(a) At a steam power plant, steam engines work in pairs, the...

(a) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are $710^{\circ} \mathrm{C}$ and $430^{\circ} \mathrm{C},$ and of the second $415^{\circ} \mathrm{C}$ and $270^{\circ} \mathrm{C}$ . If the heat of combustion of coal is $2.8 \times 10^{7} \mathrm{J} / \mathrm{kg},$ at what rate must coal be burned if the plant is to put out 950 $\mathrm{MW}$ of power? Assume the efficiency of the engines is 65$\%$ of the ideal (Carnot) efficiency. $(b)$ Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 5.5 $\mathrm{C}^{\circ}$ , estimate how much water must pass through the plant per hour.

Key Concepts

Energy Conversion in Combustion

The process of converting chemical energy stored in fuel (coal, in this case) into useful mechanical power is central to thermal power plants. This involves calculating the rate at which fuel must be burned based on its heat of combustion and the overall power output, while accounting for the system's efficiency.

Heat Transfer and Cooling

In thermal power systems, not all the input heat is converted into work, and the excess heat must be removed to maintain safe operating temperatures. Estimating the required mass flow rate of cooling water involves using the specific heat capacity of water and the allowed maximum temperature rise, ensuring effective heat disposal.

Carnot Efficiency

Carnot efficiency represents the theoretical maximum efficiency that any heat engine can achieve when operating between two thermal reservoirs. It is calculated using the temperatures of the hot and cold reservoirs (in Kelvin) and sets an idealized benchmark against which real engine performances are compared.

Real vs. Ideal Efficiency

In practical applications, real engines operate at efficiencies lower than the Carnot limit. The given efficiency factor (65% of Carnot efficiency) reflects losses and irreversible processes in the system, emphasizing the need to adjust ideal calculations to estimate actual performance.

Transcript

00:01 Part a of the problem, the efficiency for each engine is given by efficiency is equal to 0 .65 times the carnot engine efficiency, which is represented by ec.

00:15 So the efficiency for a first engine is equal to 0 .65 times the efficiency for a first car not, that is equal to 0 .65, that is equal to 0 .65 in terms of the efficiency for a first car not, that is equal to 0 .65 in terms of the temperature is given by 1 minus tl1 low temperature 1 and divided by t h high temperature 1.

00:42 That is equal to 0 .185 plugging the value for a temperature.

00:50 Similarly for efficiency 2, we will do the same procedure and the answer we get here is 0 .137.

01:02 The first engine input heat from the coal will be work is equal to efficiency 1, qh1, h1, that is equal to efficiency of engine 1 times the q coming from a coal burning, heat coming from coal burning.

01:28 And then wl1 is written as using the fossil law of thermodynamics that is a q h1 minus w1 that gives us 1 minus efficiency 1 times q of the call so for call i will like to see you only then is written as e2, qh2, qh2, that is equal to e2, e2, e2, 1 minus e2 times q of cold.

02:27 Then, excuse me, adding w1 and w2, so i will do here, excuse me, then, adding both work, w1 and w2, w2, we update expression which is e1 plus e2, minus e1 times e2, a whole multiplies with the heat from the coal.

03:00 So this is qz.

03:02 Then the rate of the burning of the coal, this is c, so this is c, rate of the burning of coal, it can be found by dividing by t, then w1 is also, w1 plus w2 is divided by t, t divided by the efficiencies e1 plus e2 minus e1 times e2.

03:35 Then the rate of a coal burn we can find it by plugging the values, which are given in the problem...

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