Mixing
Heating a box (as a thermodynamics dummy)
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I expect to build a dehydrator and am knowledgeable enough for most of it (fan, timer, power supply and so on), bar the most important part: the heating element.
If you're not familiar with electric dehydrators: you usually seek a 30-70°C (most of the time ~40°C) temperature control for a 5-12h drying timespan.
Problem is, I have absolutely no clue how to pick a heating element. The box will be made of wood (not sure about the door, will be either wood or glass) and its dimensions should be something like 30x30x50cm. The box is partially open so air can pass through (with the help of a fan).
Is there a fairly simple way to guess which amount of heating power I need? To pick a heating resistor which will "roughly" do the job at expected temperature?
resistors heat
edited Sep 6 at 8:20
winny
4,36321726
asked Sep 6 at 7:41
Kinxil
685
add a comment |Â
up vote
4
down vote
favorite
I expect to build a dehydrator and am knowledgeable enough for most of it (fan, timer, power supply and so on), bar the most important part: the heating element.
If you're not familiar with electric dehydrators: you usually seek a 30-70°C (most of the time ~40°C) temperature control for a 5-12h drying timespan.
Problem is, I have absolutely no clue how to pick a heating element. The box will be made of wood (not sure about the door, will be either wood or glass) and its dimensions should be something like 30x30x50cm. The box is partially open so air can pass through (with the help of a fan).
Is there a fairly simple way to guess which amount of heating power I need? To pick a heating resistor which will "roughly" do the job at expected temperature?
resistors heat
edited Sep 6 at 8:20
winny
4,36321726
asked Sep 6 at 7:41
Kinxil
685
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I expect to build a dehydrator and am knowledgeable enough for most of it (fan, timer, power supply and so on), bar the most important part: the heating element.
If you're not familiar with electric dehydrators: you usually seek a 30-70°C (most of the time ~40°C) temperature control for a 5-12h drying timespan.
Problem is, I have absolutely no clue how to pick a heating element. The box will be made of wood (not sure about the door, will be either wood or glass) and its dimensions should be something like 30x30x50cm. The box is partially open so air can pass through (with the help of a fan).
Is there a fairly simple way to guess which amount of heating power I need? To pick a heating resistor which will "roughly" do the job at expected temperature?
resistors heat
edited Sep 6 at 8:20
winny
4,36321726
asked Sep 6 at 7:41
Kinxil
685
I expect to build a dehydrator and am knowledgeable enough for most of it (fan, timer, power supply and so on), bar the most important part: the heating element.
If you're not familiar with electric dehydrators: you usually seek a 30-70°C (most of the time ~40°C) temperature control for a 5-12h drying timespan.
Problem is, I have absolutely no clue how to pick a heating element. The box will be made of wood (not sure about the door, will be either wood or glass) and its dimensions should be something like 30x30x50cm. The box is partially open so air can pass through (with the help of a fan).
Is there a fairly simple way to guess which amount of heating power I need? To pick a heating resistor which will "roughly" do the job at expected temperature?
resistors heat
edited Sep 6 at 8:20
winny
4,36321726
asked Sep 6 at 7:41
Kinxil
685
edited Sep 6 at 8:20
winny
4,36321726
edited Sep 6 at 8:20
winny
4,36321726
edited Sep 6 at 8:20
winny
4,36321726
4,36321726
asked Sep 6 at 7:41
Kinxil
685
asked Sep 6 at 7:41
Kinxil
685
asked Sep 6 at 7:41
Kinxil
685
685
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
Basically a thermodynamics problem with thermal conductivity, area, thickness, ΔT
q=​d​​KA(T​.hot​​−T.​cold​​)​​ [watts]
Where:
q = Conduction heat transfer (W)
K = Materials thermal conductivity (W/mK)
A = Cross sectional area (m²)
T​.Hot​​ = Higher temperature (°C)
T.Cold​​ = Colder temperature (°C)
d = Material thickness (m)
- Convert your area 30x30x50cm = 6,600 cm² to 0.66 m²
- Look up K for wood and choose d
- Compute Q heat in watts.
But if this is too slow, adding forced air evaporates faster Then the air flow volume and rate has to be pre-heated and forced thru the box at the desired temperature.
This is more complicated by the efficiency at which air temp rises thru the heat per unit volume of air. A radiator has high efficiency, a circular tube is lower coupling but allows higher flow rates.
I might suggest a 50W power resistor (s) mounted to a CPU heatsink ( tap and screw or clamp) from say an ATX PSU using 12V, using a CPU Fan with a variable speed control and same for heat control and temperature sensing to design a servo loop for the heat and fan speed. LM317's can also be mounted on heatsink for each so no heat is lost here.
simulate this circuit – Schematic created using CircuitLab
A Cheap and Dirty solution for fan control might be to use 5V on a 12V fan to reduce flow rate since evaporation rate is all day, as long as it is a good fan that starts at 4V.
For compact units choose an old Pentium heatsink. ( free at most repair stores)
answered Sep 6 at 8:24
Tony EE rocketscientist
57.4k22082
I suspect that the question is more about given a known flow rate of air thru the dryer, how much power is required to heat this air (at ambient T) to 40 C. @Kinxil needs to state an estimate the air flow of his fan.
– D Duck
Sep 6 at 8:34
Great answer. I'd just add to it that 50w resistors don't cost much, so you could overshoot by using say 4 of them since the cost would be low and there's no reason you have to drive them at the full 200W. This way you'll have more power available than most cpu heat sinks are able to dissipate.
– K H
Sep 6 at 8:53
Did the equation (0.12 thermal conductivity soft wood, 3cm thick, 0.66m², 70°C max 16°C min (worse outside case)) and with 0.12*0.03*0.66*(70-16) = 0.128W. Isn't it a bit... out of range for what we could expect (100+ W) ?
– Kinxil
Sep 6 at 9:36
The fan will prolly be controlled by PWM. The unit will have a smart controlling part anyway as I expect to experiment with non constant drying (decreasing temperature depending of either drying time already done or humidity).
– Kinxil
Sep 6 at 9:45
The air heater temp is dependant on air velocity so they must be automated together. RH can be integrated for shut-down. Just design for setpoint. Must be PWM type fan as aliasing can cause issues with some BLDC Fans
– Tony EE rocketscientist
Sep 6 at 9:58
add a comment |Â
up vote
2
down vote
I don't think you need to do any math for this. Just pick a heating element which is more than powerful enough, and connect that to a PID controller. You can have this whole thing done in a few hours, and it will still work if you decide to bump up the temp or airflow in the future.
I would aim for the max your outlets can deliver, about 1500W. If you really want to estimate the wattage, I would look for a similarly sized commercial dehydrator and see how much power that draws.
answered Sep 6 at 9:03
Drew
1,303412
They usually mention 400W but that's prolly for 70°C. They're almost made of thin plastic or stainless steel, which should dissipate quickly heat. So with 200-300W I could be fine I guess?
– Kinxil
Sep 6 at 9:32
1
To know which element is "more than powerful enough" one still has to make an estimation of what is enough. At which point it makes sense to use an element that is "just a bit more than enough", which will be cheaper.
– Dmitry Grigoryev
Sep 6 at 11:09
I guess energy consumption cost was not a design criteria for 5-12h drying timespan. ( how many kill-a-watts does it take to remove x grams of water? at a low temp.)
– Tony EE rocketscientist
Sep 6 at 13:30
Let me save you a couple hours with a thermistor and P gain and setpoint you don't need a PID loop to control slow changing air temp, just P control 0 to 100% over a 5 degree 'C range. Heat mass, latency is another thing and so is dust control. (Bin thar dun that)
– Tony EE rocketscientist
Sep 6 at 13:34
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Basically a thermodynamics problem with thermal conductivity, area, thickness, ΔT
q=​d​​KA(T​.hot​​−T.​cold​​)​​ [watts]
Where:
q = Conduction heat transfer (W)
K = Materials thermal conductivity (W/mK)
A = Cross sectional area (m²)
T​.Hot​​ = Higher temperature (°C)
T.Cold​​ = Colder temperature (°C)
d = Material thickness (m)
- Convert your area 30x30x50cm = 6,600 cm² to 0.66 m²
- Look up K for wood and choose d
- Compute Q heat in watts.
But if this is too slow, adding forced air evaporates faster Then the air flow volume and rate has to be pre-heated and forced thru the box at the desired temperature.
This is more complicated by the efficiency at which air temp rises thru the heat per unit volume of air. A radiator has high efficiency, a circular tube is lower coupling but allows higher flow rates.
I might suggest a 50W power resistor (s) mounted to a CPU heatsink ( tap and screw or clamp) from say an ATX PSU using 12V, using a CPU Fan with a variable speed control and same for heat control and temperature sensing to design a servo loop for the heat and fan speed. LM317's can also be mounted on heatsink for each so no heat is lost here.
simulate this circuit – Schematic created using CircuitLab
A Cheap and Dirty solution for fan control might be to use 5V on a 12V fan to reduce flow rate since evaporation rate is all day, as long as it is a good fan that starts at 4V.
For compact units choose an old Pentium heatsink. ( free at most repair stores)
answered Sep 6 at 8:24
Tony EE rocketscientist
57.4k22082
I suspect that the question is more about given a known flow rate of air thru the dryer, how much power is required to heat this air (at ambient T) to 40 C. @Kinxil needs to state an estimate the air flow of his fan.
– D Duck
Sep 6 at 8:34
Great answer. I'd just add to it that 50w resistors don't cost much, so you could overshoot by using say 4 of them since the cost would be low and there's no reason you have to drive them at the full 200W. This way you'll have more power available than most cpu heat sinks are able to dissipate.
– K H
Sep 6 at 8:53
Did the equation (0.12 thermal conductivity soft wood, 3cm thick, 0.66m², 70°C max 16°C min (worse outside case)) and with 0.12*0.03*0.66*(70-16) = 0.128W. Isn't it a bit... out of range for what we could expect (100+ W) ?
– Kinxil
Sep 6 at 9:36
The fan will prolly be controlled by PWM. The unit will have a smart controlling part anyway as I expect to experiment with non constant drying (decreasing temperature depending of either drying time already done or humidity).
– Kinxil
Sep 6 at 9:45
The air heater temp is dependant on air velocity so they must be automated together. RH can be integrated for shut-down. Just design for setpoint. Must be PWM type fan as aliasing can cause issues with some BLDC Fans
– Tony EE rocketscientist
Sep 6 at 9:58
add a comment |Â
up vote
5
down vote
accepted
Basically a thermodynamics problem with thermal conductivity, area, thickness, ΔT
q=​d​​KA(T​.hot​​−T.​cold​​)​​ [watts]
Where:
q = Conduction heat transfer (W)
K = Materials thermal conductivity (W/mK)
A = Cross sectional area (m²)
T​.Hot​​ = Higher temperature (°C)
T.Cold​​ = Colder temperature (°C)
d = Material thickness (m)
- Convert your area 30x30x50cm = 6,600 cm² to 0.66 m²
- Look up K for wood and choose d
- Compute Q heat in watts.
But if this is too slow, adding forced air evaporates faster Then the air flow volume and rate has to be pre-heated and forced thru the box at the desired temperature.
This is more complicated by the efficiency at which air temp rises thru the heat per unit volume of air. A radiator has high efficiency, a circular tube is lower coupling but allows higher flow rates.
I might suggest a 50W power resistor (s) mounted to a CPU heatsink ( tap and screw or clamp) from say an ATX PSU using 12V, using a CPU Fan with a variable speed control and same for heat control and temperature sensing to design a servo loop for the heat and fan speed. LM317's can also be mounted on heatsink for each so no heat is lost here.
simulate this circuit – Schematic created using CircuitLab
A Cheap and Dirty solution for fan control might be to use 5V on a 12V fan to reduce flow rate since evaporation rate is all day, as long as it is a good fan that starts at 4V.
For compact units choose an old Pentium heatsink. ( free at most repair stores)
answered Sep 6 at 8:24
Tony EE rocketscientist
57.4k22082
I suspect that the question is more about given a known flow rate of air thru the dryer, how much power is required to heat this air (at ambient T) to 40 C. @Kinxil needs to state an estimate the air flow of his fan.
– D Duck
Sep 6 at 8:34
Great answer. I'd just add to it that 50w resistors don't cost much, so you could overshoot by using say 4 of them since the cost would be low and there's no reason you have to drive them at the full 200W. This way you'll have more power available than most cpu heat sinks are able to dissipate.
– K H
Sep 6 at 8:53
Did the equation (0.12 thermal conductivity soft wood, 3cm thick, 0.66m², 70°C max 16°C min (worse outside case)) and with 0.12*0.03*0.66*(70-16) = 0.128W. Isn't it a bit... out of range for what we could expect (100+ W) ?
– Kinxil
Sep 6 at 9:36
The fan will prolly be controlled by PWM. The unit will have a smart controlling part anyway as I expect to experiment with non constant drying (decreasing temperature depending of either drying time already done or humidity).
– Kinxil
Sep 6 at 9:45
The air heater temp is dependant on air velocity so they must be automated together. RH can be integrated for shut-down. Just design for setpoint. Must be PWM type fan as aliasing can cause issues with some BLDC Fans
– Tony EE rocketscientist
Sep 6 at 9:58
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Basically a thermodynamics problem with thermal conductivity, area, thickness, ΔT
q=​d​​KA(T​.hot​​−T.​cold​​)​​ [watts]
Where:
q = Conduction heat transfer (W)
K = Materials thermal conductivity (W/mK)
A = Cross sectional area (m²)
T​.Hot​​ = Higher temperature (°C)
T.Cold​​ = Colder temperature (°C)
d = Material thickness (m)
- Convert your area 30x30x50cm = 6,600 cm² to 0.66 m²
- Look up K for wood and choose d
- Compute Q heat in watts.
But if this is too slow, adding forced air evaporates faster Then the air flow volume and rate has to be pre-heated and forced thru the box at the desired temperature.
This is more complicated by the efficiency at which air temp rises thru the heat per unit volume of air. A radiator has high efficiency, a circular tube is lower coupling but allows higher flow rates.
I might suggest a 50W power resistor (s) mounted to a CPU heatsink ( tap and screw or clamp) from say an ATX PSU using 12V, using a CPU Fan with a variable speed control and same for heat control and temperature sensing to design a servo loop for the heat and fan speed. LM317's can also be mounted on heatsink for each so no heat is lost here.
simulate this circuit – Schematic created using CircuitLab
A Cheap and Dirty solution for fan control might be to use 5V on a 12V fan to reduce flow rate since evaporation rate is all day, as long as it is a good fan that starts at 4V.
For compact units choose an old Pentium heatsink. ( free at most repair stores)
answered Sep 6 at 8:24
Tony EE rocketscientist
57.4k22082
Basically a thermodynamics problem with thermal conductivity, area, thickness, ΔT
q=​d​​KA(T​.hot​​−T.​cold​​)​​ [watts]
Where:
q = Conduction heat transfer (W)
K = Materials thermal conductivity (W/mK)
A = Cross sectional area (m²)
T​.Hot​​ = Higher temperature (°C)
T.Cold​​ = Colder temperature (°C)
d = Material thickness (m)
- Convert your area 30x30x50cm = 6,600 cm² to 0.66 m²
- Look up K for wood and choose d
- Compute Q heat in watts.
But if this is too slow, adding forced air evaporates faster Then the air flow volume and rate has to be pre-heated and forced thru the box at the desired temperature.
This is more complicated by the efficiency at which air temp rises thru the heat per unit volume of air. A radiator has high efficiency, a circular tube is lower coupling but allows higher flow rates.
I might suggest a 50W power resistor (s) mounted to a CPU heatsink ( tap and screw or clamp) from say an ATX PSU using 12V, using a CPU Fan with a variable speed control and same for heat control and temperature sensing to design a servo loop for the heat and fan speed. LM317's can also be mounted on heatsink for each so no heat is lost here.
simulate this circuit – Schematic created using CircuitLab
A Cheap and Dirty solution for fan control might be to use 5V on a 12V fan to reduce flow rate since evaporation rate is all day, as long as it is a good fan that starts at 4V.
For compact units choose an old Pentium heatsink. ( free at most repair stores)
answered Sep 6 at 8:24
Tony EE rocketscientist
57.4k22082
edited Sep 6 at 9:01
answered Sep 6 at 8:24
Tony EE rocketscientist
57.4k22082
answered Sep 6 at 8:24
Tony EE rocketscientist
57.4k22082
answered Sep 6 at 8:24
Tony EE rocketscientist
57.4k22082
57.4k22082
I suspect that the question is more about given a known flow rate of air thru the dryer, how much power is required to heat this air (at ambient T) to 40 C. @Kinxil needs to state an estimate the air flow of his fan.
– D Duck
Sep 6 at 8:34
Great answer. I'd just add to it that 50w resistors don't cost much, so you could overshoot by using say 4 of them since the cost would be low and there's no reason you have to drive them at the full 200W. This way you'll have more power available than most cpu heat sinks are able to dissipate.
– K H
Sep 6 at 8:53
Did the equation (0.12 thermal conductivity soft wood, 3cm thick, 0.66m², 70°C max 16°C min (worse outside case)) and with 0.12*0.03*0.66*(70-16) = 0.128W. Isn't it a bit... out of range for what we could expect (100+ W) ?
– Kinxil
Sep 6 at 9:36
The fan will prolly be controlled by PWM. The unit will have a smart controlling part anyway as I expect to experiment with non constant drying (decreasing temperature depending of either drying time already done or humidity).
– Kinxil
Sep 6 at 9:45
The air heater temp is dependant on air velocity so they must be automated together. RH can be integrated for shut-down. Just design for setpoint. Must be PWM type fan as aliasing can cause issues with some BLDC Fans
– Tony EE rocketscientist
Sep 6 at 9:58
add a comment |Â
I suspect that the question is more about given a known flow rate of air thru the dryer, how much power is required to heat this air (at ambient T) to 40 C. @Kinxil needs to state an estimate the air flow of his fan.
– D Duck
Sep 6 at 8:34
Great answer. I'd just add to it that 50w resistors don't cost much, so you could overshoot by using say 4 of them since the cost would be low and there's no reason you have to drive them at the full 200W. This way you'll have more power available than most cpu heat sinks are able to dissipate.
– K H
Sep 6 at 8:53
Did the equation (0.12 thermal conductivity soft wood, 3cm thick, 0.66m², 70°C max 16°C min (worse outside case)) and with 0.12*0.03*0.66*(70-16) = 0.128W. Isn't it a bit... out of range for what we could expect (100+ W) ?
– Kinxil
Sep 6 at 9:36
The fan will prolly be controlled by PWM. The unit will have a smart controlling part anyway as I expect to experiment with non constant drying (decreasing temperature depending of either drying time already done or humidity).
– Kinxil
Sep 6 at 9:45
The air heater temp is dependant on air velocity so they must be automated together. RH can be integrated for shut-down. Just design for setpoint. Must be PWM type fan as aliasing can cause issues with some BLDC Fans
– Tony EE rocketscientist
Sep 6 at 9:58
I suspect that the question is more about given a known flow rate of air thru the dryer, how much power is required to heat this air (at ambient T) to 40 C. @Kinxil needs to state an estimate the air flow of his fan.
– D Duck
Sep 6 at 8:34
I suspect that the question is more about given a known flow rate of air thru the dryer, how much power is required to heat this air (at ambient T) to 40 C. @Kinxil needs to state an estimate the air flow of his fan.
– D Duck
Sep 6 at 8:34
Great answer. I'd just add to it that 50w resistors don't cost much, so you could overshoot by using say 4 of them since the cost would be low and there's no reason you have to drive them at the full 200W. This way you'll have more power available than most cpu heat sinks are able to dissipate.
– K H
Sep 6 at 8:53
Great answer. I'd just add to it that 50w resistors don't cost much, so you could overshoot by using say 4 of them since the cost would be low and there's no reason you have to drive them at the full 200W. This way you'll have more power available than most cpu heat sinks are able to dissipate.
– K H
Sep 6 at 8:53
Did the equation (0.12 thermal conductivity soft wood, 3cm thick, 0.66m², 70°C max 16°C min (worse outside case)) and with 0.12*0.03*0.66*(70-16) = 0.128W. Isn't it a bit... out of range for what we could expect (100+ W) ?
– Kinxil
Sep 6 at 9:36
Did the equation (0.12 thermal conductivity soft wood, 3cm thick, 0.66m², 70°C max 16°C min (worse outside case)) and with 0.12*0.03*0.66*(70-16) = 0.128W. Isn't it a bit... out of range for what we could expect (100+ W) ?
– Kinxil
Sep 6 at 9:36
The fan will prolly be controlled by PWM. The unit will have a smart controlling part anyway as I expect to experiment with non constant drying (decreasing temperature depending of either drying time already done or humidity).
– Kinxil
Sep 6 at 9:45
The fan will prolly be controlled by PWM. The unit will have a smart controlling part anyway as I expect to experiment with non constant drying (decreasing temperature depending of either drying time already done or humidity).
– Kinxil
Sep 6 at 9:45
The air heater temp is dependant on air velocity so they must be automated together. RH can be integrated for shut-down. Just design for setpoint. Must be PWM type fan as aliasing can cause issues with some BLDC Fans
– Tony EE rocketscientist
Sep 6 at 9:58
The air heater temp is dependant on air velocity so they must be automated together. RH can be integrated for shut-down. Just design for setpoint. Must be PWM type fan as aliasing can cause issues with some BLDC Fans
– Tony EE rocketscientist
Sep 6 at 9:58
add a comment |Â
up vote
2
down vote
I don't think you need to do any math for this. Just pick a heating element which is more than powerful enough, and connect that to a PID controller. You can have this whole thing done in a few hours, and it will still work if you decide to bump up the temp or airflow in the future.
I would aim for the max your outlets can deliver, about 1500W. If you really want to estimate the wattage, I would look for a similarly sized commercial dehydrator and see how much power that draws.
answered Sep 6 at 9:03
Drew
1,303412
They usually mention 400W but that's prolly for 70°C. They're almost made of thin plastic or stainless steel, which should dissipate quickly heat. So with 200-300W I could be fine I guess?
– Kinxil
Sep 6 at 9:32
1
To know which element is "more than powerful enough" one still has to make an estimation of what is enough. At which point it makes sense to use an element that is "just a bit more than enough", which will be cheaper.
– Dmitry Grigoryev
Sep 6 at 11:09
I guess energy consumption cost was not a design criteria for 5-12h drying timespan. ( how many kill-a-watts does it take to remove x grams of water? at a low temp.)
– Tony EE rocketscientist
Sep 6 at 13:30
Let me save you a couple hours with a thermistor and P gain and setpoint you don't need a PID loop to control slow changing air temp, just P control 0 to 100% over a 5 degree 'C range. Heat mass, latency is another thing and so is dust control. (Bin thar dun that)
– Tony EE rocketscientist
Sep 6 at 13:34
add a comment |Â
up vote
2
down vote
I don't think you need to do any math for this. Just pick a heating element which is more than powerful enough, and connect that to a PID controller. You can have this whole thing done in a few hours, and it will still work if you decide to bump up the temp or airflow in the future.
I would aim for the max your outlets can deliver, about 1500W. If you really want to estimate the wattage, I would look for a similarly sized commercial dehydrator and see how much power that draws.
answered Sep 6 at 9:03
Drew
1,303412
They usually mention 400W but that's prolly for 70°C. They're almost made of thin plastic or stainless steel, which should dissipate quickly heat. So with 200-300W I could be fine I guess?
– Kinxil
Sep 6 at 9:32
1
To know which element is "more than powerful enough" one still has to make an estimation of what is enough. At which point it makes sense to use an element that is "just a bit more than enough", which will be cheaper.
– Dmitry Grigoryev
Sep 6 at 11:09
I guess energy consumption cost was not a design criteria for 5-12h drying timespan. ( how many kill-a-watts does it take to remove x grams of water? at a low temp.)
– Tony EE rocketscientist
Sep 6 at 13:30
Let me save you a couple hours with a thermistor and P gain and setpoint you don't need a PID loop to control slow changing air temp, just P control 0 to 100% over a 5 degree 'C range. Heat mass, latency is another thing and so is dust control. (Bin thar dun that)
– Tony EE rocketscientist
Sep 6 at 13:34
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I don't think you need to do any math for this. Just pick a heating element which is more than powerful enough, and connect that to a PID controller. You can have this whole thing done in a few hours, and it will still work if you decide to bump up the temp or airflow in the future.
I would aim for the max your outlets can deliver, about 1500W. If you really want to estimate the wattage, I would look for a similarly sized commercial dehydrator and see how much power that draws.
answered Sep 6 at 9:03
Drew
1,303412
I don't think you need to do any math for this. Just pick a heating element which is more than powerful enough, and connect that to a PID controller. You can have this whole thing done in a few hours, and it will still work if you decide to bump up the temp or airflow in the future.
I would aim for the max your outlets can deliver, about 1500W. If you really want to estimate the wattage, I would look for a similarly sized commercial dehydrator and see how much power that draws.
answered Sep 6 at 9:03
Drew
1,303412
answered Sep 6 at 9:03
Drew
1,303412
answered Sep 6 at 9:03
Drew
1,303412
answered Sep 6 at 9:03
Drew
1,303412
1,303412
They usually mention 400W but that's prolly for 70°C. They're almost made of thin plastic or stainless steel, which should dissipate quickly heat. So with 200-300W I could be fine I guess?
– Kinxil
Sep 6 at 9:32
1
To know which element is "more than powerful enough" one still has to make an estimation of what is enough. At which point it makes sense to use an element that is "just a bit more than enough", which will be cheaper.
– Dmitry Grigoryev
Sep 6 at 11:09
I guess energy consumption cost was not a design criteria for 5-12h drying timespan. ( how many kill-a-watts does it take to remove x grams of water? at a low temp.)
– Tony EE rocketscientist
Sep 6 at 13:30
Let me save you a couple hours with a thermistor and P gain and setpoint you don't need a PID loop to control slow changing air temp, just P control 0 to 100% over a 5 degree 'C range. Heat mass, latency is another thing and so is dust control. (Bin thar dun that)
– Tony EE rocketscientist
Sep 6 at 13:34
add a comment |Â
They usually mention 400W but that's prolly for 70°C. They're almost made of thin plastic or stainless steel, which should dissipate quickly heat. So with 200-300W I could be fine I guess?
– Kinxil
Sep 6 at 9:32
1
To know which element is "more than powerful enough" one still has to make an estimation of what is enough. At which point it makes sense to use an element that is "just a bit more than enough", which will be cheaper.
– Dmitry Grigoryev
Sep 6 at 11:09
I guess energy consumption cost was not a design criteria for 5-12h drying timespan. ( how many kill-a-watts does it take to remove x grams of water? at a low temp.)
– Tony EE rocketscientist
Sep 6 at 13:30
Let me save you a couple hours with a thermistor and P gain and setpoint you don't need a PID loop to control slow changing air temp, just P control 0 to 100% over a 5 degree 'C range. Heat mass, latency is another thing and so is dust control. (Bin thar dun that)
– Tony EE rocketscientist
Sep 6 at 13:34
They usually mention 400W but that's prolly for 70°C. They're almost made of thin plastic or stainless steel, which should dissipate quickly heat. So with 200-300W I could be fine I guess?
– Kinxil
Sep 6 at 9:32
They usually mention 400W but that's prolly for 70°C. They're almost made of thin plastic or stainless steel, which should dissipate quickly heat. So with 200-300W I could be fine I guess?
– Kinxil
Sep 6 at 9:32
1
1
To know which element is "more than powerful enough" one still has to make an estimation of what is enough. At which point it makes sense to use an element that is "just a bit more than enough", which will be cheaper.
– Dmitry Grigoryev
Sep 6 at 11:09
To know which element is "more than powerful enough" one still has to make an estimation of what is enough. At which point it makes sense to use an element that is "just a bit more than enough", which will be cheaper.
– Dmitry Grigoryev
Sep 6 at 11:09
I guess energy consumption cost was not a design criteria for 5-12h drying timespan. ( how many kill-a-watts does it take to remove x grams of water? at a low temp.)
– Tony EE rocketscientist
Sep 6 at 13:30
I guess energy consumption cost was not a design criteria for 5-12h drying timespan. ( how many kill-a-watts does it take to remove x grams of water? at a low temp.)
– Tony EE rocketscientist
Sep 6 at 13:30
Let me save you a couple hours with a thermistor and P gain and setpoint you don't need a PID loop to control slow changing air temp, just P control 0 to 100% over a 5 degree 'C range. Heat mass, latency is another thing and so is dust control. (Bin thar dun that)
– Tony EE rocketscientist
Sep 6 at 13:34
Let me save you a couple hours with a thermistor and P gain and setpoint you don't need a PID loop to control slow changing air temp, just P control 0 to 100% over a 5 degree 'C range. Heat mass, latency is another thing and so is dust control. (Bin thar dun that)
– Tony EE rocketscientist
Sep 6 at 13:34
add a comment |Â
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