0
$\begingroup$

I major in chemical engineering, and I'm solving a example of unit operations. The textbook is Warren L. McCabe, Unit Operations of Chemical Engineering 7th ed. Here is example 24.1. enter image description here

And this is its solution.

enter image description here

From this, it says that the density of air is 29/369 × 492/620. But it omitted the untis, so I can't understand the equation. I assume that this equation is probably ρ = PM/RT, since molecular weight of air is 28.97 g/mol when I searched for it. But the numbers 369, 492, 620 are still mystery. What is those?

$\endgroup$
0
1
$\begingroup$

If you look at Text equation 1.29, the 492/620 is Temperature in Rankine.

$$T°F = T°R - 459.67$$

Ice point is 491.67°R. 160°F = 619.67°R.

If you do the math, it does not equal 0.0641, because the 369 is wrong. The math to get 0.0641 means 369 should be 359.

1 lb-mole of an ideal gas at 0°C and 1 atm occupies 359 $ft^3$ absolute.

The molecular weight of air is 29 lb/lb-mole.

$$\frac {29 lb/lb-mole}{359 ft^3/lb-mole} \times \frac {492°R}{160°F + 460°R} = 0.0461 lb/ft^3$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.