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Skip to 0 minutes and 4 seconds We’re going to look at the design of a hand dryer that’s for use in public toilets. It generates two sheets of air that come out the hand dryer 200 metres per second, at approximately room temperature, and about 30 centimetres above their intake. Now, we’re told that the total flow rate is 32 litres per second. And the manufacturers claim it will dry your hands in about 12 seconds. And we’re asked to calculate the energy used to dry your hands, stating any assumptions made. So in just the same way that I tackled the questions last week, I’m going to draw a picture, first of all, and collect all the information together.

Skip to 0 minutes and 43 seconds And so I’m going to draw an idealised diagram of this hand dryer. I’ll try and make it roughly symmetrical. So I’ve got two jets of air coming out the top there.

Skip to 1 minute and 0 seconds Something like that. And I’ve got an axis of symmetry running down here. And the air flow in the bottom here is capital V for volume flow rate, and it’s 32 litres per second. And it’s going out the top here with a velocity. And I’m going to use lower case v for that. And that is 200 metres per second. So that’s the velocity coming out here and here, in effect. And we’re told that these exits here are 30 centimetres above the intake. So that’s the basic information that we’re told. And so I’m going to assume that the air is an ideal gas, and this is a steady flow situation. The air goes in one end and comes out the other.

Skip to 1 minute and 56 seconds It’s not accumulating in any way. And so I’m going to assume that air is an ideal gas, and use the steady flow energy equation. And so let’s write that out in its full form. So we have the mass flow rate m dot multiplied by a series of terms here. First of all, the enthalpy, h2 minus h1, plus the change in the kinetic energy. So that’s a half of v2 squared minus v1 squared, plus the change in the potential energy. So that’s g. And I’m going to use z2 and z1 as my vertical dimension. We need a square bracket around all of that, because it’s multiplied by the m dot.

Skip to 2 minutes and 58 seconds And that’s going to be equal to the net heat transfer minus the net work transfer. So not all these terms are going to have values in this particular problem, so we can set a few of them equal to 0. So first of all, this one will go to 0 because there’s no heat source here. Zero because no heat source. We’re told that the air comes out at roughly room temperature. So there’s no heat transfer opportunity going on. This volume flow rate will be going in through a large inlet. And just away from the inlet, it’ll be at approximately zero velocity.

Skip to 3 minutes and 42 seconds And so relative to the high speed it’s coming out at, we can say that this term here will be zero, as well. Zero due to large inlet.

Skip to 3 minutes and 56 seconds And then this term here will also go to zero because there’s no change in the temperature or the pressure through this system here. So this is zero due to constant temperature and pressure.

Skip to 4 minutes and 20 seconds So that doesn’t leave us with very many terms left to consider. We’ve got this one here. We’ve got the potential energy term, and we’ve got the work term here. So we can rewrite this expression as minus W dot is equal to the m dot here with the square bracket. And what we’re left with is the v2 squared upon 2, that plus g into z2 minus z1. OK? So we need to find out what this mass flow rate is, and we’re given a volume flow rate here. And so to do that, and a mass flow rate, which I’ll call m dot. The dot indicates that it’s a mass per unit time.

Skip to 5 minutes and 16 seconds It’s going to be the density of air times the volume flow rate. The density of air, it can be taken as standard as 1.2 kilogrammes per cubic metre times the 32 here. But this is in litres, and this value of density’s in kilogrammes per cubic metre. So we need to divide this by 1,000 in order to put it into metres per second. And so that gives us a value here for m dot. If get your calculators out, you’ll find that’s 0.0384 kilogrammes per second. And so now we can start substituting values into here. So we can say substituting values, we’re after W dot, the work we need to power this thing. So the m dot we’ve just found.

Skip to 6 minutes and 15 seconds That’s 0.0384 multiplied by, so now we need our velocity here, so that’s a 200 from up there. 200 squared upon 2, plus g, which is 9.81, multiplied by the height difference. And we’re told that’s 30 centimetres. So in metres, that’s 0.3. And so if we get our calculators out again and work that out, it comes out at 768.1 watts. So that’s the energy used per second. And so now we need to do it for a 12-second period while you dry your hands. So we can say for a 12-second period, total energy required will be W dot times t, or 768.1 times in 12 seconds. And that works out in round numbers to 9,200 joules.

Skip to 7 minutes and 35 seconds So that’s the amount of energy we need to run our hand dryer for the 12-second period required to dry your hands.

Hand dryer (worked example)

A hand-dryer, designed for use in public toilets, generates two 200m/s sheets of air at approximately room temperature and at about 30cm above the air intake. The total flow rate is 32 litres/second and the manufacturers claim it will dry your hands in 12 seconds. Calculate the energy used by the dryer to dry your hands, stating any assumptions made.

Try solving this question yourself and then watch the video or check your answer against the solution PDF at the bottom of the page.

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Energy: Thermodynamics in Everyday Life

University of Liverpool