November 18th, 2014, 03:58 AM  #1 
Newbie Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0  Evaporation from changing surface
Hey everybody. I have folowing problem. I want to calculate mass of stream which is evaporating from surface, which is changed in time. Basic formula for this problem is: qs=OxSx(xsx) [kg/s] where O is coeficient of evaporation S is evaporation surface xs is humidity ratio in satured air x is humidity ratio in the air I am filling vassel, which has shape like cone with cut off the top, with hot water. Inlet diameter of cone is 15mm, outlet diameter is 255mm and hight is 200mm. Flow rate is 300l/h. For better understanding of my problem, I attached following link: Evaporation from Water Surfaces How I should modify this furmula to solve my problem? Thank you very much. 
November 18th, 2014, 04:11 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,087 Thanks: 700 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
The formula you have seems like the correct one to use. I would just use the surface area of the aperture of your vessel for the calculation and not worry about the change in surface area caused by filling it up with water, at least as a first approximation. If you definitely know that the rate of evaporation is changing with time because the vessel is filling up with water, then you can calculate the crosssectional area as a function of height up the vessel and calculate the water level as a function of time to obtain the surface area as a function of time. 
November 18th, 2014, 05:49 AM  #3 
Newbie Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0 
Thank you. Sorry, but I forgot the temperatute of the water is 50°C. I am sure the rate of evaporation is changing with time. I do not understant why I should calculate first the crosssectional area as a function of height up the vessel if I interesting for water surface? I appreciate your answer, but first I want to understand how I can calculate the surface area as a function of time? Sorry for my English. 
November 19th, 2014, 02:52 AM  #4  
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,087 Thanks: 700 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
No worries dude. Quote:
The formula quoted for the rate of evaporation uses the surface area of water exposed to the air. If this crosssection is constant, your rate of evaporation would be constant. If you have a sort of "beaker" that is filling up with water, the surface area exposed to air changes as the beaker fills up. This will change the rate of evaporation. If you're not sure what I mean, think of it this way. Would you expect more evaporation from a bucket of water or a vial of water? If you have a coneshaped beaker, then the more water that is inside the beaker, the lower the surface area exposed to air at the top.  
November 20th, 2014, 05:27 AM  #5 
Newbie Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0 
Problem is more difficult. I know, if I have vessel with shape like cilinder it is easy to solve evaporation from surface area of water, because surface area of water does not change. But, how is flow rate of evaporation affected by changing surface area? At the begining I want to simplify my problem. First thing is changing surface area of water. For second, it is changing humidity ratio in air, because vessel has big internal volume for a such small outlet hole. I want send you the describe picture, but I do not how? I am new member is there some posibility send it? 
November 20th, 2014, 05:50 AM  #6 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,087 Thanks: 700 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
I would just attach it to a post in this thread. If you click on "advanced" and scroll down to additional options, you can put attachments there. If you have a URL, you can also click on the "attach image" button and post the URL there.

November 20th, 2014, 02:28 PM  #7 
Newbie Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0 
I found it. Thank you

November 21st, 2014, 02:58 AM  #8 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,087 Thanks: 700 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Okay...interesting. Unfortunately the diagram does not make it obvious which surface area is exposed to air and where air/water meets. Is the exposed area at the very top part of the image?

November 21st, 2014, 05:58 AM  #9 
Newbie Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0 
Surface area of water is expose to air from bottom to top. Bottom level and top level limited from area of interesting (attached picture above). Surface area is changing as you filling up vessel from bottom to top with flow.

November 25th, 2014, 11:21 AM  #10 
Newbie Joined: Nov 2014 From: Slovakia Posts: 10 Thanks: 0 
Thank you, but I am not sure about this formula. First: 300L/h=0,0833mm3/s (this is not a big deal) Second: How can I add units (s2+m2) [between first brackets] this do not make a sense Sure in t=0 is surface area of water S=3,14xr2 but I think, it will be better devide volume of vessel to small volumes. First will be cilinder (cilinder represent inlet pipe). Second will be cone with cut off top (in this area where surface of water will be changing by time), thirt will be cilinder or pipe and etc. What do you think about it? Maybe I am wrong. 

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