Air is expanded in an adiabatic turbine of 85 percent isentropic efficiency from an inlet state

Air is expanded in an adiabatic turbine of 85 percent...

Key Concepts

Energy Balance in Turbines

The energy balance in turbines involves applying the steady flow energy equation, which relates the change in enthalpy of the working fluid to the work produced or consumed. In turbines, the drop in enthalpy of the working fluid is directly converted into work, making this balance essential for calculating work outputs.

Thermodynamic Property Evaluation

Evaluating thermodynamic properties such as temperature, pressure, and enthalpy is crucial for analyzing processes in turbines and other devices. This evaluation is typically achieved using property tables, equations of state, or computer-aided methods, and it enables engineers to accurately predict the performance and efficiency of thermodynamic systems.

Isentropic Efficiency

Isentropic efficiency measures the performance of a turbine by comparing the actual performance to that of an ideal isentropic process. It is defined as the ratio of the actual work output to the work output that would be obtained under isentropic conditions, and it is used to account for real-world inefficiencies in energy conversion devices.

Adiabatic Process

An adiabatic process is one in which no heat is transferred into or out of the system. In such a process, the change in internal energy of the working fluid is solely due to work interactions. This concept is fundamental in thermodynamics and is particularly relevant in devices like turbines where rapid expansion or compression often minimizes heat transfer.

Isentropic Process

An isentropic process is a reversible adiabatic process, meaning that it occurs without any change in the entropy of the working fluid. This idealized process serves as a benchmark for evaluating real processes, helping to quantify the effects of irreversibilities and losses in practical devices.

Transcript

00:01 So here in this problem we are given that air is expanded from pressure p1 we are given as 2 .2 megapascal and temperature t1 is 300 degrees celsius to the outlet temperature as outlet pressure as p2 as 200 kilo pascal and we have to determine the outlet temperature t2 and the work produced by the turbine per unit mass of air.

00:23 So power output we have to calculate.

00:27 Now to solve this problem will be requiring some input data that is a the value of cp constant specific heat at constant pressure and the value of k so cp we have 1 .013 kilojoule over kg degrees celsius and this one is 1 .395 and we have obtained these value from the table a to a corresponding to an average temperature of around 400 kelvin so now next i will be applying the energy balance equation that is energy in minus is energy out is equal to change in energy of the system and since the this is a steady system so delta e will be equals to zero so we can direct to write that energy n is equals to energy out from here energy in we have m .h1 and energy out we have the work output this one is actual and plus m dot h2 so work output from actual work output from here comes out to be as m dot h1 minus h2 and if we calculate this the work output the per work produce per unit mass of air so we can say that the work actual work output will be equals to h1 minus h2 that is cpt1 minus t2 now if we calculate the isentropic work output this one was actual if we calculate isentropic work output it will be cp t1 minus t2s now we can calculate this t2 s by using the isentropic relation that is t2s will be equals to t1 times p2 divide by p1 to the power k minus 1 divide by k from here t2s will be be equals to t1 we have 300 degrees celsius but we have to convert this thing to kelvin so we'll get 573 kelvin times p2 divided by p1 so p2 we have 1 200 k2 we have 2 200 k we have 0 the 1 .395 so k minus 1 will be 0 .395 and divided by 1 .395 so after solving this we'll get the value of t 2 s as nearly 290 .6 kelvin now we can calculate the isentropic work output by using this formula and before this we can do one all one thing also that the work output actual this is actual is equal to isentropic efficiency of the turbine times the isentropic work output so efficiency of the isentropic efficiency of the turbine we are given in the question as 0 .85 so you're going to plug here to calculate the actual work output efficiency we are given as 0 .85 times the work output isentropic case cp times t1 minus t2s so cp we have a 1 .013 kilojoule over kg degree celsius times t1 minus t2s so t1 we have 573 kelvin minus t2 s we have obtained us 290s we have obtained us 2 90 .6 kelvin from here we get the actual work output as nearly 243 .2 kilojoule over kg so this is the actual work output...

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