00:00:04.569 --> 00:00:08.879 In this screencast we are going to set up and solve an energy balance on the condenser, 00:00:08.879 --> 00:00:12.410 and we will use sensible and latent heat to help us with the solution. 00:00:12.410 --> 00:00:17.410 If we want to calculate the cooling duty that is required to condense and cool acetone from 00:00:17.410 --> 00:00:23.109 100 C to 25 C at atm pressure, and we know the heat of vaporization for acetone at its 00:00:23.109 --> 00:00:27.300 normal boiling point is 30.2 kJ/mol. 00:00:27.300 --> 00:00:31.730 So by definition the normal boiling point is the boiling point at atm pressure. 00:00:31.730 --> 00:00:36.590 The process is illustrated here in the flow chart below, and we can see that we have both 00:00:36.590 --> 00:00:40.340 a change in temperature, as well as a change in phase. 00:00:40.340 --> 00:00:54.879 Both at which are listed in specific enthalpy, and we are trying to solve for Q. 00:00:54.879 --> 00:01:01.061 So our open system energy balance in this case reduces to Q is equal to delta H, which 00:01:01.061 --> 00:01:05.539 assumes that there is no change in kinetic or potential energy, and we can also write 00:01:05.539 --> 00:01:11.250 that q is equal to n dot, which is the molar flow rate time the change in specific enthalpy, 00:01:11.250 --> 00:01:14.520 which is the enthalpy on a per mole basis. 00:01:14.520 --> 00:01:18.730 Now if we had information about the specific enthalpy of acetone at any given temperature 00:01:18.730 --> 00:01:22.040 and pressure as well we do as water in the case that we have the steam table. 00:01:22.040 --> 00:01:26.710 This would be very easy we could look up the specific enthalpy of acetone at the outlet. 00:01:26.710 --> 00:01:30.950 Look at the specific enthalpy of acetone at the inlet and take the difference and substitute 00:01:30.950 --> 00:01:33.020 into our energy balance. 00:01:33.020 --> 00:01:37.229 Without that information available to use we have to get more creative and set up a 00:01:37.229 --> 00:01:43.220 hypothetical path that allows for the calculation of the specific enthalpy for this process. 00:01:43.220 --> 00:01:47.610 So the key property that allows us to set up a hypothetical path is that enthalpy is 00:01:47.610 --> 00:01:52.170 a state function, which means with a change an enthalpy it only depends on the initial 00:01:52.170 --> 00:01:53.979 and final condition. 00:01:53.979 --> 00:01:59.340 In this case our initial we have acetone in the vapor phase our temperature is 100 degC, 00:01:59.340 --> 00:02:06.899 and our pressure is in atm, and we are going to the liquid phase at 25 C, and we are keeping 00:02:06.899 --> 00:02:09.510 pressure constant at 1 atm. 00:02:09.510 --> 00:02:14.560 So our overall change in enthalpy for the process we can call delta H, and we can set 00:02:14.560 --> 00:02:19.840 up any number of hypothetical steps that is necessary to get from the same initial condition 00:02:19.840 --> 00:02:21.269 to the final condition. 00:02:21.269 --> 00:02:26.199 A key piece of information that is going to help us in this processes is what temperature 00:02:26.199 --> 00:02:27.570 does the phase change occur. 00:02:27.570 --> 00:02:30.880 We have the heat of vaporization at it's normal boiling point. 00:02:30.880 --> 00:02:33.739 So we need to know what temperature does that occurs at. 00:02:33.739 --> 00:02:39.010 In this case the normal boiling point for acetone is 56 degrees. 00:02:39.010 --> 00:02:43.650 We can look up this information in a variety of different text books or chemical engineering 00:02:43.650 --> 00:02:44.959 hand books. 00:02:44.959 --> 00:02:47.600 So we need to get to the temperature at which the phase change occurs. 00:02:47.600 --> 00:02:53.359 We can get there in one step by changing the temperature and keeping the phase constant. 00:02:53.359 --> 00:02:57.841 So we have an intermediate step here were we have the vapor at the boiling point, which 00:02:57.841 --> 00:03:02.729 is 56 degC, and we are keeping the pressure constant at an atm. 00:03:02.729 --> 00:03:04.940 At the boiling point we are going to change phase. 00:03:04.940 --> 00:03:06.900 So we have a second step. 00:03:06.900 --> 00:03:12.930 We are changing phase to the liquid phase at the same temperature, and our third step 00:03:12.930 --> 00:03:18.959 we are changing the temperature of the liquid from 56 to 25 while keeping the phase constant. 00:03:18.959 --> 00:03:21.200 So we have 3 hypothetical steps here. 00:03:21.200 --> 00:03:26.910 Each of these have there own change in specific enthalpy, which we will call delta H1, 2, 00:03:26.910 --> 00:03:28.310 and 3. 00:03:28.310 --> 00:03:32.880 And by the definition of the state function the overall enthalpy here is going to be the 00:03:32.880 --> 00:03:34.480 sum of these 3 steps. 00:03:34.480 --> 00:03:41.739 So we can coperate that into our energy balance, and write that Q is equal to n dot times delta 00:03:41.739 --> 00:03:46.550 H1, plus delta H2, plus delta H3. 00:03:46.550 --> 00:03:50.100 Two of our three step involve a change in temperature, and any time that we are adding 00:03:50.100 --> 00:03:51.880 heat. 00:03:51.880 --> 00:03:56.459 That causes a change in temperature we are referring to sensible heat. 00:03:56.459 --> 00:04:01.750 Sensible heat and the change in specific enthalpy associated with sensible heat is related to 00:04:01.750 --> 00:04:03.459 heat capacity. 00:04:03.459 --> 00:04:08.350 So we can calculate a change in specific enthalpy over a specific temperature interval. 00:04:08.350 --> 00:04:11.790 If we integrate over our temperature of integral. 00:04:11.790 --> 00:04:13.570 So from T1 to T2. 00:04:13.570 --> 00:04:19.100 We integrate the heat capacity, which is a function of temperature with respect to temperature, 00:04:19.100 --> 00:04:23.850 and the heat capacity and the dependence of temperature is generally expressed as a polynomial 00:04:23.850 --> 00:04:24.850 function. 00:04:24.850 --> 00:04:27.880 There are several different polynomials function that can be used. 00:04:27.880 --> 00:04:35.300 In table B2 in F and R the expression for the heat capacity that is used is; for acetone 00:04:35.300 --> 00:04:39.370 we can look up the values for a through d from table B2. 00:04:39.370 --> 00:04:44.570 So from table B2 we can see that the coefficient for acetone vapor and liquid are different. 00:04:44.570 --> 00:04:48.600 For both the vapor and the liquid the coefficients are given in the tables below, which we will 00:04:48.600 --> 00:04:51.590 use in our sensible heat calculations. 00:04:51.590 --> 00:04:54.790 In our first step we are cooling acetone vapor. 00:04:54.790 --> 00:04:57.690 From the inlet condition at 100 to the boiling point. 00:04:57.690 --> 00:05:04.750 So we can calculate delta H1 by integrating from 100 to the boiling point of 56. 00:05:04.750 --> 00:05:07.420 The heat capacity were we have 4 coefficients. 00:05:07.420 --> 00:05:11.311 If we evaluate this integral and put in our coefficients and limits of integration we 00:05:11.311 --> 00:05:19.540 will find that the sensible heat associated with this process is -3.2 kJ/mol. 00:05:19.540 --> 00:05:23.430 Our second step involve the change in phase, which I will hold off of for now, and go to 00:05:23.430 --> 00:05:26.810 the other sensible heat calculation, which is the cooling of the liquid. 00:05:26.810 --> 00:05:28.810 We want to calculate delta H3. 00:05:28.810 --> 00:05:34.630 We are integrating the heat capacity from the boiling point of 56 to the outlet condition 00:05:34.630 --> 00:05:38.700 at 25, and here we have 2 coefficients. 00:05:38.700 --> 00:05:44.780 So we are integrating a plus bT with respect to temperature, and if we evaluate this integral 00:05:44.780 --> 00:05:49.450 and plug in the coefficients we find that the sensible heat for this step is negative 00:05:49.450 --> 00:05:51.080 4.06 kJ/mol. 00:05:51.080 --> 00:05:55.990 The last change in enthalpy that we need to calculate which is delta H2. 00:05:55.990 --> 00:05:58.500 Which is associated with the phase change. 00:05:58.500 --> 00:06:03.600 Anytime that we are adding heat and we are changing phase we are referring to the latent 00:06:03.600 --> 00:06:04.600 00:06:04.600 --> 00:06:10.240 We know some information about the phase change, and that heat that is absorbed when acetone 00:06:10.240 --> 00:06:14.150 is vaporized at the boiling point, and at the beginning we were given the heat of vaporization 00:06:14.150 --> 00:06:23.220 as 30.2 kJ/mol, and that is specifically at the normal boiling point at 56 deg and atm 00:06:23.220 --> 00:06:24.220 pressure. 00:06:24.220 --> 00:06:25.220 So in this case we are condensing. 00:06:25.220 --> 00:06:30.030 We are not vaporizing, but we can still use this information if we consider that enthalpy 00:06:30.030 --> 00:06:31.310 is a state function. 00:06:31.310 --> 00:06:37.420 If we consider the path of heat of vaporization represents the transition from a liquid to 00:06:37.420 --> 00:06:44.090 a vapor, and a state function by definition only depend on the initial and final condition. 00:06:44.090 --> 00:06:49.090 So if we reverse our final and initial condition, and we consider our path in this case to be 00:06:49.090 --> 00:06:53.680 the transition from a vapor to a liquid then the change in enthalpy associated with this 00:06:53.680 --> 00:06:58.680 path must be equal to the opposite of the heat of vaporization. 00:06:58.680 --> 00:07:04.450 To calculate delta H2 it is going to be equal to the opposite of the heat of vaporization 00:07:04.450 --> 00:07:11.150 at the same temperature, which is 56, and so that is -30.2 kJ/mol. 00:07:11.150 --> 00:07:16.870 So scrolling down I have redrawn our hypothetical path, and I have shown the results from our 00:07:16.870 --> 00:07:19.190 sensible and latent heat calculation. 00:07:19.190 --> 00:07:22.110 Which we can use to calculate our Q. 00:07:22.110 --> 00:07:28.430 So we showed that energy balance is Q is equal to n dot times the sum of each specific enthalpy 00:07:28.430 --> 00:07:30.260 change for our steps. 00:07:30.260 --> 00:07:31.580 So plugging in our values. 00:07:31.580 --> 00:07:39.020 We know that the molar flow rate is 100 mol/sec, and we have 3 change in enthalpies that we 00:07:39.020 --> 00:07:40.300 calculated. 00:07:40.300 --> 00:07:47.930 So that gives us a final value for our cooling duty of negative 3808 kJ/s. 00:07:47.930 --> 00:07:52.810 Here the negative makes sense because we are drawing energy from the system and we can 00:07:52.810 --> 00:07:57.230 see from our hypothetical path that most of the cooling duty that is required is associated 00:07:57.230 --> 00:07:58.199 with a change in phase.
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