1:30 or enough air, to facilitate a 60 squarefeet SOG grow😏 I'm here to understand, how one liter of water can be collected, without investing 0,6 kW/h.
Great explanation. But, you found out mass flow rate of air to be 12.57 kg/s in the first calculation but used it as 12.57 g/s for the calculation of the mass flow rate of water. If in g/s, it should have been 12570 g/s. Which means about 276.54 kg/s should be the answer.
No I think you are wrong. He actually used “g/kg” but should have used “kg/s” and then his answer would have the correct unit of “g/s”. So the answer is correct but he just used wrong unit in the beginning.
And a reality check - 276kg/s is like a small river, that each building in the city would produce when cooling air. That would be massive amounts of water!
Thank you for sharing this vedio, you explained very well. But i wanna ask one thing also can we calculate the heat load as per given below calculation. Heat load= sensible heat of air + condensation load of water vapour Q= Ma Cp (40-8)+ Mw ¥
Remember tho Heat never removes any water, it only changes the state of it. It only Compressors running, doing refrigeration, that sends water down the drain. Good work on a confusing subject
Sir the day after tomorrow is my terminal exam of HVAC your videos about sychrometric chart are Very helpful ... Thanks for spreading the knowledge 😊👍❤️🌺💐💯🙌
it’s simple just put the evaporator coil the closet to the fan and put the condenser coil the closest to the intake that is dumb the company’s put the condenser coil closet to the fan just sucks up all the hot air.
In an air washer installation, 100 m3/min of air at 5 oC DBT and 80% relative humidity has to be heated and humidified to 25 oC and 45% of relative humidity by the following processes; (a) preheating, (b) adiabatic saturation in a re-circulated air washer up to 95% relative humidity and then (c) reheat to final state. Calculate: 1. total heat added in kW 2. water mass to be provided to the washer and 3. humidifying efficiency of the air washer. Can you help me solve the problems
This process takes a zig-zag shape on the Psychrometric chart. There are 4 states in this process. State 1 and state 4 are fully defined, except the pressure. We usually either assume 1 bar or 1 atmosphere, and constant throughout the system, if not otherwise specified. I'll assume 1 bar. State 1: initial condition of air State 2: preheated air State 3: adiabatic saturation after state 2 State 4: continue heating to the final state of air delivered Knowns for states 1 and 4: T[1] = 5C r[1] = 0.8 w[1] = humrat(AirH2O, T=T[1], r=r[1], P=100[kPa]) "w[1] = 0.004372" V_dot[1] = 100 [m^3/min] /convert(min, sec) "V_dot[1] = 1.667 [m^3/sec]" rho[1] = density(AirH2O, T=T[1], r=r[1], P=100[kPa]) "rho[1] = 1.244 [kg/m^3]" m_dot[1] = rho[1]*V_dot[1] "m_dot[1] = 2.073 [kg/s]" T[4] = 5C r[4] = 0.45 w[4] = humrat(AirH2O, T=T[4], r=r[4], P=100[kPa]) "w[4] = 0.008998" m_dot[1] = m_dot[2] Conservation of mass constrains states 1 and 2, and between states 3 and 4, such that there is no change in the humidity ratio across each of these pairs of states. We know the humidity ratio will be higher at state 3 than at state 2, because water was added. This allows us to lock-in state 3, since we know its relative humidity is 100%, and its humidity ratio equals that of state 4. We now have states 1, 3, and 4, fully defined. Knowns for state 3: w[3] = w[4] r[3] = 0.95 T[3] = Temperature(AirH2O, r=r[3], w=w[3], P=100[kPa]) "T[3] = 13.03 [C]" bw[3] = WetBulb(AirH2O, T=T[3], w=w[3], P=100[kPa]) "bw[3] = 12.57 [C]" Now, we relate states 2 and 3. Because this is an adiabatic saturation process, we need the wet bulb temperature of state 2 to equal the dry bulb temperature at state 3. We follow the wet bulb line that starts at state 3, until it intersects with the horizontal line from state 1. At that intersection, we will lock-in state 2. Constraints for state 2: bw[2] = bw[3] "Constant wet bulb temp across adiabatic saturation" bw[2] = WetBulb(AirH2O, T=T[2], w=w[2], P=100[kPa]) "Solves for T[2] = 24.3[C]" To find m_dot[3], we need to use conservation of dry air from state 2. The dry air flow rate will be equal at all points, and is determined by dividing the total mass flow rate by (1 + w) at each state. This means: m_dot[2]/(1+w[2]) = m_dot[3]/(1+w[3]) "Solves for m_dot[3] = 2.083 [kg/s]" Equate to find m_dot[4]: m_dot[4] = m_dot[3] Now find enthalpy at each state: h[i] = Enthalpy(AirH2O, T=T[i], w=w[i], P=100 [kPa]) Now we find each thing we were looking for: Item 1, Total heat added: Q_dot_net = m_dot[1]*(h[2] - h[1]) + m_dot[3]*(h[4] - h[3]) Q_dot_net = 66 kW Item 2, Mass flow rate of humidification: m_dot_water = m_dot[3] - m_dot[2] m_dot_water = 9.55 grams/sec Item 3, Humidifying efficiency = actual drop in dry bulb temperature, over ideal drop in dry bulb temperature to wet bulb temperature. eta_hum = (T[2] - T[3])/(T[2] - bw[2]) eta_hum = 96%
actually process of determining the point 3 can be made very simple by drawing the line connecting point 1 and point 2. then find out mass flow rate at 1 and mass flow rate at 2. and use ratio law to get the dry bulb temperature at 3. intersecting dry bulb temperature with the process line will give you exact point 3.
Thanks! After searching for mixed air videos as long as I have , I don’t know why it took so long for this one to be in my recommend videos. This would have made the last few weeks easier. Please don’t stop sharing what you know. Thanks
I did not see the answers to you. Here are my understanding. Assuming you have Air handlers in the rooms or spaces to be cooled: 1) If you have outside air (OA) damper and OA is cooler and lower Rh than inside room air (IA), you can open up damper to mix two air streams (IA and OA), by release some IA; 2) If AHU allows adjustment of supply air temperate, set the supply air temperature below saturation temperature. Thus the moisture will be condensed out in AHU. Before you purge the near-saturated air to the space, heat up a little bit to avoid condensation in the room or space.
I work at a hospital and they use steam through a coil to reheat the air, the air in the building is dry for the most part year round. In the winter however we have issues with the air being too dry due to the chiller being on and the dry outside air being mixed in. I have seen the humidity in the lower teens, not good during flu season when people are sick. We have humidification in some parts, but not most.
"My humidity ratio is going to decrease, and I'm going to go... 0:52."
1:30 or enough air, to facilitate a 60 squarefeet SOG grow😏 I'm here to understand, how one liter of water can be collected, without investing 0,6 kW/h.
Hi, can I work with you? I have over 18 years of experience in HVAC and MEP services, and I really learned from you.
would that mean i need 620 KW heater capacity?
Thumbs up!
God bless you x2
Great explanation. But, you found out mass flow rate of air to be 12.57 kg/s in the first calculation but used it as 12.57 g/s for the calculation of the mass flow rate of water. If in g/s, it should have been 12570 g/s. Which means about 276.54 kg/s should be the answer.
No I think you are wrong. He actually used “g/kg” but should have used “kg/s” and then his answer would have the correct unit of “g/s”. So the answer is correct but he just used wrong unit in the beginning.
And a reality check - 276kg/s is like a small river, that each building in the city would produce when cooling air. That would be massive amounts of water!
What will be the minimum temperature of air after cooling by water. Does the minimum temperature will be achieved at saturation condition?
What is role of added water temp. Does the outlet air temp will be the same for any temprature of added water?
Thank you for sharing this vedio, you explained very well. But i wanna ask one thing also can we calculate the heat load as per given below calculation. Heat load= sensible heat of air + condensation load of water vapour Q= Ma Cp (40-8)+ Mw ¥
Remember tho Heat never removes any water, it only changes the state of it. It only Compressors running, doing refrigeration, that sends water down the drain. Good work on a confusing subject
Awesome!
Be mindful of avoiding that smacking sound after you swallow.
Sir the day after tomorrow is my terminal exam of HVAC your videos about sychrometric chart are Very helpful ... Thanks for spreading the knowledge 😊👍❤️🌺💐💯🙌
Easily understands the concept of the psycrhometric chart and cooling. Best video for this concept
Great
Thank you, from brazil
Thank youu
Once again Excellent. your very clear you stick to the point amazing. Never bored watching your content thank you!!!!
Amazing!!! Would love a video on how to size a reheat coil for dehumidification if you don’t already have one. Thank you!
ua-cam.com/video/0VTiMd7clMg/v-deo.html
Just when I needed this tutorial the most, Thanks a ton. :D
Thank you very much for sharing this. I really appreciate it. Good luck 👍🏼
14:23 i love how he turned an equal symbol into (2/
ua-cam.com/video/0VTiMd7clMg/v-deo.html .
u look like andy buckley
Thanks dude, great video
ua-cam.com/video/0VTiMd7clMg/v-deo.html ..
What study of engineering encompasses this subject? Mechanical? Environmental?
My architectural engineering (building engineering) program covered this at Drexel University. Penn state has a similar program
Refrigeration system in Thermodynamics
HVAC industry using a mechanical engineering degree
@@lavenarch9570 ua-cam.com/video/0VTiMd7clMg/v-deo.html
@@vu4991 ua-cam.com/video/0VTiMd7clMg/v-deo.html
thanks great explanation
it’s simple just put the evaporator coil the closet to the fan and put the condenser coil the closest to the intake that is dumb the company’s put the condenser coil closet to the fan just sucks up all the hot air.
Thank you for clear explanation.
In an air washer installation, 100 m3/min of air at 5 oC DBT and 80% relative humidity has to be heated and humidified to 25 oC and 45% of relative humidity by the following processes; (a) preheating, (b) adiabatic saturation in a re-circulated air washer up to 95% relative humidity and then (c) reheat to final state. Calculate: 1. total heat added in kW 2. water mass to be provided to the washer and 3. humidifying efficiency of the air washer. Can you help me solve the problems
This process takes a zig-zag shape on the Psychrometric chart. There are 4 states in this process. State 1 and state 4 are fully defined, except the pressure. We usually either assume 1 bar or 1 atmosphere, and constant throughout the system, if not otherwise specified. I'll assume 1 bar. State 1: initial condition of air State 2: preheated air State 3: adiabatic saturation after state 2 State 4: continue heating to the final state of air delivered Knowns for states 1 and 4: T[1] = 5C r[1] = 0.8 w[1] = humrat(AirH2O, T=T[1], r=r[1], P=100[kPa]) "w[1] = 0.004372" V_dot[1] = 100 [m^3/min] /convert(min, sec) "V_dot[1] = 1.667 [m^3/sec]" rho[1] = density(AirH2O, T=T[1], r=r[1], P=100[kPa]) "rho[1] = 1.244 [kg/m^3]" m_dot[1] = rho[1]*V_dot[1] "m_dot[1] = 2.073 [kg/s]" T[4] = 5C r[4] = 0.45 w[4] = humrat(AirH2O, T=T[4], r=r[4], P=100[kPa]) "w[4] = 0.008998" m_dot[1] = m_dot[2] Conservation of mass constrains states 1 and 2, and between states 3 and 4, such that there is no change in the humidity ratio across each of these pairs of states. We know the humidity ratio will be higher at state 3 than at state 2, because water was added. This allows us to lock-in state 3, since we know its relative humidity is 100%, and its humidity ratio equals that of state 4. We now have states 1, 3, and 4, fully defined. Knowns for state 3: w[3] = w[4] r[3] = 0.95 T[3] = Temperature(AirH2O, r=r[3], w=w[3], P=100[kPa]) "T[3] = 13.03 [C]" bw[3] = WetBulb(AirH2O, T=T[3], w=w[3], P=100[kPa]) "bw[3] = 12.57 [C]" Now, we relate states 2 and 3. Because this is an adiabatic saturation process, we need the wet bulb temperature of state 2 to equal the dry bulb temperature at state 3. We follow the wet bulb line that starts at state 3, until it intersects with the horizontal line from state 1. At that intersection, we will lock-in state 2. Constraints for state 2: bw[2] = bw[3] "Constant wet bulb temp across adiabatic saturation" bw[2] = WetBulb(AirH2O, T=T[2], w=w[2], P=100[kPa]) "Solves for T[2] = 24.3[C]" To find m_dot[3], we need to use conservation of dry air from state 2. The dry air flow rate will be equal at all points, and is determined by dividing the total mass flow rate by (1 + w) at each state. This means: m_dot[2]/(1+w[2]) = m_dot[3]/(1+w[3]) "Solves for m_dot[3] = 2.083 [kg/s]" Equate to find m_dot[4]: m_dot[4] = m_dot[3] Now find enthalpy at each state: h[i] = Enthalpy(AirH2O, T=T[i], w=w[i], P=100 [kPa]) Now we find each thing we were looking for: Item 1, Total heat added: Q_dot_net = m_dot[1]*(h[2] - h[1]) + m_dot[3]*(h[4] - h[3]) Q_dot_net = 66 kW Item 2, Mass flow rate of humidification: m_dot_water = m_dot[3] - m_dot[2] m_dot_water = 9.55 grams/sec Item 3, Humidifying efficiency = actual drop in dry bulb temperature, over ideal drop in dry bulb temperature to wet bulb temperature. eta_hum = (T[2] - T[3])/(T[2] - bw[2]) eta_hum = 96%
Hi, why not just use the Cp of air to calculate the heating? Q=mxCpxDetlaT
actually process of determining the point 3 can be made very simple by drawing the line connecting point 1 and point 2. then find out mass flow rate at 1 and mass flow rate at 2. and use ratio law to get the dry bulb temperature at 3. intersecting dry bulb temperature with the process line will give you exact point 3.
Thank you very much for clear explanation. Only 1 small correction: in the last calculation the 12,57 are in Kg/sec.
final calculated result is 276.54 g/sec
@@kennethlim6960 exactly, but it is a clear description of method. thanks a lot
Thanks! After searching for mixed air videos as long as I have , I don’t know why it took so long for this one to be in my recommend videos. This would have made the last few weeks easier. Please don’t stop sharing what you know. Thanks
It's Perfect, thank you
contact for dehumidifier requirement or consulting TAME AIR TECHNOLOGY NEW DELHI 9891167386, 8130338384
In My plant tem 38C and rh is 90% with 75 ton chiller ac plant how I reduce rh I want 55% humidity with 23c temperature
How about the volume, width height length, of the plant ?
@@gozit3516 plant or ahu ?
@@yuvraj-gaming. the plant dimension not ahu.
@@gozit3516 30×40×20 feet
I did not see the answers to you. Here are my understanding. Assuming you have Air handlers in the rooms or spaces to be cooled: 1) If you have outside air (OA) damper and OA is cooler and lower Rh than inside room air (IA), you can open up damper to mix two air streams (IA and OA), by release some IA; 2) If AHU allows adjustment of supply air temperate, set the supply air temperature below saturation temperature. Thus the moisture will be condensed out in AHU. Before you purge the near-saturated air to the space, heat up a little bit to avoid condensation in the room or space.
Excellent tutorial, thanks.
I work at a hospital and they use steam through a coil to reheat the air, the air in the building is dry for the most part year round. In the winter however we have issues with the air being too dry due to the chiller being on and the dry outside air being mixed in. I have seen the humidity in the lower teens, not good during flu season when people are sick. We have humidification in some parts, but not most.
Excellent. Thanks
best
bless you
is 12.57 kg/s
One of the best
One of the best educational videos I found. Clear and concise. Thanks for the video.
Thank you so much.
Nice vid, thanks
Thanks 😌
the letter are not clear yet