A stationary gas-turbine power plant operates on a simple...

A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the compressor at $95 \mathrm{kPa}$ and $290 \mathrm{~K}$ and the turbine at $760 \mathrm{kPa}$ and $1100 \mathrm{~K}$. Heat is transferred to air at a rate of $35,000 \mathrm{~kJ} / \mathrm{s}$. Determine the power delivered by this plant (a) assuming constant specific heats at room temperature and $(b)$ accounting for the variation of specific heats with temperature.

Key Concepts

Specific Heats (Constant vs. Variation with Temperature)

The specific heat of a gas is the amount of energy required to raise the temperature of a unit mass by one degree. In thermodynamic analyses, assuming constant specific heats (often at a reference temperature) simplifies calculations. However, in more precise analysis, the variation of specific heats with temperature is considered, as it can significantly affect the accuracy of energy balance computations, especially when the temperature range is wide.

Steady-Flow Energy Equation

The steady-flow energy equation is a fundamental principle employed in analyzing the energy exchanges in devices with a continuous flow of mass, such as compressors and turbines. It relates the energy added or removed (in the form of heat and work) to the change in enthalpy of the flowing fluid, and is essential for quantifying work outputs and heat transfer rates in power cycles.

Mass Flow Rate and Power Output Calculation

In power cycle analysis, determining the mass flow rate of the working fluid is crucial for relating the rate of heat addition to the work produced. By using energy balance equations, one can estimate the mass flow rate from the known heat transfer rate and temperature changes, and subsequently calculate the power output delivered by the system. This connection between heat input, fluid properties, and work output is a key concept in thermodynamic cycle analysis.

Brayton Cycle

The Brayton cycle is a thermodynamic cycle that describes the workings of a constant-pressure heat engine, such as a gas turbine. In this cycle, air is compressed, heat is added at constant pressure, the resulting high-energy gas expands in the turbine to produce work, and finally, the exhaust is released. This cycle serves as the basis for analyzing gas turbine performance and efficiency.

Air-Standard Assumptions

Air-standard assumptions simplify the analysis of gas turbine cycles by modeling air as an ideal gas with constant or known properties. This approach neglects complications such as fuel combustion details and heat losses, allowing for the use of idealized processes to obtain a theoretical performance estimate in the cycle’s analysis.

Transcript

00:01 Our question is, a stationary gas turbine power plant operates on a simple ideal burden cycle with air as the wooden fruit.

00:10 The air enters the compressor a 95 kilo possible in 290 kelvin and the turbine at 760 kilo possible and 1100 kelvin.

00:18 Heat is transferred to air at a rate of 35 ,000 kilojoules per second.

00:23 Determine the power delivered by this plant.

00:27 First, we have to find the power delivered by this.

00:34 Plant assuming constant heat specific heats at room temperature and accounting for the variation of specific heats with the temperature so we have tns diagram of this question as there is the heat input and this is the heat output there are two temperatures that is 1 ,100 kelvin and this is 290 kelvin so in the part one that is assuming constant specific heats t2 x is equals to t1 into p2 by b1 color k minus 1 plus 2 by k this is equals to 290 into 8 0 .4 by 1 .4 so this will be equal to 525 0 .310.

02:42 T4 ever refers to p3, k -9x5 divided by p3 power k minus 1 divided by k, that is equals to 1100 into 1 by 8, 0 .4 that is 607 .2 total.

03:07 So thermal efficiency is supposed to 1 minus q -in that is equals to 1 minus kw alpha qn, that is equals to 1 minus c b t 4 minus t 1 divided by c b t3 minus d3 minus equal this is equals to 1 minus t 4 minus 3 2 that is equals to putting the values 1 minus 607 .2 minus 2 902 divided by 1100 minus 525 this is equals to 0 .448.

04:04 So our net outward work is equals to thermal efficiency and heat input.

04:13 This is equals to 0 .448 into 35 ,000 kilowatt.

04:24 The answer turns out to be 15 ,6, 18 kilowatt...

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