# Homework Help: Condensation of your Breath

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1. Feb 28, 2016

### Raptor112

1. The problem statement, all variables and given/known data
On a winter day, suppose your breath has a temperature of 30°C and a dew point of 28°C,and it mixes with the ambient air of temperature –10°C and a dew point of –11°C. Will you see your breath? Assume you are at sea level, and that your breath and the environment mix in proportions of 2 parts (by mass) breath to 1 part (by mass)
environment. Show all your working.

2. Relevant equations
$\frac{p_1}{p_2} = e ^{\frac{L}{R_s}(\frac{1}{t_2}- \frac{1}{t_1})}$
$Relative Humidity = \frac{actual vapour pressure}{saturated vapour pressure} *100$
3. The attempt at a solution
Found the relative humidity of the breath and air by using, for air:
$t_1 = -10, t_2 = -11$
and for breath
$t_1 = 30, t_2 = 28$
which gives
Relative humidity of air =114%
Relative humidity of breath= 108%
using $L =2.5 *10^6 Jkg^-1$
$R_s = 461 J K^-1 kg^-1$

2. Feb 28, 2016

### BvU

Hello Raptor,

Ineresting: in both cases $t_1>t_2$ yet one of the two is over 100% and the other is below ?

Willl you see your breath? was the exercise question, so you till have some steps to take, right ?

What kind of assistance do you need, at what point ?

3. Feb 28, 2016

### Raptor112

In both cases the humidity is over 100%.

Assuming my humidity calculation is correct, not sure how to proceed from there? How does the relative humidity of both air and breath tell us if the breath will condensate or not?

4. Feb 28, 2016

### BvU

My mistake... ? Can you show your working ?
(not that I don't trust your calculations -- although I do get a lightly different answer), but I wonder about your relevant equation: I expect the relative humidity to go down as $t_1$ increases...

Last edited: Feb 28, 2016
5. Feb 28, 2016

### Raptor112

Breath:
$\frac{actual vapour pressure}{satrated pressure} = exp [\frac{2.5*10^6}{461} (\frac{1}{28+273.15} - \frac{1}{30+273.15)}] *100 = 113$
Air:
$\frac{actual vapour pressure}{satrated pressure} = exp [\frac{2.5*10^6}{461} (\frac{1}{-11+273.15} - \frac{1}{-10+273.15)}] *100 = 108$

6. Feb 28, 2016

### Staff: Mentor

This is a problem in combining two masses of air having different temperatures and mass fractions of water into a single equilibrium mixture of uniform temperature and water mass fraction. It is easiest to do this using Psychometric tables and charts for air/water. Are you allowed to use these tables?

Incidentally, the relative humidity of both streams is less than 100% to begin with.

Chet

7. Feb 28, 2016

### Raptor112

8. Feb 28, 2016

### Staff: Mentor

Well, be that as it may, that's not what your problem statement says. And, since it's a made-up problem, we're not going to let ourselves get bogged down on that, are we?

The equilibrium vapor pressure of water at 28 C is 3780. Pa and the equilibrium vapor pressure of ice at -11C is 237.74 Pa. There is no need to calculate the relative humidity in this problem.

Now, are you allowed to use tables or graphs of the physical properties or air containing water vapor in this assignment, are are you expected to do the calculations from scratch? We are going to have to calculate the enthalpy per unit mass of each of these streams, and also the enthalpy per unit mass of the combined stream.

Chet

9. Feb 29, 2016

### Raptor112

As Psychometric tables were not covered, I think this has to be caluclated from scratch.

10. Feb 29, 2016

### BvU

You're still in good shape. Start doing the work on mixing these two streams as Chet proposes.

11. Feb 29, 2016

### Raptor112

Found the mixed air temperature to be 16.67 degrees with pressure of 1911.04 Pa....

12. Feb 29, 2016

### Staff: Mentor

Please show us your work. Also, if you did this correctly, what is your conclusion as to whether you see your breath?