An electric generating station is designed to have an...

An electric generating station is designed to have an electric output power of $1.40 \mathrm{MW}$ using a turbine with twothirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at $110^{\circ} \mathrm{C}$. (a) Find the rate at which the station exhausts energy by heat as a function of the fuel combustion temperature $T_{h^{*}}$ (b) If the firebox is modified to run hotter by using more advanced combustion technology, how does the amount of energy exhaust change? (c) Find the exhaust power for $T_{h}=800^{\circ} \mathrm{C}$. (d) Find the value of $T_{h}$ for which the exhaust power would be only half as large as in part (c). (e) Find the value of $T_{h}$ for which the exhaust power would be one-fourth as large as in part (c).

An electric generating station is designed to have an electric output power of $1.40 \mathrm{MW}$ using a turbine with twothirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at $110^{\circ} \mathrm{C}$. (a) Find the rate at which the station exhausts energy by heat as a function of the fuel combustion temperature $T_{h^{*}}$ (b) If the firebox is modified to run hotter by using more advanced combustion technology, how does the amount of energy exhaust change? (c) Find the exhaust power for $T_{h}=800^{\circ} \mathrm{C}$.
(d) Find the value of $T_{h}$ for which the exhaust power would be only half as large as in part (c). (e) Find the value of $T_{h}$ for which the exhaust power would be one-fourth as large as in part (c).

Key Concepts

Effect of High Combustion Temperatures

Increasing the fuel combustion (or firebox) temperature generally improves the potential efficiency of a thermal engine by increasing the temperature difference with the cold reservoir. This higher temperature can reduce the relative amount of energy wasted in the exhaust, although the exact relationship depends on the engine's operational characteristics and design limits.

Energy Balance and Rate of Energy Transfer

The energy balance involves accounting for the input energy from fuel combustion, the useful work extracted as electric power, and the energy expelled as waste heat. Calculating the rate of energy transfer is essential in designing power plants and understanding how variations in operating temperatures affect overall performance.

Exhaust Heat and Heat Rejection

Exhaust heat is the portion of energy that is not converted into work and is instead rejected to a cooler reservoir. Managing and quantifying this rejected heat is critical in the evaluation of engine performance and efficiency, as it directly impacts the overall energy conversion process.

Carnot Efficiency

Carnot efficiency represents the maximum theoretical efficiency that a heat engine can achieve operating between two temperatures. It is determined solely by the temperatures of the hot and cold reservoirs, setting an upper limit for any real engine's performance and guiding the design and improvement of thermal cycles.

Thermal Efficiency in Real Engines

Real engines operate at efficiencies below the Carnot limit due to practical irreversibilities and losses. For instance, when a turbine operates at a fraction (such as two-thirds) of the Carnot efficiency, it reflects realistic performance metrics and guides the calculation of both useful work output and energy waste in the system.

Transcript

00:01 Here is in part a, the actual efficiency is 2 third the carnot efficiency.

00:08 Now, read this as the equation.

00:11 Now ve engineer, modulus of qh, which is equals to w engineer divided by models of carnot efficiency plus w work engineer which is equal to 2 by 3 multiply good.

00:32 1 minus tc divided by th which is equal to 2 by 3 thh 2 by 3 thh minus tc divided by capital thh it is capital th capital tch capital tc all the t's represent absolute temperature then we will have qc modulus of qc plus w engineer divided by engine burke of engine is equal to 1 .5 t .h divided by t .h minus tc which is equals to which tends to qc divided by work engineer which is equal to engine 1 .5 of t h divided by t h minus tc and the whole is minus 1.

01:33 From here we get the value 1 .5 t h minus t h plus tc divided by t h minus tc now from here we will get the value of modellus of qc divided by delta t tends to w engineer divided by delta t which tends to multiply with 0 .5 t h plus tc divided by t h minus tc now from here we get this value the value of qc by q by delta t is equal to 1 .40 multiply with 0 .5 t h plus 383 divided by t h minus 383 where delta q by t where q c by delta t is in megawatt and t is in calvin now in part b the exhaust the exhaust power decreases as the firebox temperature increases now in part c we have the value of modulus uc efficiency of carnot divided by delta t which is equals to 1 .40 megawatt multiply with 0 .15 t h plus 383 kelvin divided by th minus 383 kelvin.

03:37 Now from here we get 1 .40 megawatt multiply with when we will put the value 0 .5 into 1073 calvin plus 383 calvin this is divided by 1073 kelvin minus 383 kelvin after solving we get 1 .87 megawatt.

04:04 Now in part d we require here we require modulus of qc divided by delta t which is equals to 1 by 2 1 .87 megawatt now we can put the value 1 .40 megawatt multiply with 0 .5 t h and plus 383 kelvin divided by th minus 383 kelvin.

04:41 From here we get 0 .5 th plus 383 kelvin divided by th minus 383 kelvin which is equal to 0 .6...

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