An isentropic Rankine cycle using steam is shown in Figure 13.68. For the given data, determine

An isentropic Rankine cycle using steam is shown in Figure...

Key Concepts

Rankine Cycle

The Rankine cycle is a thermodynamic cycle used to convert heat into mechanical work, typically in steam power plants. The cycle involves four main components: a boiler where water is heated to produce steam, a turbine where high-pressure steam expands and does work, a condenser where steam is converted back into water, and a pump that pressurizes the liquid before it is sent back to the boiler. This cycle provides the fundamental framework for analyzing energy conversion and efficiency in thermal systems.

Isentropic Process

An isentropic process is one in which the entropy remains constant, meaning that it is both adiabatic and reversible. In the context of the Rankine cycle, both the expansion in the turbine and the compression in the pump are ideally treated as isentropic, which simplifies the analysis by allowing the use of entropy values from the steam tables to determine other state properties.

Steam Quality

Steam quality is defined as the ratio of the mass of vapor to the total mass of a mixture of vapor and liquid in a two-phase region. It is crucial in the analysis of steam turbines because it affects the energy content and efficiency of the cycle; determining the quality of steam at the turbine exhaust helps in evaluating the turbine performance and in avoiding issues like blade erosion due to moisture.

Cycle Thermal Efficiency

Cycle thermal efficiency in the Rankine cycle is the ratio of the net work output to the heat input. It provides a measure of how effectively the cycle converts thermal energy into useful work. Factors such as the high and low pressures, the temperatures at different stages, and the quality of steam all influence the thermal efficiency of the cycle.

Mass Flow Rate Determination

Determining the mass flow rate is essential for ensuring that the cycle produces the desired net power output. This calculation typically involves using an energy balance that relates the net work per unit mass to the total power generation requirement. By knowing the net work output per unit mass and the target power output, the required mass flow rate through the system can be accurately computed.

Transcript

00:02 In this problem, we're going to analyze a combination gas steam cycle.

00:08 So here is the gas cycle up here, from 5 to 7 to 8, back to 5.

00:15 And then at 9 is where we have a heat exchanger.

00:18 So we're extracting the heat from the hot exhaust gases.

00:29 And essentially, well, i mean, basically, we're actually, you know, in this cycle, we're not really, we're just kind of pulling in new air all the time.

00:38 So there's really not a, this is not really a process from here to here.

00:44 So basically we're just extracting the heat as it comes out from here.

00:49 And then it's not a closed loop for a gas, gas turbine, because we're just using air as the working fluid.

00:56 So we don't basically have.

00:59 Have a closed loop where in the steam cycle we usually have.

01:03 We almost always have a closed loop here, although we wouldn't necessarily have to.

01:11 Then let's see here.

01:13 So if we look at the gas side, we can use ideal gas laws.

01:18 So from six to seven, we can figure out the temperature using hydro gas relationship.

01:28 And so we have, we have a lot of temperature, have because we know the pressures at six and at seven, which i didn't write down.

01:42 So let me see what the gas side.

01:47 Let's see here.

01:51 Well, you know the pressure ratio is 16.

01:53 So we know that's what we knew.

01:56 P6 over p5 is 16.

02:00 So that's the pressure ratio ratio.

02:03 So we can figure out the temperature here by knowing the pressure difference between what is.

02:09 Oh, sorry, from here to here, i'm looking at the wrong thing.

02:12 From here to here is the pressure difference of 16.

02:15 And it's isotropic so we can just use the isotropic with ideal gas formulation and using, you know, these heat capacity ratio.

02:23 So in the end, we get a temperature up here of, sorry, right here, of 662 .5 kelvin up from 300.

02:33 Under kelvin here.

02:35 This is kelvin.

02:36 This is also in kelvin.

02:38 So these temperatures here are in kelvin.

02:41 Let's see here.

02:42 So we can figure out the heat transfer into the air because it's just the mass fluoride of the air, which we know 14 kilograms per second, times the difference in enthalpy from here to here.

02:58 And the difference in enthalpy would just use the constant heat capacity, which may not be such a great idea because we've got a pretty big swing and temperature here, but we'll use a maybe an average.

03:09 What did i use here actually for the heat capacity? i didn't write the value down.

03:14 Let me find the problem in my notes.

03:18 Let's see here.

03:22 77.

03:26 Okay, i used the heat capacity of 1 .005.

03:32 So you seep equals 1 .005.

03:37 Kilojoules per kilogram kelvin.

03:39 So that i think is pretty much just room temperature air, if i remember, right? so that's probably not a great estimate for, you know.

03:50 I mean, it's all right for, you know, it's 300 kelvin, but up here at 600, up here to, you know, 1 ,500, maybe not so good.

04:00 But it's an estimate.

04:02 We could do better by using your tables or using, you know, reduced, you know, adjusting it using reduced pressures and temperatures and stuff.

04:13 Now, in the compressor from 5 to 6, so here is our compressor.

04:21 So basically instead of a pump, we have a compressor.

04:25 And again, we can figure out the change in enthalpy by using, you know, constant heat capacity.

04:30 And we get it, i guess i never said, the heat in is about 11 ,000, or 11 .78.

04:39 Megawatts of heat coming in here.

04:42 And then the work out, work into the compressor is about 5 .1 megawatts.

04:50 So we can see that actually, you know, in this cycle, in the gas cycle, the compression work is considerable, where in the steam cycle, the pump work is is very, is very insignificant.

05:06 Again, we have an is entropic process here so we can figure out the temperature at 8 and knowing the pressure ratio from 7 to 8, which is also, well, it's 16 from here to here again.

05:26 So again, the pressure ratio ratio between 7 and 8 is 16.

05:32 So here we use 1 over 16, and then the heat, the exponent here...

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