00:00:04.549 --> 00:00:12.879 In this screencast, we're going to look at a concentric heat exchanger system that has 00:00:12.879 --> 00:00:23.390 lubricating oil that needs to be cooled by water. In this system we're going to be given 00:00:23.390 --> 00:00:31.159 the outlet temperatures of both the oil and the water, and the inlet temperatures of the 00:00:31.159 --> 00:00:40.059 oil and the water, as well as the mass flow rate of the oil. And what we need to figure 00:00:40.059 --> 00:00:51.519 out is, what is the mass flow rate of the water going to be to cool this oil by a specific 00:00:51.519 --> 00:01:01.519 amount? So let's start with drawing the picture. This is what's known as a concentric heat 00:01:01.519 --> 00:01:10.450 exchanger. what you have is an inner tube where fluid flows, and in this case we're 00:01:10.450 --> 00:01:19.690 going to say that it's the oil, and around it is another tube where another fluid flows, 00:01:19.690 --> 00:01:29.020 and in this case it's going to be water. We're going to consider this a counter-current system, 00:01:29.020 --> 00:01:39.140 where the oil and water flow in different directions. So, if our mass flow rate of our 00:01:39.140 --> 00:01:51.920 oil is 0.1 kilograms per second, what does the mass flow rate of the water have to be 00:01:51.920 --> 00:02:06.080 in order to achieve this outlet temperature of the oil of 55 degrees C? Heat exchangers 00:02:06.080 --> 00:02:14.190 are generally adiabatic, which means that all the heat that's lost by one of the fluids 00:02:14.190 --> 00:02:22.360 is gained by the other fluid. In other words, there is no heat loss to the surroundings. 00:02:22.360 --> 00:02:31.530 So, our governing equation is going to be that the heat transfer rate, or q dot, is 00:02:31.530 --> 00:02:40.450 equal to the mass flow rate of the fluid, times the heat capacity of the fluid, times 00:02:40.450 --> 00:02:50.480 the change in temperature. Because the system is adiabatic, what we can say is that the 00:02:50.480 --> 00:03:01.040 mass flow rate of the oil, times its heat capacity, times the difference in temperature 00:03:01.040 --> 00:03:10.959 coming out versus coming in, has to equal the mass flow rate of the water, times its 00:03:10.959 --> 00:03:23.180 heat capacity, times the difference in temperature of the water. The heat capacities are values 00:03:23.180 --> 00:03:34.599 that can be looked up, and in this case the heat capacity of the oil is 2131 joules per 00:03:34.599 --> 00:03:49.260 kilogram degrees Celsius, and the heat capacity of the water is equal to 4178 joules per kilogram 00:03:49.260 --> 00:03:58.880 degrees Celsius. So let's set up this equation here, and write it in terms of what we want 00:03:58.880 --> 00:04:07.930 to find, which is the mass flow rate of the water. So the mass flow rate of the water 00:04:07.930 --> 00:04:18.690 is going to be that mass flow rate of the oil, times its heat capacity, times the change 00:04:18.690 --> 00:04:27.960 in temperature, and this is going to be all divided by the heat capacity of the water 00:04:27.960 --> 00:04:40.219 times its change in temperature. And let's fill in some numbers, and when we do the calculations, 00:04:40.219 --> 00:04:52.080 we find that the necessary mass flow rate of the water is 0.33 kg/s. Note that we didn't 00:04:52.080 --> 00:04:59.529 even have to calculate the heat transfer in this particular problem, but we easily could 00:04:59.529 --> 00:05:19.409 using either one of the fluids. So our q dot will equal 9590 joules per second. And just 00:05:19.409 --> 00:05:25.939 as a check, you might want to do the same thing for the water. And when we calculate 00:05:25.939 --> 00:05:42.330 this, we end up with 9650 joules per second. So why is there a difference between the two? 00:05:42.330 --> 00:05:51.309 This kind of difference comes from rounding the numbers, in particular this mass flow 00:05:51.309 --> 00:05:52.349 rate of the water.
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