The real question in real world!

by xyz
Tags: real, world
Apr16-03, 02:12 PM
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I do need your help...I'm doing my test during this spring break.

Many current electrical power plants (especially in the eastern U.S.)use coal as the fuel to heat water at 25 degrees Celsius into steam at 190 degrees Celsius,which then expands and drives a turbine generator. The "use" steam is condensed by passing it through coils in contact with river water at 20 degrees Celsius. Burning coal produces 7800 calories of heat per gram. Assume the plant efficiency is typical of older coal plants, about 15%.1calorie is 4.186 joules..

question c is "One of the many problems with this kind of power plant is its Thermal Pollution, as the waste heat warms the river. What is the minimum rate that river water must flow past the coils so that its temperature does not exceed 25 degrees Celsius?"
I guess it ask the minimum mass of river water must flow past the coils from 20 degrees Celsius to 25 degree Celsius.
...Thanks for any replies...
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FZ+ is offline
Apr16-03, 07:27 PM
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Hmmm..... I don't think the question is complete. You need to know details as to the power output of the plant ie. the amount of heat energy dissipated in water per second or whatever. The efficiency percentage helps, but we don't know the energy input to begin with, as well as how much waste energy ends up in the cooling coils.

What were the other parts of the question?
Apr17-03, 12:04 AM
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hummm...u are professional. Yes..Sorry,I forgot.
There are two questions, the frist question is "how many metric tons (1 metric ton=1000kg) of coal are used each day to produce 150MW of power to a city? This question i have figured out. The key is we know W=150MW*24*60*60s, Q_h=W/15%=86400000MJ,each day.the mass of coal each day is 86400000MJ/32650.8J/g=2650ton. The second question is above.

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Apr17-03, 06:47 PM
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The real question in real world!

Ok... so you know the work output of the plant.(150MW) Let's assume all of the waste heat goes into the water.

From the information, you know that 150MW is only 15% of the total power output. So, 85% is waste. The total power is 150/0.15 = 1000MW, and hence we have a heat dissipation rate of 850MW, or 850MJ per second.

Use Q = mass * specific heat capacity * 5 (the change in temp of the water), substituting in 4.12 Jkg^-1K^-1 as the specific heat capacity of water and 850000000 J as Q. Now, you calculate the mass of water per second from this, and hence work out the rate of water flow, in kg per second. Any problems?
Apr17-03, 07:08 PM
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I calculate that there are at least 40612kg river water per second must flow past the coils...

Thanks a lot[;)]

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