Learning undergraduate engineering thermodynamics might be less painful with a blog. I hope that students, faculty and interested observers will share their thoughts on the laws of thermodynamics, phase and chemical equilibrium and many related topics.
Please post any questions or discussion related to this problem as comments on this message. Feel free to answer other students' questions. I will check the blog M-F and once on the weekend. Dr. B
A student emailed this question to me before I had this blog entry posted.
"I do not understand what problem 9 is asking or where to begin. Are the particles that enter as liquid drying into solids and exiting through the dry particle path? And what does this have to do with the problem?"
The feed that enters the spray dryer is a slushy stream of liquid water with a lot of solid particles floating in the water. A suspension of solids in a liquid is called a slurry. As this slurry falls through the spray drying tower, it comes into contact with hot, dry air and ALL of the water evaporates. The humid air leaves the top of the tower. The completely dry solids leave at the bottom of the tower.
The key to part (a) is that the molar flow rate of dry air (bone dry air or BDA) that enters the column must be the same as the molar flow rate leaving the column. Use mole fractions, humidity and partial and partial pressures to determine the molar flow rate of BDA into the tower. Then, use the IG EOS to convert the molar flow rate into a volumetric flow rate in m^3/min.
The solids are more important in part (b). Here, the keys are that ALL of the water is removed from the slurry, leaving dry solids AND that water makes up 30% of the mass that enters the spray dryer in the slurry.
I still don't really understand how molar fractions help to determine what fraction of 310 m^3/min is air. Are we supposed to assume that water vapor and air have the same molar volume?
anonymous: I suggest you treat the humid air leaving the spray dryer as an IG. This lets you use Pw = yw * Ptot. This will help when you use the definition of relative humidity, because Pw is int he numerator. It also allows you to use the IG EOS to convert the volumetric flow rate into a total molar flow rate.
I still don't really follow. I see how what you've described will yield a molar fraction for water. But I'm still at a loss about how to get the total molar outflow rate -- thats the part that doesn't make sense (I also see how, once you get that molar outflow rate, you can apply the IG EOS).
anon: PV=nRT, right ? Well, P*Vdot = ndot*R*T, where Vdot is the volumetric flow rate and ndot is the total molar flow rate. You are given Vdot for the humid air. Solve for ndot. ndot * yw = nw = molar flow rate of water OUT of the spray dryer in the humid air. This must also be the molar flow rate of water INTO the spray dryer in the slurry because the entering air and the dry solids leaving contain no water.
7 comments:
A student emailed this question to me before I had this blog entry posted.
"I do not understand what problem 9 is asking or where to begin. Are the particles that enter as liquid drying into solids and exiting through the dry particle path? And what does this have to do with the problem?"
The feed that enters the spray dryer is a slushy stream of liquid water with a lot of solid particles floating in the water. A suspension of solids in a liquid is called a slurry. As this slurry falls through the spray drying tower, it comes into contact with hot, dry air and ALL of the water evaporates. The humid air leaves the top of the tower. The completely dry solids leave at the bottom of the tower.
The key to part (a) is that the molar flow rate of dry air (bone dry air or BDA) that enters the column must be the same as the molar flow rate leaving the column. Use mole fractions, humidity and partial and partial pressures to determine the molar flow rate of BDA into the tower. Then, use the IG EOS to convert the molar flow rate into a volumetric flow rate in m^3/min.
The solids are more important in part (b). Here, the keys are that ALL of the water is removed from the slurry, leaving dry solids AND that water makes up 30% of the mass that enters the spray dryer in the slurry.
I hope this is helpful.
I still don't really understand how molar fractions help to determine what fraction of 310 m^3/min is air. Are we supposed to assume that water vapor and air have the same molar volume?
anonymous:
I suggest you treat the humid air leaving the spray dryer as an IG. This lets you use Pw = yw * Ptot. This will help when you use the definition of relative humidity, because Pw is int he numerator. It also allows you to use the IG EOS to convert the volumetric flow rate into a total molar flow rate.
I still don't really follow. I see how what you've described will yield a molar fraction for water. But I'm still at a loss about how to get the total molar outflow rate -- thats the part that doesn't make sense (I also see how, once you get that molar outflow rate, you can apply the IG EOS).
im so lost... period
anon:
PV=nRT, right ?
Well, P*Vdot = ndot*R*T, where Vdot is the volumetric flow rate and ndot is the total molar flow rate. You are given Vdot for the humid air. Solve for ndot. ndot * yw = nw = molar flow rate of water OUT of the spray dryer in the humid air. This must also be the molar flow rate of water INTO the spray dryer in the slurry because the entering air and the dry solids leaving contain no water.
Another:
Ouch. Check the solution tomorrow. Maybe that is the best way to go at this point.
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