hi..i have a question in different subject if you can help in evaporative cooling...is it useful to decrease the temp of water spry or water washer ex: if out air 36c,60%rh..on psy with 90% evaporative efficiency.h ll get a 28c db and 90%rh.. so if i enter a 7c cold water to the washer .will i get a cold air temp....with knowing that in 100%humidity the db is about 27c thanks
Why is the method not valid when 1-2 crosses the 100% RH curve? If your point lies in the super saturated zone can't you just follow the line of constant enthalpy to find your state point, then calculate how much water condensed out?
As an example, consider 35C air at 95% relative humidity mixing with 10C air at 90% relative humidity, with 1kg/s mass flow rates each. If you blindly follow this method, you'll end up finding a fictional state of 20C and 108% relative humidity for this air above the saturation curve, by doing a weighted-average. Following the isenthalp to the saturation curve would give you 21C. The correct answer is saturated air at 23C, with 6.4 g/s of condensate. The reason it is hotter than 21C that your method would find, is that the energy from condensing the water, heating it slightly above the weighted average of 20C, where it would have an equilibrium, if it didn't touch the saturation line. The proper way to calculate mixing of air states that would cross the saturation line, is to consider the path that the air would take on this chart. The hot air will move horizontally to the left until it hits the saturation line. Then the hot air would follow the saturation line until the equilibrium point. Then you would find the line that is tangent to the saturation curve, and joins the initial state of the cold air (this is the hard part to find, in my example this happens at 14C). This is the path that this mixing process will follow. The mixed state will occur either A) on the line from state 2 to the tangent point on the saturation curve, B) on the saturation curve, or C) on the horizontal from the saturation curve to state 1. It will be at a point where the enthalpy loss of the hot air plus the enthalpy loss of the condensate, equals the enthalpy gain of the cold air.
Dear sir when do we use control volume assumptions in thermodynamics
hi..i have a question in different subject if you can help
in evaporative cooling...is it useful to decrease the temp of water spry or water washer
ex: if out air 36c,60%rh..on psy with 90% evaporative efficiency.h ll get a 28c db and 90%rh..
so if i enter a 7c cold water to the washer .will i get a cold air temp....with knowing that in 100%humidity the db is about 27c
thanks
Why is the method not valid when 1-2 crosses the 100% RH curve? If your point lies in the super saturated zone can't you just follow the line of constant enthalpy to find your state point, then calculate how much water condensed out?
ua-cam.com/video/7DX6O9PgKrc/v-deo.html
As an example, consider 35C air at 95% relative humidity mixing with 10C air at 90% relative humidity, with 1kg/s mass flow rates each. If you blindly follow this method, you'll end up finding a fictional state of 20C and 108% relative humidity for this air above the saturation curve, by doing a weighted-average. Following the isenthalp to the saturation curve would give you 21C.
The correct answer is saturated air at 23C, with 6.4 g/s of condensate. The reason it is hotter than 21C that your method would find, is that the energy from condensing the water, heating it slightly above the weighted average of 20C, where it would have an equilibrium, if it didn't touch the saturation line.
The proper way to calculate mixing of air states that would cross the saturation line, is to consider the path that the air would take on this chart. The hot air will move horizontally to the left until it hits the saturation line. Then the hot air would follow the saturation line until the equilibrium point. Then you would find the line that is tangent to the saturation curve, and joins the initial state of the cold air (this is the hard part to find, in my example this happens at 14C). This is the path that this mixing process will follow. The mixed state will occur either A) on the line from state 2 to the tangent point on the saturation curve, B) on the saturation curve, or C) on the horizontal from the saturation curve to state 1. It will be at a point where the enthalpy loss of the hot air plus the enthalpy loss of the condensate, equals the enthalpy gain of the cold air.
How do you do Determine point3 ?
ua-cam.com/video/7DX6O9PgKrc/v-deo.html