A natural gas-fired electric power plant has an efficiency of 30 percent with 25 percent wasteheat being released to the atmosphere as stack heat and the other 75% taken away incooling water drawn from a nearby river (used to cool and condense down the hot steamexiting the generator turbines). To eliminate heating the bodies of water surrounding thepower plant, the evaporation of the coolant water is what cools the steam; the water vaporgenerated by this evaporation is released to the atmosphere. Thus, the amount of heatdissipated will be proportional to the amount of water evaporated and the latent heat ofvaporization.1ttEvaporative coolingStack heat25% of waste heatSteamPower plant30% efficientMakeup waterRetum to bolilerColing towerDetermine how many gallons per minute of cooling water (marked "makeup water" in theillustration) is required for every 1000 MW of electric power generated by this facility. Theriver water is 15°C and all of the cooling water evaporates at 15°C. [The fact that theefficiency is less than 100% means the electric power output to be computed here is lessthan the total amount of heat created by the combustion of the fuel source.]

Approach: We will first calculate the actual amount of power dissipated by water's evaporation. then…

Answered: A natural gas-fired electric power…

Similar questions

A heat engine operates between a hot reservoir at 500°C and a cold reservoir at 25°C. It has an efficiency of 40%. Calculate the amount of heat absorbed from the hot reservoir when the engine expels 1500 J of heat to the cold reservoir.
A certain heat engine does 9.5 kJ of work and 8.00 kJ heat transfer occurs to the environment in a cyclical process. (a) What was the heat transfer (in kJ) into this engine? kJ (b) What was the engine's efficiency (in percent)? %
A heat engine running backward is called a refrigerator if its purpose is to extract heat from a cold reservoir. The same engine running backward is called a heat pump if its purpose is to exhaust warm air into the hot reservoir. Heat pumps are widely used for home heating. You can think of a heat pump as a refrigerator that is cooling the already cold outdoors and, with its exhaust heat QH, warming the indoors. Perhaps this seems a little silly, but consider the following. Electricity can be directly used to heat a home by passing an electric current through a heating coil. This is a direct, 100% conversion of work to heat. That is, 19.0 kW of electric power (generated by doing work at the rate 19.0 kJ/s at the power plant) produces heat energy inside the home at a rate of 19.0 kJ/s. Suppose that the neighbor's home has a heat pump with a coefficient of performance of 3.00, a realistic value. NOTE: With a refrigerator, "what you get" is heat removed. But with a heat pump, "what you…
Assume that the turbines at a coal-powered power plant were upgraded resulting in an improvement in efficiency of 3.14%. Assume that prior to the upgrade the power station had an efficiency of 38%, and that the heat transfer into the engine in one day is still the same at 2.63 × 1014 J. (a) How much more electrical power is produced due to the upgrade? 95581 kW (b) How much less heat transfer occurs to the environment due to the upgrade? 1548000000000 x J
With 2.56×106 J of heat transfer into this engine, a given cyclical heat engine can do only 1.50×105 J of work.(a) What is the engine’s efficiency? (b) How much heat transfer to the environment takes place?
An engine has a hot-reservoir temperature of 980 K and a cold-reservoir temperature of 550 K. The engine operates at three-fifths maximum efficiency. What is the efficiency of the engine?
please don't copy existing solutions