me 405 professor john m. cimbala lecture 04 · 2019-09-04 · me 405 professor john m. cimbala...
TRANSCRIPT
![Page 1: ME 405 Professor John M. Cimbala Lecture 04 · 2019-09-04 · ME 405 Professor John M. Cimbala Lecture 04 . Today, we will: • Continue to discuss various forms of concentration](https://reader030.vdocument.in/reader030/viewer/2022040514/5e6e012b2d25523d4c33a2a4/html5/thumbnails/1.jpg)
ME 405 Professor John M. Cimbala Lecture 04
Today, we will:
• Continue to discuss various forms of concentration and conversions between them • Do some example problems – concentrations and conversions • Discuss relative humidity • Do an example problem – relative humidity • Begin a discussion about the human respiratory system
Recall, we defined mass concentration as jj
mc =
V = mass of species j per unit volume.
Typical units are 3mgm
, 3g
ftµ
, gLµ
, etc.
Similarly, we define molar concentration as molar,j
j
nc =
V = number of mols of species j per
unit volume. Typical units are 3molm
, 3molft
, molL
, etc.
![Page 2: ME 405 Professor John M. Cimbala Lecture 04 · 2019-09-04 · ME 405 Professor John M. Cimbala Lecture 04 . Today, we will: • Continue to discuss various forms of concentration](https://reader030.vdocument.in/reader030/viewer/2022040514/5e6e012b2d25523d4c33a2a4/html5/thumbnails/2.jpg)
![Page 3: ME 405 Professor John M. Cimbala Lecture 04 · 2019-09-04 · ME 405 Professor John M. Cimbala Lecture 04 . Today, we will: • Continue to discuss various forms of concentration](https://reader030.vdocument.in/reader030/viewer/2022040514/5e6e012b2d25523d4c33a2a4/html5/thumbnails/3.jpg)
Example Given:
• The bulk volume flow rate of an air/ammonia mixture is 1000 ACFM through a duct. • The air contains 5.0 PPM of ammonia vapor (Mammonia = 17.0). • The temperature is 200.oC (473.15 K) and the pressure is 90. kPa.
(a) To do: Calculate the ammonia mass concentration. Solution: (b) To do: Calculate the emission rate of ammonia into the atmosphere in g/hr. Solution:
![Page 4: ME 405 Professor John M. Cimbala Lecture 04 · 2019-09-04 · ME 405 Professor John M. Cimbala Lecture 04 . Today, we will: • Continue to discuss various forms of concentration](https://reader030.vdocument.in/reader030/viewer/2022040514/5e6e012b2d25523d4c33a2a4/html5/thumbnails/4.jpg)
![Page 5: ME 405 Professor John M. Cimbala Lecture 04 · 2019-09-04 · ME 405 Professor John M. Cimbala Lecture 04 . Today, we will: • Continue to discuss various forms of concentration](https://reader030.vdocument.in/reader030/viewer/2022040514/5e6e012b2d25523d4c33a2a4/html5/thumbnails/5.jpg)
Example Given: A hot summer day: • T = 95.0oF (35.0oC) • P = 99.6 kPa • RH = 90.0% (90% relative humidity)
To do: Calculate the mole fraction of water vapor in the air (in units of PPM).
Solution:
![Page 6: ME 405 Professor John M. Cimbala Lecture 04 · 2019-09-04 · ME 405 Professor John M. Cimbala Lecture 04 . Today, we will: • Continue to discuss various forms of concentration](https://reader030.vdocument.in/reader030/viewer/2022040514/5e6e012b2d25523d4c33a2a4/html5/thumbnails/6.jpg)
Example Given: The same hot summer day as in the previous example: • T = 95.0oF (35.0oC) • P = 99.6 kPa • RH = 90.0% (90% relative humidity) • Now the temperature drops rapidly to 86.0oF (30.0oC) • At the same time, the pressure drops to 98.5 kPa
To do: Calculate the new relative humidity of water vapor in the air and discuss.
Solution:
![Page 7: ME 405 Professor John M. Cimbala Lecture 04 · 2019-09-04 · ME 405 Professor John M. Cimbala Lecture 04 . Today, we will: • Continue to discuss various forms of concentration](https://reader030.vdocument.in/reader030/viewer/2022040514/5e6e012b2d25523d4c33a2a4/html5/thumbnails/7.jpg)