HOMEWORK 5 part B

Answers to Chapter 5, problems 1 and 7 are given in Homework 4 (as originally assigned).

Chapter 6

1. There are three methods of heat flow through a storm window: radiation directly through the window, conduction throught the glass, air and frame, and convection of air along the inside and outside surfaces (combined with direct heat conduction through the central part of the window). A sketch should show all of these methods.

5. The fundamental features of passive solar homes are good insulation, solar collection via south facing windows and thermal storage to even out day and night temperatures.

The three fundamental types of passive solar home design differ in how the solar energy is collected and stored. Their common feature is that no fans or pumps are required to move air or heat around, thus they require no electricity or fuel during normal operation. Direct gain homes allow sun from south facing windows to fall on massive rock or concrete floors for heat storage. Indirect gain homes allow sun to fall on a heat storage wall called a "Trombe" wall. Indirect gain can also be achieved via an attached greenhouse with its own heat storage.

6. See separate sketch for a solar "A-frame" design. For two stories, it is hard to avoid a fan to move heat around as much warm air will collect near the ceiling. I opted for a separate underground heat storage vault which would be filled with dry rock (a few tons).

Ventilation ducts and a fan would move air from the ceiling to the storage vault, and move cool air back into the ground floor if required. Note: the vault can be vented to the outside, so the occupant has great control over the flow of warm or cool air with this design.

Problem 2. p 194

5000 watts x 30 hours/month = 150 kwh x $0.09 / kwh = $13.50 saved a month by using the "experimental" clothes drier (a clothes line). Clothes smell much better when dried outside (providing it doesn?t rain for a couple of days!).

Problem 5. p 195

A flat plate solar energy collector has 40% efficiency, which means only 40% of the energy received is converted into useful energy. The "insolation" is 800 Btu/sq. ft/day, which is the total energy the sun supplies in one day.

If the rate at which the collector is to supply energy is 30,000 Btu/h, the total per day is 30,000 Btu/h x 24h/day = 720,000 Btu/day. Of course, the sun doesn't shine all day but that is taken into account in the definition of "insolation".

At 40% efficiency, the collector would deliver only 800 x 0.4 = 320 Btu/sq foot/day so it would have to be 720,000 Btu/320 Btu per sq ft = 2,250 square feet!

A square collector of that size is about 47 feet on a side, which is enormous.
 

Problem 6.

We are given that 10 cubic feet of water will store 620 Btu / degree F. We want to know how many cubic feet of rock will store 620 Btu/degree F.

The specific heat of rock is 0.2 Btu/pound (for water, it is 1.0 Btu/pound!). To store 620 Btu, we need 620/0.2 = 3,100 pounds of rock.

How many cubic feet is this? The density is 170 lb/cubic foot, so the number of cubic feet is 3,100 pounds/170 pounds per cubic foot = 18.2 cubic feet (compared to 10 cubic feet for water). So, even though rock stores much less heat per pound than water does (has much lower specific heat), it has more mass per volume, so the total volume is not that much different.

Another way of working this problem is to find that if rock stores 0.2 Btu/pound, it stores 0.2 Btu/pound * 170 pound /cu ft = 34 Btu/cu ft. Then we need 620 Btu/34 Btu/cu ft = 18.2 cu ft of rock.

In fact, a bin of rock actually has a significant advantages over a tank of water for heat storage! Rock does not move in the bin, so convection is not a problem. It does not evaporate, so humidity is not a problem. It can't leak out of bins and does not support growth of plants, insects or bacteria.