A certain industrial process requires a steady 0.5 kg / s of air at 200 m / s, at the condition

A certain industrial process requires a steady 0.5 kg / s of...

A certain industrial process requires a steady 0.5 $\mathrm{kg} / \mathrm{s}$ of air at $200 \mathrm{m} / \mathrm{s}$, at the condition of $150 \mathrm{kPa}$ $300 \mathrm{K},$ as shown in Fig. $\mathrm{P} 9.132 .$ This air is to be the exhaust from a specially designed turbine whose inlet pressure is $400 \mathrm{kPa}$. The turbine process may be assumed to be reversible and polytropic, with polytropic exponent $n=1.20$ a. What is the turbine inlet temperature? b. What are the power output and heat transfer rate for the turbine? c. Calculate the rate of net entropy increase, if the heat transfer comes from a source at a temperature $100^{\circ} \mathrm{C}$ higher than the turbine inlet temperature.

A certain industrial process requires a steady 0.5 $\mathrm{kg} / \mathrm{s}$ of air at $200 \mathrm{m} / \mathrm{s}$, at the condition of $150 \mathrm{kPa}$
$300 \mathrm{K},$ as shown in Fig. $\mathrm{P} 9.132 .$ This air is to be the exhaust from a specially designed turbine whose inlet pressure is $400 \mathrm{kPa}$. The turbine process may be assumed to be reversible and polytropic, with polytropic exponent $n=1.20$
a. What is the turbine inlet temperature?
b. What are the power output and heat transfer rate for the turbine?
c. Calculate the rate of net entropy increase, if the heat transfer comes from a source at a temperature $100^{\circ} \mathrm{C}$ higher than the turbine inlet temperature.

Key Concepts

Polytropic Process

A polytropic process is one where the pressure and volume of a working fluid are related by the equation PV^n = constant. This concept is used to describe non-isentropic expansions or compressions where heat transfer may occur, and the exponent n characterizes how much the actual process deviates from an ideal isentropic process.

Turbine Energy Balance

Turbine energy balance involves applying the first law of thermodynamics to turbomachinery to relate the energy changes of the fluid, including changes in enthalpy and kinetic energy. This concept is crucial for determining the work output of the turbine by analyzing the inlet and exit conditions of the fluid under steady?state operation.

Mass Flow Rate Considerations

Mass flow rate is central to process calculations in steady-state devices like turbines since it bridges the per unit mass energy changes to overall energy rates. It is used to scale the specific energy differences into power outputs, ensuring that the energy balance accounts for the actual amount of fluid moving through the system.

Kinetic Energy Effects

In flows where the velocity of the working fluid is significant, the kinetic energy term becomes important in the overall energy balance. Understanding how changes in velocity contribute to energy differences allows for a more accurate computation of turbine performance and the resulting work output.

Entropy Generation and the Second Law

Entropy generation quantifies the irreversibility of a process. It is calculated when heat transfer occurs between reservoirs at different temperatures. This concept is critical for assessing the degree of inefficiency in a turbine system by comparing the entropy change in the working fluid to the entropy change imposed by external heat sources.

Transcript

00:01 In this problem, we have an industrial process that requires half a kilogram per second of air at 200 meters per second, and the conditions at the conditions of 150 kilopascals and 300 calvin.

00:19 So to get that, this air is to be exhausted from a specially designed turbine, whose inlet pressure is 400 kilopascals.

00:28 The turbine process may be assumed to be reversible and polytropic with a polytropic exporter of 1 .2 okay so we have air and we want to know the turbine inlet temperature the power output and heat transfer for the turbine and the rate of net entropy increase if the heat source of the heat transfer comes from a source at 100 degrees c oh no 100 degrees high and higher than the inlet temperature so again the reservoir source is 100 degrees above whatever inlet temperature we find.

01:07 So for a polytropic process, we know we, and an ideal gas, we have this formula rating the temperatures and pressures.

01:17 And so we can solve for the temperature, and that is 353 .0 .3 kelvin.

01:29 Our energy equation, it should be a plus here.

01:37 So our energy equation is the, you know, the energy flux here from the enthalpy, the internal energy and the pressure and the flow, and then the velocity, this is then the pressure here, and then the kinetic energy, and then we have the heat energy, and then coming out here, we have the energy flux out, and the work out.

02:03 Now, we know, because it's a polytropic process, we know the specific work.

02:10 Is related as far as the pressures go, right? and then we have the change in specific kinetic energy.

02:24 So this is the change as the air was compressed.

02:32 We know everything here, and we know we're going to assume that v1 was roughly zero, so that the inlet velocity was roughly zero, or very negligible compared to the exit velocity.

02:46 At least in squared, squaring, and you'll get a big difference.

02:52 Anyway, the specific work is 71 .8 kilojoules per kilogram.

02:59 Now, we know the mass flow rate.

03:01 That should be a dot on that.

03:04 You know the mass flow rate, and so that gives us the power to be 35 .9 kilowatts of power that we need to be putting into, to the putting into the turbine to pressurize and speed up this air...

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