A battle has erupted over a local power plant owner\'s decision to raise the pow

Question

A battle has erupted over a local power plant owner's decision to raise the power delivered by the plant from 198 MW to 564 MW (The delivered power accounts for energy losses). The local population is increasing, so the area needs more power. However, the plant expends its heated waste water astir into the Frustration River; local citizens are concerned that the heated water will raise water temperatures to the point where the endangered river smelt will die from oxygen depletion. Local biologists called into the fray report that the smelt can only withstand an average increase of 3.44 degree C. The river is 222 ft wide at the point where the waste water is dumped, but the depth and cross section is not well known. Farther downstream, however, the water flows through a narrow man-made concrete channel of rectangular cross-section, which has a width of 88.7 ft and depth of 47.2 ft. The water flows through this channel at 10.4 mph. The power plant claims a planned efficiency of 32.50 percentage for its new power plant design. Based on this information, what is the maximum power that the power plant can output without endangering the river smelt? Based on your results, which advice would you offer city planners about the proposed upgrade? The power plant should be able to deliver 564 MW of power without endangering the smelt. The smelt population will be endangered if the power plant increases its delivered power to 564 astir Although termed "waste water," the water in this case is not sewage; rather, the water is no longer usable to create power.

Explanation / Answer

A battle has erupted over a local power plant owner's decision to raise the power delivered by the plant from 171 MW to 488 MW (The delivered power accounts for energy losses). The local population is increasing, so the area needs more power. However, the plant expends its heated waste water* into the Frustration River; local citizens are concerned that the heated water will raise water temperatures to the point where the endangered river smelt will die from oxygen depletion. Local biologists called into the fray report that the smelt can only withstand an average increase of 3.44C.  The river is 192 ft wide at the point where the waste water is dumped, but the depth and cross section is not well known. Farther downstream, however, the water flows through a narrow man-made concrete channel of rectangular cross-section, which has a width of 76.7 ft and depth of 47.2 ft. The water flows through this channel at 10.4 mph. The power plant claims a planned efficiency of 32.50% for its new power plant design. Based on this information,What is the maximum power that the power plant can output without endangering the river smelt?

88.7 ft = 27.036 m
47.2 ft = 14.387 m
velocity = 10.4 mph = 4.649 m/sec

Volumetric flow rate = 27.036*14.387*4.649 = 1808.307 m^3 / sec

Mass flow rate = volumetric flow rate * 10^3 =1808.307 * 10^3 kg / sec

Temperature change = 3.44 C, therefore heat rate required =1808.307 * 10^3 * 4186 * 3.44 = 26039331.47 Joule / sec = 26039.331 MW

On the other hand, efficientcy = 32.5 %

so.. total power * ( 1 - 32.5/100) = 26039.331 MW

so.. total power = 38576.79 MW
and power transmitteed =38576.79 -26039.331 = 12537.46 MW

Maximum power transmitted =12537.46 MW for temperature to increase 3.44 C

the power can be increased to 564 MW

the maximum power that can be delivered without harming smelt = 198 +12537.46 MW = 12735.46 MW

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