 Problem two. Water flows into and out of the rigid tank shown. The water at the inlet is 400 degrees Celsius and 100 kilopascals while the water at the outlet is 200 degrees Celsius and 2500 kilopascals. The inlet diameter is 2 centimeters, the outlet diameter is 4 centimeters, and you measure the velocity at the inlet and outlet to both be 3 meters per second. Determine the mass flow rate at the inlet and outlet and the time rate of change of the mass contained within the tank. So in this rigid tank, I know that if I were to call the inlet state 1, which the diagram has done for me, and the outlet state 2, then T1 is 400 degrees Celsius, which would be 673.15 Kelvin. P1 would be 100 kilopascals or 1 bar and the velocity at the inlet was 3 meters per second and the diameter at the inlet was 2 centimeters. Which would be 0.02 meters. Then at state 2, I know the temperature at the outlet is 200 degrees Celsius, which is equal to 473.15 Kelvin. I know P2 is 2,500 Kelvin. I know the velocity at the exit is also 3 meters per second. And that the diameter at state 2 was 4 centimeters. Which would be 0.04 meters. Any other given information? Well, I know that this is water, and I know it's a rigid tank. So rigid tank here is going to mean that there's no change in volume. Awesome. I know a lot of information about the inlet and the outlet, but I don't know the mass flow rate. So the first thing I'm going to try to do is calculate the mass flow rate and the mass flow rate calculation is going to come from the fact that I know the velocity, I know the diameter, and I know enough information to figure out the specific volume. If I were to consider either the inlet or the outlet, I have water flowing through it and that water is going to have a flow profile that looks something like this maybe. So the velocities higher the center, very low close to the edges, but it doesn't matter because what I have is the average velocity. And then I use the fact that I know the diameter to calculate the area. It's the area of a circle. So once I know the velocity and the area, I can calculate a volumetric flow rate. So the volumetric flow rate of a cylindrical tube in a cylindrical tube would be the velocity times the cross-sectional area. This is average velocity and then the mass flow rate would come from the fact that I could take the density of our substance multiplied by our volumetric flow rate. So the more convenient form of that here would be volumetric flow rate divided by specific volume. So I could calculate the mass flow rate by taking average velocity times cross-sectional area divided by a specific volume and that specific volume will come from the tables. I know this is water and I know two independent intensive properties at both states therefore I can look up anything else I need. So at state one I know T1 is 400 degrees Celsius and P1 is 100 kilopascals. I'll use that to determine the specific volume and then I'll use that to determine the mass flow rate. I had state two. I also know the temperature and pressure. So T2 was what was it? 200 degrees Celsius and P2 was a much higher 2500 kilopascals. That'll give me V2 which will allow me to calculate mass flow rate too. So jumping over to the tables if I go back to my appendix I'm going to want SI units which is page 925 and I want properties of saturated water by temperature or pressure. So if we go over to 927 we have properties of water by temperature. So the first thing I have to do here is fix the phase of my water. So I could either look up the pressure and compare my temperature to the saturation temperature or I could look up the temperature and compare my pressure to the saturation pressure. So just for fun here, I'm going to look up my pressure. So I'm actually going to use the properties of saturated water listed by pressure. Remember that these two tables are the same thing. It's just listed by a different piece of information. So at state one I have a pressure of 100 kilopascals and I know that 100 kilopascals is one bar. So I go to my tables and look up a pressure of one bar. I can see that I have a saturation temperature of 99.63. So my temperature is 400 degrees Celsius which is above my saturation temperature. Therefore the phase at state one is a superheated vapor. So T1 is greater than T sat at P1. Therefore superheated vapor. That also means that I'm going to be using a different table. I need to find my superheated vapor tables. Which is conveniently on the next page. So for the superheated vapor tables I need to look up a pressure. These sub-tables are for an individual pressure and once I find my sub-table I can look up my temperature and get a value. So my pressure was 100 kilopascals which is one bar which is right here. This sub-table is let me zoom in a little bit. It might make it easier to read. Okay too much zoom in. So one bar is right here and then 400 degrees Celsius is my temperature. So specific volume was my first column in this sub-table. So the specific volume corresponding to a pressure of one bar and a temperature of 400 degrees Celsius for water would be 3.103. Scroll up read that the units are meters cubed per kilogram. So 3.103, 3.103, v1 is 3.103 cubic meters per kilogram. And that was from table A4. Now I'm going to repeat the same process for state 2. Zoom back out a little bit. Okay so I can look up either the temperature or pressure on the saturated water tables and then compare the other property to the saturated property. So I'm going to look up 2500 kilopascals which would be 25 bar. So at 25 bar I have a saturation temperature of 224 degrees Celsius. My temperature at state 2 is 200 degrees Celsius. So my temperature is below the saturation temperature. Therefore my phase is a compressed liquid. Oh by the way I forgot to switch over. So here on the textbook 25 bar has a saturation temperature of 224 degrees Celsius. There you didn't miss anything. See I just highlighted to a 24. Anyway my phase is a compressed liquid. So T2 was less than T sat at P2. Therefore compressed liquid. Which means I'm going to have to use the compressed liquid tables. Keep forgetting to switch. Okay I just wrote this part down. Tables, compressed liquid tables are on table A5. So conveniently the first sub table, let me zoom in. For compressed liquid tables again I have sub tables for each pressure so I can look up my pressure first, find my temperature, and get a value. So my first sub table is at 25 bar. So my temperature was 200 degrees Celsius. I read off that the specific volume corresponding to 200 degrees Celsius and 25 bar is 1.1555. And this is not in units of cubic meters per kilogram. Note that the units specified here are V times 10 to the third. So this value is the specific volume with three times 10 to the third. So in order to get it back to the actual specific volume I have to take this number and multiply it by 10 to the negative third. Which is moving the decimal place over three times. So my number was 1.1555. Which means that my specific volume at state two would be zero point. Zero zero one one five five five and that was table A5. So now that I had the specific volumes I could calculate a mass flow rate. So the mass flow rate at state one would be my velocity which was three meters per second. Three meters per second multiplied by my cross-sectional area. So because it's a circle and I know the diameter instead of using pi r squared I'm actually going to be substituting that radius is equal to diameter over two. So this is going to be I bring the diameter out and square it. This is pi over four times diameter squared. It's just a little bit more convenient. One fewer calculation. So three meters per second times pi over four times diameter squared. So the diameter at state one was zero point zero two meters quantity squared. And then I divide that number by my specific volume which I just learned was three point one zero three cubic meters excuse me cubic meters per kilogram. So my meters and square meters are going to cancel cubic meters which is going to give me an answer in kilograms per second which is what I want. So the mass flow rate at state one is going to be three times pi over four times point zero two squared divided by three point one oh three. So three times pi over four times zero times zero point zero two squared times one over three point three point one oh three gives me an answer of point zero zero zero three zero four. So my mass flow rate at state one is zero point zero zero zero three zero four kilograms per second which is my mass flow rate in I can just copy that over here zero point zero zero zero three zero four. Then mass two is much the same process I'm just going to have the same velocity so three meters per second times pi over four times this four centimeters this time. So zero point zero four meters squared divided by my specific volume which was zero point zero zero one one five five five cubic meters per kilogram. So while I'm typing in numbers for forever think about how such a small denominator will affect my answer. Would you expect a mass flow rate at two to be greater than or less than state one? Three times pi over four times the quantity I typed that incorrectly this time but not that awesome squared times one over zero point zero zero one one five five five. So my mass flow rate at state two then is three point two six two five eight. So all of those of you at home that said bigger than congratulations much bigger than three point two six two five eight. I don't need nearly that many decimal points. I'm a bad example but regardless I have answers to part A and B now I can move on to part C. So part C is asking me for the time rate of change of the mass contained within the tank. So for part C what I'm actually looking for is dm dt and that's going to come from a mass balance. So just like with energy if I know how much mass is entering and I know how much mass is leaving I know what the change in mass is. So if you saw two people go into a house and then three people come out you know that the rate of change excuse me you know that the change in number of people in the house was negative one. So if I divide all three terms here by a small amount of time I get dm dt and m dot in and m dot out. So I can figure out my time rate of change of mass by just taking my mass flow rate in which was zero point zero zero zero three zeros yes three zeros let me get my calculator out of the way zero zero zero three zero four kilograms per second minus three point two six two five eight kilograms per second. So I get an answer of negative three point two six two two eight kilograms per second. So for every second that this is operating the water in the tank drops by three and a quarter kilograms. So because my outflow is so large the inlet of water is almost insignificant and that's problem two.