Energy & Environment II Answers in Red

WQ02

Final Exam

100 points

Closed book and notes

10-page personal “crib sheet” permitted. This must be submitted with exam.

QUESTION #1 (10 points)

Once in a while, on a day during the period from about mid-June to early-July, the global solar flux (G_th) received in Seattle can reach as much as 950 watts/m².

A. Estimate the amount of diffuse solar flux (G_dh) contained within the 950 watts/m² of global solar flux.

This is a very sunny condition for Seattle. The diffuse percentage of the global solar flux should be small – about 15-20%.

Taking the mean of 17.5% gives G_dh @ 950 x 0.175 = 165 w/m².

B. Estimate the amount of beam solar flux (G_b^*) associated with the 950 watt/m² of global solar flux. Cleary show your calculation steps and state any assumptions used.

The amount of beam solar flux received by the horizontal surface on the ground is the difference between the global and diffuse solar fluxes. Thus, G_b^*cosq_z @ 950 – 165 @ 785.

The time period in question encompasses summer solstice. For solar noon on the longest day of the year in Seattle, q_z = 47.5 – 23.5 = 24 degrees. Thus, G_b^* @ 785/cos24 = 860 w/m²

C. Estimate the transmissivity of the atmosphere (t) for this situation. Cleary show your calculation steps and state any assumptions used.

For the time of year of this problem, the extraterrestrial beam solar flux is about equal to the solar constant divided by 1.035, that is: 1367/1.035 @ 1320 w/m². Thus, the transmissivity is: t @ 860/1320 = 0.65.

QUESTION #2 (16 points)

You wish to build a house in western Washington that is energy efficient and uses solar energy as intelligently as possible. Your building site is not shaded by trees and hills. State the important features of your house. Clearly state the general benefit of each feature. Then clearly state the particular relevancy (or benefit) of each feature of your house in context to the climate of western Washington. That is, state why each feature of your house would be particularly appropriate in western Washington.

There are several “correct” answers to this question. Four of my “favorites” are given below.

A. Feature #1:

Well insulated walls, ceiling, windows, and foundation, which increases the overall “R-value” of the house.

General benefit of feature #1:

These features reduce the heat loss of the house. Thus, the house requires less purchased energy to maintain reasonable warmth during the heating season.

Why is feature #1 especially appropriate in western Washington?

Although western Washington winters lack significant snow and sub-freezing temperatures (except in March 2002), most houses require heating from about late September through as late as early June. Thus, good insulation and a high overall “R-value” for the house are appropriate and beneficial.

Walls and ceilings with high R values should be used. Windows with high R-values should be used. Windows with less R-value and thus more transmissivity might not be so beneficial, since there is little sun beam radiation to harvest in the winter.

B. Feature #2:

Daylighting

General benefit of feature #2:

Daylighting, that is natural lighting through windows, skylights, and solar tubes, reduces or eliminates the amount of electricity needed for operation of lamps during the day.

Why is feature #2 especially appropriate in western Washington?

For much of the year, from October to as late as May, Seattle is a cloudy and gray place. Thus, many houses require artificial lighting even during the day. Letting in the light from outdoors can significantly reduce the need for artificial lighting – this is true even on cloudy and overcast days – in fact diffuse daylight can be preferable to beam sunlight, since the latter leads to glare on computer screens and other surfaces.

For the daylighting one does not need to use large south facing windows – though these could be used. Of course, there would be passive solar heating from such windows, but the benefit would not be as great as in places with sunny, cold winters, such as the Rocky Mountain and High Plains states.

C. Feature #3:

Summer shading

General benefit of feature #3:

This blocks direct sunlight from entering the house in the summer, thereby helping to maintain a pleasantly cool interior on hot days.

Why is feature #3 especially appropriate in western Washington?

Although Seattle winters are gloomy, summers, especially from about mid-July to mid-September, are bright, with very good sunlight. A house without shading might become uncomfortably warm (or hot) on sunny summer days. Therefore, overhangs to block the summer sun rays from entering south-facing windows are recommended. Broadleaf shade trees planted on the south side of the house may also be useful for shading.

D. Feature #4:

Natural ventilation and cooling

General benefit of feature #4:

In conjunction with feature #3, this feature maintains a comfortable interior in the summer, without the need for mechanical/electrical air conditioning.

Why is feature #4 especially appropriate in western Washington?

As with feature #3, this feature is appropriate because of the sunny summers. Allowing cool air to flow through the house, and using the natural “coolth” of the ground, are cost effective, passive cooling (or nearly passive) methods for maintaining reasonable indoor temperatures.

QUESTION #3 (14 points)

In HW#10, you determined the size of a forest, sustainably harvested, that would provide fuel for a 50 MW wood-fired steam-electric power plant. The answers found fell in the range of 100 to 400 km².

Instead of this approach, suppose you locate 100 km² of cleared land or scrub land in the same state, and you decide to seek financing for a solar PV electricity generating facility on the land.

A. Estimate the rated electrical power (as MW) of the solar PV facility. Clearly state your assumptions, including the efficiency assumed for the solar PV panels. Clearly show your calculation steps.

100 km² = 10,000 meters x 10,000 meters = 10⁸ m². Not all of this area can be fitted with solar PV panels. Space is needed for walking and driving through the facility, so that maintenance and inspection can be conducted, and with the panels tilted, there must be space between them to prevent shading. I assume a 40% utilization of the area. I assume high efficiency panels, that is, 15% efficiency panels. Thus, the panels are rated at 150 watts per sq meter (ie, 1000 watts/m² x 0.15). Thus, the rated electrical power of the facility is: 150 w/m² x 10⁸ m² x 0.4 = 6,000,000,000 watts = 6 GW.

(Note this is 20 times the current annual production of solar PV panels in the world!)

B. Estimate the average electrical power (as MW, average for the year) of the solar PV facility. Clearly state your assumptions, including the capacity factor assumed for the system. Clearly show your calculation steps.

Assuming the facility is placed in a reasonably sunny location, a 20% annual capacity factor is assumed. Thus, the average electrical power output will be 6 GW x 0.2 = 1.2 GW. (This is about the rated output of a large coal or nuclear steam-electric generating station.)

C. Estimate the amount of electrical energy (as megawatt-hours) produced per year by the facility.

Solar PV does not require much downtime for maintenance. Let’s assume continuous operation (8760 hrs per year). Then the electrical energy production is: 1200 MW x 8760 hrs/yr = 10,500,000 MWh/yr.

With a bit of down time thrown in, let’s say 10,000,000 MWh/yr.

D. Estimate the capital cost of the facility. Clearly state your assumptions.

The retail cost of high efficiency solar PV panels is $5-6/watt-rated. The support structures will add some cost, and the land and its improvements will need to be paid. Additionally, power electronics will be required to condition and manage the electricity. At some point the dc output of the panels will need to be converted to ac, though this might be done after transmission. Total cost of the facility could be about $10/watt-rated. However, with the huge volume of PV panels (and balance of system components) used, and with the possibility of subsidies and price reductions in the cost of solar PV, it should be possible to get the cost under $5/watt-rated. Let’s go for $2-3/watt-rated (though this is a best guess on my part). This gives a capital cost of 6,000,000,000 watts rated x $2-3/watt-rated = $12-18 billion. Rounding off, let’s say $10-$20 billion.

E. Estimate the simple cost of the electricity produced by the facility (as cents per kwh).

Assuming a 30 years lifetime for the facility, the simple cost of the electrical energy is:

[$10-$20 x 10⁹ x 100 ¢/$] / [10 x 10⁶ MWh/hr x 1000 kwh/MWh x 30 yr] = 3.3-6.7 ¢/kwh.

QUESTION #4 (10 points)

What could the UW do to use renewable energy, aside from the hydroelectricity it purchases from Seattle City Light? What embodiments of renewable energy would make good sense and would make a real difference for integrating renewable energy into the UW campus? In preparing your answer, keep in mind that for much of the school year clouds hang over the UW campus, and that the UW is a financially limited institution.

There are several reasonable responses to this question – however, we have to keep in mind the clouds and lack of money. Thus, we need to think of approaches that could pay for themselves or could receive support from the federal government or from foundations. Some possibilities are:

Biofuels for university vehicles might make sense, with the biofuels produced from local (and campus) garbage and food service wastes. Biodiesel might make sense for the UW’s diesel vehicles.

The UW is using increasing amounts of air conditioning. Solar driven air conditioning systems might make sense, though the economics could be less than favorable. It is possible subsidies for the solar equipment would be required.

Green buildings might be the best opportunity – that is, buildings designed for daylighting, for passive solar heating (though because of Seattle’s climate we would want to carefully assess this), and for passive (or near passive) ventilation.

The UW could purchase green power, such as electricity from wind and biomass, though since green power sells at a premium, we might need to look into special financing for this.

QUESTION #5 (20 points)

The tip speed ratio of a wind turbine is defined as:

l = tangential velocity of tip / ambient wind speed

Multi-bladed wind turbines (as used for pumping water on farms) have their best efficiency (of about 30%) when l @ 1.

Two-bladed horizontal axis wind turbines (as used for generating electricity) have their best efficiency (of about 45%) when l @ 6.

The definition of efficiency for the wind turbines is:

Efficiency = power of turbine / power of wind based on turbine area

A. Explain why there is a significant difference in the l values for best efficiency. That is, why does the multi-bladed farm turbine perform very poorly at l @ 6, and why does the two-bladed HAWT perform very poorly at l @ 1?

If l is too low, the turbine is not rotating fast enough to use the air flowing though the disk area of the turbine. On the other hand, if l is too large, the following blade encounters “bad” air. That is, it encounters the wake of the preceding blade.

B. Assume an ambient wind speed of U_o = 10 m/s and a turbine rotational speed of N = 45 revolutions per minute (rpm). Complete the table on the next page. Clearly show all calculations.

VARIABLE	MULTI-BLADED FARM TURBINE	TWO-BLADED HORIZONTAL AXIS TURBINE
Uo	10 m/s	10 m/s
N	45 rpm = 0.75 rps	45 rpm = 0.75 rps
l	1	6
Efficiency	30%	45%
Tangential velocity of the tip of blade (m/s)	VT-tip = l Uo = 1 x 10 m/s = 10 m/s	VT-tip = l Uo = 6 x 10 m/s = 60 m/s
Radius of turbine (m)	VT-tip = w R =2pRN Thus: R = VT-tip/2pN R = 10/2p0.75 = 2.12 m	VT-tip = w R =2pRN Thus: R = VT-tip/2pN R = 60/2p0.75 = 12.73 m
Power coefficient, CP	CP = efficiency = 0.30	CP = efficiency = 0.45
Torque coefficient, CT	0.30	CT = CP/l = 0.45/6 = 0.075
Turbine power (kw)	P = ½ rUo^3ACP A = pR^2 = p 2.12^2 A = 14.12 m^2 P = ½(1.2)10^3(14.12)0.3 P = 2542 watts = 2.54 kw	138 kw
Angle f at tip (degrees): tanf = U1/VT	34 degrees	6 degrees
Estimate of angle of attack at tip (degrees)	In our HW problem, we found the angle of attack to be close to the stall angle, about 16 degrees	In our HW problem, we found the angle of attack to be about 8 degrees
Blade set angle at tip (degrees)	g = f - a g = 34 - 16 @ 20 degrees	g = f - a g = 6 - 8 @ 0 degrees

QUESTION #6 (20 points)

Chapter 5 of the text gives us a roadmap to use in selecting the most appropriate turbine for a hydroelectric application. This roadmap is a plot of total head versus volume flow rate. It shows the regimes for the Pelton wheel, the Francis turbine, and the propeller-type turbine.

Suppose Q = 10 m³/s. Then the roadmap tells us the following:

· If the total head is greater than about 150 m, use a Pelton wheel.

· If the total head is less than about 150 m and greater than about 5 m, use a Francis turbine.

· If the total head is less than about 5 m, use a propeller-type turbine.

The text also tells us about specific speed, N_s, which is defined as:

N_s = n P^1/2 / H^5/4 @ 500 h^1/2 (r/R) (v_B /v_w)

where n = rotational speed of turbine as revolutions per minute

P = turbine power as kw

H = available (or effective) head as meters

h = efficiency of turbine

r = radius of water stream (of jet) entering turbine

R = radius of turbine

v_B = velocity of turbine blade tip

v_w = velocity of water steam (or jet) entering turbine

Complete the table on this and the following page. Clearly show your calculations. Clearly state any assumptions.

VARIABLE	PELTON WHEEL	PROPELLER TYPE
Total head	150 m	5 m
Q	10 m^3/s	10 m^3/s
H (m)	Account for friction: H @ 0.9 x 150 = 135 m	Account for friction: H @ 0.9 x 5 = 4.5 m
vw (m/s)	vw = (2gH)^1/2 vw = (2 x 9.81 x 135)^1/2 vw = 51.5 m/s	vw = (2gH)^1/2 vw = (2 x 9.81 x 4.5)^1/2 vw = 9.4 m/s
VARIABLE	PELTON WHEEL	PROPELLER TYPE
vB (m/s)	From the theory learned regarding Pelton wheels, for best conversion of the jet KE to the mechanical energy of the turbine vB = ½ vw = 51.5/2 =25.8 m/s	From the discussion in Box 5.8 of the text for a small propeller-type turbine (a Kaplan turbine), vB = 2 vw = 2 x 9.4 = 18.8 m/s
r (m)	Q = pr^2vw Thus, r = (Q/pvw)^1/2 r = (10/p 51.5)^1/2 r = 0.25 m	Q = pr^2vw Thus, r = (Q/pvw)^1/2 r = (10/p 9.4)^1/2 r = 0.58 m
R (m)	For Pelton wheels, the ratio of the jet radius to turbine radius is about 1/10. Thus, R @ 2.5 m	For Kaplan turbines. By Box 5.8, the ratio of the jet radius to turbine radius is about 1. Thus, R @ 0.58 m
P (kw)	P = h r g Q H P @ 0.8(1000)9.81(10)135 = 10,600,000 watts = 10,600 kw	P = h r g Q H P @ 0.8(1000)9.81(10)4.5 = 350,000 watts = 350 kw
Ns	24	950
n (rpm)	105 rpm	330 rpm

QUESTION #7 (10 points)

Consider biomass having a chemical formula of C₁₂H₂₂O₁₁. Write the complete chemical equations for conversion of this biomass by the following processes.

A. Direct Combustion with the chemically correct (ie, stoichiometric) amount of air.

C₁₂H₂₂O₁₁ + (12+22/4-11/2)O₂ + 3.773(12+22/4-11/2)N₂ Þ

12CO₂ + 11H₂O +3.773(12+22/4-11/2)N₂

C₁₂H₂₂O₁₁ + 12O₂ + 45.3N₂ Þ12CO₂ + 11H₂O +45.3N₂

B. Alcoholic Fermentation.

C₁₂H₂₂O₁₁ + H₂O Þ 4C₂H₅OH + 4CO₂

C. Anaerobic Digestion.

C₁₂H₂₂O₁₁ + H₂O Þ 6CH₄ + 6CO₂