Energy & Environment I NAME Answers

HW#5

Due Friday, November 2, 2001

Answers are in red

  1. Summarize the efficiencies of the important heat engines.

    2. We have studied several important heat engines:

    1. Rankine steam cycle, as used in fossil-fueled (mainly coal) and nuclear power plants. The average efficiency is about 33%. The range for fossil-fueled power plants is about 25 to 45% (45% is extremely good). The range for nuclear power plants is narrow, about 30 to 35%. The Rankine-steam cycle rejects heat at low T (which is good), but it doesn’t add heat at an especially high T. Max for fossil-fuel systems is 600 deg C, while for nuclear systems it is about 300 deg C.
    2. Gas turbine engine, as used for jet propulsion for aircraft, for ship propulsion, and for land-based electricity generation and for mechanical applications (especially driving pipeline compressors). For the non-jet applications, efficiency is about 25 to 40%. Average is in the range of 35%. This cycle adds heat at a high T (which is good), but suffers from a high T of heat rejection.
    3. Combined cycle. Combined cycles can be constructed in several ways. However, the combined cycle of great importance today is the gas turbine-Rankine combined cycle (ie, the combined cycle combustion turbine). Over half of new electrical generation stations use this cycle. Efficiency is 50 to 60%. This cycle takes advantage of the high T of heat added to the gas turbine cycle, and the low T of heat rejected from the steam cycle.
    4. Gasoline engine. The efficiency of this for US driving conditions is about 20%. It suffers from a high T of heat rejection and other factors, such as high mechanical friction and significant pump work (because of the air throttle restriction).
    5. The diesel engine is somewhat more efficient than the gasoline engine, because of its higher compression ratio, fuel-lean running, and low pump work.
  2. Write the equation for the COP of a heat pump (in heating mode).

    3. COPheating = heat added to warm space/ work of compression

    COPheating = QH/ W = (QC + W)/ W = 1 + QC/ W

    The COP of a heat pump (in heating mode) is 2.0. What is the efficiency of this heat pump when it is switched to air conditioning mode?

    If COPheating = 2, then the COPcooling = QC/W must be about 1.

  3. Give the percentages of residential energy used for:

    4. Table 20.1 of Bodansky is very useful here.

    Heating and air conditioning: 24%

    Water heating: 15%

    Lighting: 12%

    Refrigerators: 21%

    Cooking: 6%

    Other appliances: 22%

    These percentages are valid for 1986. It would be good to look up the new values to see if there has been any shift.
  4. Define lumen:

The lumen is a measure of the ability of a light source to evoke the sensation of brightness (of luminousity) as observed by the human eye. The eye is most sensitive to light at the wavelength of 0.555 micro-meters. The range of the sensitive of the eye to electromagnetic radiation is 0.38 to 0.76 micro-meters. A monochromatic light source (of 0.555 micro-meters) of 1 watt radiant power is defined to give 683 lumens of light. Real light sources emit much of their radiation at wavelengths invisible to the human eye – mostly they emit in the long-wavelength or infrared part of the electromagnetic spectrum. Thus, the visible light emitted from these devices is small compared to 683 lumens per watt.

For incandescent lighting, what percentage of the input electrical energy is converted to the energy of the visible light?

According to Bodansky: A typical incandescent lamp gives about 16 lumens of visible light for every watt of total radiant flux (power). (The total radiant power is essentially the electrical power input to the lamp.) This gives an efficiency of 16 x 100/683 = 2.3%. This corresponds to a lamp with a tungsten filament operating at a temperature of 2845 K.

Hinrichs and Kleinbach, in Table 3.1, give an efficiency of about double this: 5%.

A three-way light bulb recently purchased lists the following data:

50 watts operation gives 640 lumens.

100 watts gives 1580 lumens.

150 watts gives 2220 lumens.

Calculation of the efficiencies indicates:

640 lumens x 100/ (683 lumens/watt x 50 watts) = 1.9%

1580 x 100/ (683 x 100) = 2.3%

2220 x 100/ (683 x 150) = 2.2%.

For comparison, a compact fluorescent bulb recently purchased had on its package: 1100 lumens for 15 watts.

The efficiency is:

1100 x 100/ (683 x 15) = 11%.