A steam turbine receives steam at 6 MPa, 800^∘ C. It has a heat loss of 49.7 kJ / kg and an ise

A steam turbine receives steam at 6 MPa, 800^∘ C. It has a...

Key Concepts

Isentropic Process

An isentropic process is a thermodynamic process during which the entropy of the system remains constant. It represents an idealized scenario where the process is both adiabatic (no heat transfer) and reversible, serving as a benchmark for evaluating the performance of real systems in which energy losses occur.

Isentropic Efficiency

Isentropic efficiency is a key performance metric used to compare the actual work output of a device, such as a turbine, with the maximum possible (ideal) work output under isentropic conditions. It quantifies the impact of irreversibilities like friction and heat loss, and it guides efforts to improve system design and performance.

Reversible Work

Reversible work refers to the theoretical maximum work that can be extracted from a process if it were carried out in a completely ideal and reversible manner, without any energy losses. This concept is used as a standard comparison to assess the performance of actual devices, highlighting the gap created by real-world inefficiencies.

Actual Work

Actual work is the measured work output from a real-world process that takes into account all practical losses such as friction, turbulence, and heat leakage. It is lower than the ideal reversible work due to these irreversibilities, and it is a critical parameter for evaluating the efficiency of thermodynamic devices.

Energy Balance Analysis

Energy balance analysis involves accounting for all energy inputs, outputs, and losses within a system to ensure that the conservation of energy principle is satisfied. In the context of turbine analysis, it includes the assessment of energy carried by the working fluid, the work produced, and any heat losses, thereby providing a comprehensive understanding of the system's performance.

Transcript

00:01 In this question it is given that steam at a pressure of 6 megapascal and the temperature is 800 degrees centigrade enters in a steam turbine.

00:12 Heat loss from the turbine is 49 .7 kilojoules per kilogram and the isotropic efficiency is equals to 90 % or we can write 0 .9.

00:29 The exit pressure and temperature are 15 kilo pascal and 20 degrees centigrade respectively.

00:40 We are required to calculate the actual work and the reversible work between the inlet and the exit.

00:47 So let's see how to solve this question.

00:50 First of all, let's write the properties of the steam corresponding to state 1 from the table.

00:57 So we can write s -i is equal to 7 .3.

01:02 6 .6 kilojoules per kilogram kelvin and h i is equals to 4 ,132 .74 kilojoules per kilogram.

01:19 So these are the properties for state i and now at state e we can write s s is equals to s i so this will be equals to 7 .6566 kilojoules per kilogram kelvin.

01:45 Now let's apply the formula of entropy to calculate dryness fraction.

01:51 So we can write ses is equals to sfes plus xes into sfges.

02:09 Now substitute all the values from the the table and the calculated values so we can write 7 .6566 and this will be equals to 0 .7548 these values have been taken from the table plus x e comma s multiplied by 7 .2536 so when we further calculate we get dryness fraction e.

02:42 Comma s is equals to 0 .9515 and now let's apply the formula to calculate specific enthalpy for this state and this will be equals to hfes plus xes into hfges now substitute all the values so we will have hes is equals to 225 .91 plus 0 .9515 multiplied by 2 ,373 .14.

03:31 When we further calculate we get 2 ,484 kilojoules per kilogram.

03:40 Now let's apply the formula to calculate.

03:43 Turbine work.

03:45 So wt will be equals to h .i minus h .e.

03:52 Comma s.

03:54 Now substitute all the values so we will have 4 ,132 .74 minus 2 ,484.

04:03 When we further calculate, finally we get 1 ,648 .74 kilojoules per kilogram...

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