm{air}} g \Delta T}\right]^{1 / 3}}{\left[\frac{\mu_{\text {air }} T_m}{g \Delta T C_{P, \text { air }} k_{\text {air }}^2 \rho_{\text {air }}^2}\right]^{1 / 3}}$ where $T_m$ is the mean temperature of the cover and air and all other variables are defined in the chapter. The properties of air are evaluated at $T_m$. All variables are in MKS units and the temperature is in Kelvins. Compare this relation to eqn (7.27) for various wind velocities (from $0 \rightarrow 4.5 \mathrm{~m} / \mathrm{s}$ ) using a range of temperature differences (from $5 \rightarrow 45 \mathrm{~K}$ ) and assume the outside air temperature is $300 \mathrm{~K}$. Comment on your results.

Nick Auwerda

Numerade Educator

07:36

Problem 7

Assume water is the working fluid for a solar tower. Use eqn (7.35) to determine what the maximum power can be for a water tower compared to one where air is the working fluid.

Prabhat Tyagi

Numerade Educator

08:31

Problem 8

Repeat Example 7.3 where the working fluid is water and not air.

Saman Zulfiqar

Numerade Educator

04:57

Problem 9

Repeat Example 7.4 where the working fluid is water and not air.

Himanshu Kushwaha

Numerade Educator

01:29

Problem 10

Determine the power generated in the Manzanares, Spain solar tower when the wind is blowing and do not include the effect of insolation. In other words, assume values of the wind velocity from 0 to $50 \mathrm{~m} / \mathrm{s}$ and determine how much wind would contribute to the power output from the solar tower.

Chai Santi

Numerade Educator

01:43

Problem 11

It has been proposed to make a solar tower on the side of a mountain (S.V. Panse et al. Energy Conv. Man. 52 (2011) 3096), which is an interesting concept since the cover and tower are integrated into one. The cover is placed on the side of the mountain and the air channeled up to the top where turbines are located. Assume the mountain is $500 \mathrm{~m}$ high and has a slope of $25^{\circ}$ from the horizontal with a circular base, the cover encircles one-quarter of the base and follows the conical contour to the apex (i.e. the cover, 'covers' one-quarter of the conical, mountain surface area) and is $2 \mathrm{~m}$ above the mountain surface. The useful insolation is $500 \mathrm{~W} / \mathrm{m}^2$, this is the rate of energy transfer that is absorbed by the air, and the inlet air temperature is $300 \mathrm{~K}$ while the outlet area at the top of the mountain is $100 \mathrm{~m}^2$; again the one-quarter of the mountain circumference will be covered by the cover at the top of the cover. Determine the mass flow rate, temperature rise (i.e. the operating point) and maximum work that could be generated in the inclined solar tower. Assume the actual work output is $20 \%$ of the maximum to report the amount of work that could be generated; is this a lot? Since the cover area is so large, heat transfer must be considered, let the wind velocity be $4 \mathrm{~m} / \mathrm{s}$. Also, you must consider the assumptions that go into the derivation of eqn (7.8). The temperature just inside the cover will not be vastly different to just outside the cover making it difficult to estimate the draft into the chimney. In addition, it was assumed the air had a constant temperature within the chimney which will certainly not be true. So, assume the area in eqn (7.8) is $\sqrt{A_2 A_3}$ while the air density in the chimney is $\sqrt{\rho_2 \rho_3}$, making the equation $\dot{m}=\sqrt{\rho_2 \rho_3} \sqrt{2 \frac{\Delta T}{T_1} g H} \sqrt{A_2 A_3}$ You may also have to make other engineering assumptions!

Dominador Tan

Numerade Educator

03:58

Problem 12

Consider Exercise 7.11 from an optical point of view. Assume the solar zenith angle $\theta_Z$ is $25^{\circ}$; find the projected area of the cover at solar noon to estimate the actual area that will be exposed to the Sun. Of course, the curved surface will reflect light differently than a flat surface, however, the point of this exercise is to ascertain how much of the covered area is useful in terms of heating air. Assume the mountain is in the northern hemisphere and the conical segment faces to the south.

Sarah Lewites

Numerade Educator

04:34

Problem 13

Air conditioning used to cool buildings in the USA accounts for $5 \%$ of electricity use. Suppose you live in the desert and want to power your air conditioning unit during the day with a solar tower and require $3500 \mathrm{~W}$ to do so. Assume $P_D(0)=800$ $\mathrm{W} / \mathrm{m}^2$ and $T_1=40{ }^{\circ} \mathrm{C}$; design the solar tower assuming local building codes only allow a $50 \mathrm{~m}$ high chimney.

Vipender Yadav

Numerade Educator

05:28

Problem 14

It has been proposed by the balloon manufacturer and adventurer, Per Lindstrand, to make an inflatable tower that is held up by balloons. The design calls for a $1 \mathrm{~km}$ high tower with a $7 \mathrm{~km}$ radius collector, which would create $130 \mathrm{MW}$ of power and over a year would produce $281 \mathrm{GWh}$ of electricity. Consider these numbers and indicate if they are correct. Incidentally, it is stated that the inflatable tower would only cost $$\text { \$US }$$ 20 -million while a conventional tower of similar dimensions would cost $$\text { \$US }$$ 750-million, a considerable saving.

Keshav Singh

Numerade Educator

02:53

Problem 15

Explain why you would use water rather than any other common material to absorb insolation under the cover of a solar tower.

Jonathan Everett

Numerade Educator