Thermodynamics problem with relative humidity

[luysion]

Homework Statement

After taking a deep breath, the mass of air expelled when I exhale is approximately 2.5g. The intake air is at 18deg dry bulb temperature at 60% relative humidity, and the air exhaled is at 100% relative humidity at a temperature of 32deg. The net mass of water expelled during one breathing cycle is (in mg)

a)49
b)58
c)65
d)73
e)81

Homework Equations

The Attempt at a Solution

I used the psychometric chart, and I found the M.C for both cases and calculated the mw for inhaling and exhaling, found the difference, but my answer was not the same as the answer given.

could someone please show me to how to do this question cheers

 

[anathema]

Hello,

Thank you for sharing your attempt at solving this problem. It is great to see that you have already tried using the psychometric chart to solve this problem. However, I believe there may be a mistake in your calculation. Here is how I would approach this problem:

First, let's find the mass of water in the intake air. We can do this by finding the specific humidity (ω) using the given dry bulb temperature and relative humidity. Using the psychrometric chart, we can see that at 18°C and 60% relative humidity, the specific humidity is approximately 0.006 kg/kg.

Next, we can use the formula mw = ω * ma, where mw is the mass of water, ω is the specific humidity, and ma is the mass of air. In this case, the mass of air we are inhaling is 2.5g (or 0.0025 kg). Plugging in these values, we get mw = 0.006 * 0.0025 = 0.000015 kg.

Now, let's find the mass of water in the exhaled air. Using the same formula, we can see that the mass of water in the exhaled air is given by mw = ω * me, where me is the mass of exhaled air. We know that the total mass of air we are exhaling is 2.5g, but we also know that this includes the mass of water vapor. So, we need to subtract the mass of water vapor from the total mass to get the mass of dry air. The mass of water vapor can be found using the psychrometric chart again, at 32°C and 100% relative humidity, the specific humidity is approximately 0.025 kg/kg. Therefore, the mass of dry air is 2.5g - (0.025 * 2.5g) = 2.4375g (or 0.0024375 kg). Plugging this into the formula, we get mw = 0.025 * 0.0024375 = 0.0000609375 kg.

Finally, to find the net mass of water expelled during one breathing cycle, we simply subtract the mass of water in the intake air from the mass of water in the exhaled air. This gives us 0.0000609375 - 0.000015 = 0