Energy & Environment I

HW#3

Due Friday, October 19, 2001

ANSWERS IN RED.
    1. Propane: determine the heating value (kJ/kg or Btu/pound or Btu/gallon) of the fuel in the propane tank you bought for your back-yard grill. Then determine the chemical energy content (kJ or Btu) of the fuel in the tank?

      2. Information on commercial grade propane (which is not pure propane) is available at http://www.lpg.com. On page www.lpg.com/non-flash/tech/tech-a.cfm, one finds technical information on commercial grade propane. The heating value of the vapor (what you burn) is given as 91,502 BTU/gallon, and the latent heat of vaporization is given as 773 BTU/gallon. The heating value of the liquid commercial grade propane is the sum of these two numbers: 92,275 BTU/gallon. The density of the commercial grade propane is given as 4.20 pounds/gallon (at 60 degrees F). Elsewhere on this web site one learns that propane tanks for backyard grills nominally hold 20 pounds of commercial grade propane. Thus, the energy in one tank of propane (which is predominantly liquid propane) is 92,275 BTU/gallon x 20 pounds / (4.20 pounds/gallon) = 439,405 BTU.

      (For the engineers: do you think the heating value numbers given above, as well as below for coal, express the higher heating value or the lower heating value? What about for natural gas? Is the 1033 BTU per std cu ft value given on p. 3:6 by Bodansky the higher or lower heating value?)
    2. Coal: what is heating value of coal? This is normally expressed as BTU per short ton, or as kJ per tonne. Do you think this would vary a lot with the source and type of the coal (ie, the location of the coal mine)?

      3. Bodansky, on p. 9:10 tells us about coal. There are 4 major ranks of the coal, from lignite to anthracite. This is noted in Bodansky’s Table 9.1. With increasing rank, the carbon content of the coal increases, and the oxygen content decreases. The heating values given by Bodansky are as follows:

    COAL RANK

    HEATING VALUE (million BTU/short ton)

    Lignite

    £ 16.6

    Sub-bituminous

    16.6 ® 23.0

    Bituminous

    21.0 ® 28.0

    Anthracite

    ³ 26.0

    As seen from the table, the heating value varies significantly with the type of coal. It also varies with the location of the coal mine. Coals in the Eastern and Midwestern US are generally bituminous, whereas coals from the Great Plains tend to be sub-bituminous. Lignite is found in North Dakota and Texas.

    (Note: high rank, ie, low-volatile, bituminous coal has the highest heating value of all the coals. Bituminous coals are preferred over the other coals for burning in steam-electric power plants. Sub-bituminous coals from the Great Plains are preferred by some power-plants because of their relatively low sulfur content. Anthracite is difficult to burn because of its low volatility. Anthracite is important in metallurgy. Lignites are difficult to burn because of high moisture, high mineral matter content, high alkali content, and the relatively low heating value. Nonetheless, they are burned for generating electricity. The furnaces tend to be oversized and designed to withstand the high mineral matter content and corrosion.)
  1. A steam-cycle power plant and a gas turbine engine have identical efficiencies when separately operated: 33%. However, when combined, the waste heat of the gas turbine engine serves as the heat source for the steam-cycle power plant. What is the value of the efficiency of this combined cycle? Note: the only source of fuel input is the gas turbine engine. Each engine has work output. Thus the efficiency equation for the combined cycle is: h = (WGT + WST)/QH-GT.

If the energy of the fuel (QH-GT) = 100 kJ

Then the work output of the GT (WGT) = 33 kJ

And the heat rejected from the GT = 67 kJ

This is the heat added to the steam cycle = 67 kJ

Since the steam cycle is 33% efficient, its work output (WST) = 0.33 x 67 = 22 kJ

Then the heat rejected from the steam cycle = 45 kJ

This is the heat rejected from the combined cycle = 45 kJ.

Thus, the combined cycle efficiency = (33 kJ + 22 kJ)/100 kJ = 0.55 (55 %)