The change in energy is not determined solely by the difference in pressure the temperature of the steam is also changing as it expands through the machine. Turbines operate by extracting energy from the expansion of steam through the device. Let's make sure I understand the variables first: That being said, there is enough to calculate the delta P (but not a static inlet or outlet pressure as you pointed out) without having all the other information. Good old thermo.Īs for the second question, I'm thinking that perhaps the pressure is a change in pressure, or what I always call the delta P. I would have done the problem 100% the same. Good job on the first question, SteamKing. What's this equation, where is it derived from? It's in the answers for a question, the question asked to calculate the work done by the air in a cylinder as it expands out. The property info comes from the attached tables: The isentropic work (or more accurately, the power) is therefore W = m*(hi - ho) = 0.5 kg/s * (3423.1 - 2321.7) kJ/kg The enthalpy of this mixture can be calculated using the saturated liquid properties: S = 6.8826 kJ/kg*K = sf + x * sfg, where x is the quality of the vaporĭoing the algebra and solving for x = 0.870, or a vapor quality of 87% Then h = 3423.1 kJ/kg and s = 6.8826 kJ/kg*Kįor P = 30 kPa, we need to find the properties for s = 6.8826 kJ/kg*K, Since Mollier diagrams are hard to find on the web, you can also use steam tables. However, since you have specified a turbine operating isentropically, then using a Mollier diagram (a special diagram of enthalpy versus entropy), the state line of this turbine will be a single vertical line running from the inlet point down to the exhaust pressure of 30 kPa. Unless you know the enthalpy of the inlet and exhaust steam, yes, you need to refer to steam tables.įor your exhaust condition, knowing the pressure is often not enough information. And that's your final answer.Firstly, lets assume the expansion is isentropic.įrom this, how do we find the power generated and final steam quality?ĭo we look at steam tables, W = m(flow) (h_1 - h_2) - is that helpful? So you actually get 1,882 foot pounds, Oops. So I need to divide by 12 to convert these inches in the inch pounds two ft in foot pounds. But I need to check units these are inch pounds, which are not units of work, units of work again, our foot pounds.
So now you're just plugging in and what you get is 22,588. So using just my normal immigration rules, I'm going to plug in and I will get let's see here, V. And the only thing that's left Is the to the -1.4 DV. Which means that we have 100,953 that's gonna come out. Okay, so uh I can plug into my integral which is the work is the integral from the initial volume to the final volume of P as a function of V. No matter how pressure and volume change, this will always be true. Okay, So that tells me and I know that that is always p times feed to the 1.4. And I know that my initial volume is a 100 And she's cute to the 1.4. And then I can turn P into a function of the, so I know that my initially Pressure is £160 print squared. So what I can do is I can plug in here based on my initial values and I can find what K. So um first I need P to be a function of V. Okay, so we're working with inch pounds right now basically. Now it is extremely important that we notice we are in inches but work is in units of foot pounds, not inch pounds foot pounds. And I know that V one equals 100 cubic inches And I know that V two or final volume is 800in cute. Uh I know that the initial pressure P one is 100 and 60 uh huh, pounds per square inch. I know that work is the integral from the first volume to the second volume of P as a function of the devi right. That's always going to be the same number.
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