Question

в котле паровой машины температура равна 160 С, а температура холодильника равно 10 С. определите максимальную работу, которую можно получить от машины, считая ее идеальной, если в топке, КПД которой равно 60%, сожжено 200кг угля с удельной теплотой сгорания 2,9*10 в 7Дж/кг

Answer 1

Дано:

Тн = 160 С = 433 К

Тх = 283 К

n (топки) = 60%

λ = 3 * 10^7 Дж/кг

m = 200 кг

А - ?

Решение:

n = A/Qн

n = Тн - Тх/Тн

Q = λm

Найдем кпд тепловой машины:

n = 433 - 283/433 ~ 0.35%

Найдем теплоту, которая выделится при полном сгорании угля, учитывая кпд топки.

Q = λnm = 2 * 10^2 * 3 * 10^7 * 0.6 = 36 * 10^8 Дж

Теперь найдем работу.

n = A/Qн => A = nQн

A = 35 * 10^2 * 36 * 10^8 = 1260 * 10^6 Дж = 1.26 ГДж.

Ответ: А = 1.26 ГДж.

 

Answer 2

Calculation of Maximum Work Output

To determine the maximum work that can be obtained from the steam engine, we can use the Carnot efficiency formula. The Carnot efficiency is given by the equation:

Efficiency = 1 - (Tc/Th)

Where: - Tc is the temperature of the cold reservoir (in Kelvin) - Th is the temperature of the hot reservoir (in Kelvin)

In this case, the temperature of the steam in the boiler is 160°C, which is equivalent to 433.15 Kelvin, and the temperature of the refrigerator is 10°C, which is equivalent to 283.15 Kelvin.

Using these values, we can calculate the Carnot efficiency:

Efficiency = 1 - (283.15/433.15) = 0.345

The Carnot efficiency represents the maximum possible efficiency of an ideal heat engine operating between these two temperatures. To calculate the maximum work output, we need to know the amount of heat energy supplied to the engine.

Calculation of Heat Energy Supplied

To calculate the heat energy supplied to the engine, we need to know the mass of coal burned and its specific heat of combustion.

Given: - Mass of coal burned (m) = 200 kg - Specific heat of combustion (Q) = 2.9 * 10^7 J/kg The heat energy supplied (Qs) can be calculated using the formula:

Qs = m * Q

Substituting the given values:

Qs = 200 kg * 2.9 * 10^7 J/kg = 5.8 * 10^9 J

Calculation of Maximum Work Output

The maximum work output (W) can be calculated using the formula:

W = Efficiency * Qs

Substituting the values:

W = 0.345 * 5.8 * 10^9 J = 2 * 10^9 J

Therefore, the maximum work that can be obtained from the steam engine, assuming it is ideal, is 2 * 10^9 Joules.

Please note that this calculation assumes ideal conditions and does not take into account any losses or inefficiencies in the steam engine.

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