Answer and Explain #6 please The Tennessee Valley Authority\'s coal-burning Para

Question

Answer and Explain #6 please The Tennessee Valley Authority's coal-burning Paradise power plant produces 1000 MW of power and has an overall thermal efficiency of 39%. The overall thermal efficiency is defined as the work output divided by the absolute value of the heat of the combustion of the fuel. [Since only 85 to 90% of the heat of combustion is transferred to the steam, the overall thermal efficiently is not the same as the efficiency defined by e = |w |/q_H.] The typical enthalpy of combustion of coal is -10000 British thermal units (Btu) per pound, where 1 Btu equals 1055 J. How many pounds of coal does the Paradise plant bum in (a) one minute; (b) one day; (c) one year?

Explanation / Answer

Ans. #1. Let the total (absolute) heat produced per second = X joule

Given,

            Usable Energy production rate = 1000 MW

                                                = 1000 x 106 W                                 ; [mega = 106]

                                                = 1000 x 106 J/s                                ; [W = J/s]

                                                = 109 J/s

                        Thus, the plant produces 109 J usable energy per second.

Overall thermal efficiency = 39 %

So,

            39 % of X J = 109 J

            Or, X = 109 / 0.39                                                                 ; [39% = 0.39]

            Hence, X = 2.56 x 109

Therefore, total/ absolute heat produced per second = 2.564 x 109 J

Note: The information provided in brackets [] is not required at all for calculation.

However, it tells that (all values per second)-

            I. Only 85- 90% of absolute (total) heat (2.56 x 109 J per second) produced from coal combustion is transferred to water. That is, say at 90% efficiency, only 2.304 x 109J (out of total 2.56 x 109 J) is transferred to water to form steam, the rest is wasted.

           II. Out of 2.304 x 109J heat transferred to water, only 109 J (= 1000 MW) is actual harvested as work done by steam, the rest is wasted.

Thus, the overall efficiency (step I + II) remaining to be 39 %.

#2. Given,

            Enthalpy of combustion of coal = - 10000 Btu/ pound

                                                            = [104 x (1055 J)] / pound

                                                            = 1.055 x 107 / pound

Pounds of coal burnt per second =

Absolute heat produced per second / Enthalpy of combustion of coal

                        = 2.564 x 109 J / (1.055 x 107 / pound)

                        = 2.430 x 102 pounds

Thus, rate of coal combustion = 2.430 x 102 pounds/ s

#3. Pounds of coal burnt per minute =

Rate of coal combustion x (number of seconds in a minutes)

= (2.430 x 102 pounds/ s) x 60 s

= 1.45 x 103 pounds

#4. Pounds of coal burnt per day =

Rate of coal combustion x (number of seconds in a day)

= (2.430 x 102 pounds/ s) x 86400 s

= 2.09 x 107 pounds

#4. Pounds of coal burnt per year =

Rate of coal combustion x (number of seconds in a year)

= (2.430 x 102 pounds/ s) x 31536000 s

= 7.6 x 109 pounds

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