A dehumidifier removes water vapor from air and has been refer...

Question

A dehumidifier removes water vapor from air and has been referred to as a “refrigerator with an open door.” The humid air is pulled in by a fan and passes over a cold coil, whose temperature is less than the dew point, and some of the air’s water condenses. After this water is extracted, the air is warmed back to its original temperature and sent into the room. In a well-designed dehumidifier, the heat that is removed by the cooling coil mostly comes from the condensation of water vapor to liquid, and this heat is used to re-warm the air. Estimate how much water is removed in 1.0 h by an ideal dehumidifier, if the temperature of the room is 25°C, the water condenses at 8°C, and the dehumidifier does work at the rate of 550 W of electrical power.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with coefficient of performance. So in this problem, a condensation unit in a chemical plant operates for four hours at a power consumption of 700 watts. The surrounding temperature is 40 °C while condensation occurs at 15 °C. Assuming that the unit operates using an ideal carno cycle. How much of the condensation by-product can be collected. Given that the latent heated vaporization of water is 2260 kilojoules per kilogram. We're getting four possible choices as our answers. Choice A we have 29 kg. For choice B, we have 45 kg for choice C, we have 51 kg. For choice D we have 65 kg. Now, in this problem, we were told that we have a cardinal cycle and we're given the two temperatures. So it's gonna operate at the 40 degrees Celsius and the 15 °C. So let's start off by looking at our coefficient of performance. So the coefficient of performance which will label as cop for choc cycle is going to be equal to. And if you recall your formula that is gonna be TL divided by the quantity of th minus TL where TL here is the temperature of the low temperature reservoir and th is going to be the temperature of the high temperature reservoir. Now, we also have another formula for the coefficient of performance that is true for any um refrigerator or condensation unit. And so recall that formula that the coefficient of performance is going to be equal to RQL divided by WQL is the heat that's being transferred into or out of the low temperature reservoir. And W here is the work that's being done by the unit. So what we could do is combine these two formulas together because I'm ultimately wanting to know how much heat has been um taken out of the unit, which is gonna be my QL. And that's gonna allow me to calculate how much water has been collected. So we're gonna combine these two formulas together. So we'll have QL divided by W is going to be equal to TL divided by the quantity of th minus TL. So we can solve this for QL, we'll have QL is equal to W multiplied by TL divided by the quantity of th minus TL. Now, we were not given how much work this unit um does, but we were given the power consumption of the unit and how long it operates. And so we can actually use um that information to calculate how much work is being done. So the work being done by, the unit is just gonna be equal to the energy that's being consumed. And recall that that is just our power multiplied by time. So we'll have W is equal to P multiplied by T. And so we'll plug that into our formula or the heat. And so we'll have QL is equal to P multiplied by T that's multiplied by TL, divided by T minus to. So at this point, we can actually plug in values. So we'll have QL equal to. And for power, we are told that in the problem, that is the 700 watts for the time that it gets multiplied by. And that was given the problem as well. That is four hours. But I'm gonna need to convert this into seconds because a watt is a joule per second. And so if I want my energy in joules, I didn't even have my time in seconds. So I'll multiply this by the conversion factor of 3600 seconds divided by one hour. And this gets multiplied by the temperature of our low temperature reservoir, which in this case is going to be the 15 °C. However, my temperature needs to be in Kelvin. So to convert it into Kelvin, I'm gonna multiply my power and time by the quantity I'll have the 15 and to convert that 15 °C to Kelvin, I'll add 273 to it. That will give you my temperature in Kelvin, then this gets divided by the quantity of th minus TL. And so for th this is gonna be the 40 °C and so on the denominator here, I'll have the 40 minus RT L which is the 15 °C. So 40 minus 15. And since I'm taking a difference here instead of using degrees Celsius, that's also going to be in units of Kelvin. If I had done the conversion, I'd add 273 to both quantities. But since I'm subtracting, that conversion factor would get um canceled out. So this is sort of a shortcut. If you're taking the temperature difference, then you can just um calculated in degrees Celsius. And that change in degrees Celsius is the same as the change in um Kelvin because eight °C is the same size as a Kelvin. That's a little bit of a shortcut. So at this point, I have just values on the right hand side of the equation. So I can plug them into my calculator and I get QL is equal to 1.16 multiplied by 10 to the eight jules. And here I've just rounded to you three significant figures. Now, ultimately, what I want is the mass of the water. So this is the amount of heat that's been extracted. And so I can relate that um to the mass by using a formula for the latent heat of vaporization. To recall for that formula. We, we will have QL is equal to M multiplied by LVM. Here is gonna be the mass of the condensed water and LV is by latent heat of vaporization. So I need to solve this for the mass. So I'll just um divide over the LV. So I'll have M is equal to QL, divided by LV. And have values for both of those, we have M is equal to the QL. We just calculated that that is the 1.16 multiplied by 10 to the eight jewels. And I guess divided by the latent heat of vaporization or water, which we were given the problem, that was the 2260. And this had units of kilojoules per kilogram. But I actually need this in joules per kilogram. So to convert that, I'll multiply this um 2260 by 10 to the three, they'll give me units of Js per kilogram. And so when I plug those values into my calculator, I'll get a mass M that is equal to 51 kg. And here I've just rounded to two significant figures. And so when I check it, my answer choices, this corresponds to answer choice C. So just a quick little recap of what we've done here. We started off by using our definitions for the coefficient performance, both for that of a kno cycle and for a general refrigeration condensation unit. When we combined these two formulas together, we were able to calculate how much heat was being extracted out of the condensation unit. We then use that heat to find the mass of the condensed water used in our formula for the latent heat of vaporization. So I hope that this has been useful and I'll see you in the next video.

Key Concepts

Dew Point

Heat Transfer

Power and Energy Calculations

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