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.
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 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 ¥
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.
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%
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!
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.
One of the best educational videos I found. Clear and concise. Thanks for the video.
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.
Easily understands the concept of the psycrhometric chart and cooling. Best video for this concept
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
Thank you for clear explanation.
"My humidity ratio is going to decrease, and I'm going to go... 0:52."
thanks great explanation
Nice vid, thanks
One of the best
It's Perfect, thank you
best
Thank you, from brazil
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.
Thanks dude, great video
ua-cam.com/video/0VTiMd7clMg/v-deo.html
..
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 ¥
bless you
God bless you x2
Thumbs up!
Thank youu
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.
Excellent. Thanks
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
14:23 i love how he turned an equal symbol into (2/
ua-cam.com/video/0VTiMd7clMg/v-deo.html
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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%
is 12.57 kg/s
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!
Be mindful of avoiding that smacking sound after you swallow.
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.
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