Is DC heating faster than AC heating?What material and current do I need for a diy plant heater?Design of a nichrome heating element - achieving predetermined power dissipationRunning a 230V heating element using 120VEfficient DC battery power heaterHeating up water with electrical currentUsing a copper wire as a heatercalculating the size of a heating resistorThermo Electric Can Heater / Cooler Heating vs Cooling SpeedHeating resistor power limitSuggestions for improvement of a water heating control system
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Is DC heating faster than AC heating?
What material and current do I need for a diy plant heater?Design of a nichrome heating element - achieving predetermined power dissipationRunning a 230V heating element using 120VEfficient DC battery power heaterHeating up water with electrical currentUsing a copper wire as a heatercalculating the size of a heating resistorThermo Electric Can Heater / Cooler Heating vs Cooling SpeedHeating resistor power limitSuggestions for improvement of a water heating control system
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$begingroup$
If I have a 120V @ 50Hz AC heater rated at 750W, and I run it using 120V DC will the DC heat it faster, and why? (Assuming I could get a clean 120V DC power source).
heat
asked 9 hours ago
user77232user77232
1175 bronze badges
$endgroup$
add a comment |
$begingroup$
If I have a 120V @ 50Hz AC heater rated at 750W, and I run it using 120V DC will the DC heat it faster, and why? (Assuming I could get a clean 120V DC power source).
heat
asked 9 hours ago
user77232user77232
1175 bronze badges
$endgroup$
$begingroup$
Since the duty cycle of power is 50% and assuming that the heating element is somthing like nichrome, the inertia of this element can not respond that fast. Using proper adjustable PWM is different. In any case you can calculate the thermal time constant of the material
$endgroup$
– GR Tech
6 hours ago
$begingroup$
Note that while the heating element will probably not care, the fan motor might not work with DC. In any case, it's probably not a good idea to run these things out of spec due to the fire hazard…
$endgroup$
– wrtlprnft
54 mins ago
add a comment |
$begingroup$
If I have a 120V @ 50Hz AC heater rated at 750W, and I run it using 120V DC will the DC heat it faster, and why? (Assuming I could get a clean 120V DC power source).
heat
asked 9 hours ago
user77232user77232
1175 bronze badges
$endgroup$
If I have a 120V @ 50Hz AC heater rated at 750W, and I run it using 120V DC will the DC heat it faster, and why? (Assuming I could get a clean 120V DC power source).
heat
heat
asked 9 hours ago
user77232user77232
1175 bronze badges
asked 9 hours ago
user77232user77232
1175 bronze badges
asked 9 hours ago
user77232user77232
1175 bronze badges
asked 9 hours ago
user77232user77232
1175 bronze badges
asked 9 hours ago
user77232user77232
1175 bronze badges
1175 bronze badges
$begingroup$
Since the duty cycle of power is 50% and assuming that the heating element is somthing like nichrome, the inertia of this element can not respond that fast. Using proper adjustable PWM is different. In any case you can calculate the thermal time constant of the material
$endgroup$
– GR Tech
6 hours ago
$begingroup$
Note that while the heating element will probably not care, the fan motor might not work with DC. In any case, it's probably not a good idea to run these things out of spec due to the fire hazard…
$endgroup$
– wrtlprnft
54 mins ago
add a comment |
$begingroup$
Since the duty cycle of power is 50% and assuming that the heating element is somthing like nichrome, the inertia of this element can not respond that fast. Using proper adjustable PWM is different. In any case you can calculate the thermal time constant of the material
$endgroup$
– GR Tech
6 hours ago
$begingroup$
Note that while the heating element will probably not care, the fan motor might not work with DC. In any case, it's probably not a good idea to run these things out of spec due to the fire hazard…
$endgroup$
– wrtlprnft
54 mins ago
$begingroup$
Since the duty cycle of power is 50% and assuming that the heating element is somthing like nichrome, the inertia of this element can not respond that fast. Using proper adjustable PWM is different. In any case you can calculate the thermal time constant of the material
$endgroup$
– GR Tech
6 hours ago
$begingroup$
Since the duty cycle of power is 50% and assuming that the heating element is somthing like nichrome, the inertia of this element can not respond that fast. Using proper adjustable PWM is different. In any case you can calculate the thermal time constant of the material
$endgroup$
– GR Tech
6 hours ago
$begingroup$
Note that while the heating element will probably not care, the fan motor might not work with DC. In any case, it's probably not a good idea to run these things out of spec due to the fire hazard…
$endgroup$
– wrtlprnft
54 mins ago
$begingroup$
Note that while the heating element will probably not care, the fan motor might not work with DC. In any case, it's probably not a good idea to run these things out of spec due to the fire hazard…
$endgroup$
– wrtlprnft
54 mins ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
When you say 120V @ 50Hz AC you are implicitly saying 120Vrms.
The RMS voltage is qualitatively defined as the voltage which will give the same resistive heating (averaged out over time) as a DC voltage of the same number. Therefore, by the definition of RMS, the heating will be the same because the RMS voltages are the same.
If you said 120Vpeak or something different, then things would be different.
answered 9 hours ago
DKNguyenDKNguyen
5,4431 gold badge6 silver badges25 bronze badges
$endgroup$
1
$begingroup$
So the utility supply is rated at 120Vrms which is actually 170V peak?
$endgroup$
– user77232
9 hours ago
2
$begingroup$
@user77232 Indeed.
$endgroup$
– DKNguyen
9 hours ago
add a comment |
$begingroup$
No, because 120V RMS is the AC voltage that produces the same heat as 120 VDC. Theoretically it should be no difference at all.
answered 9 hours ago
Marko BuršičMarko Buršič
11.2k2 gold badges9 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
For heating, the only thing that matters is active power. If the load is a resistor, active power is $ V^2/R $ with V being the RMS voltage.
The RMS value of 120V DC is 120V RMS, which will give the same active power in a resistor than 120V AC RMS.
AC power will be pulsed, but a 750W resistor will have enough thermal mass to smooth it out, so there is no difference.
answered 9 hours ago
peufeupeufeu
26.4k2 gold badges39 silver badges78 bronze badges
$endgroup$
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you say 120V @ 50Hz AC you are implicitly saying 120Vrms.
The RMS voltage is qualitatively defined as the voltage which will give the same resistive heating (averaged out over time) as a DC voltage of the same number. Therefore, by the definition of RMS, the heating will be the same because the RMS voltages are the same.
If you said 120Vpeak or something different, then things would be different.
answered 9 hours ago
DKNguyenDKNguyen
5,4431 gold badge6 silver badges25 bronze badges
$endgroup$
1
$begingroup$
So the utility supply is rated at 120Vrms which is actually 170V peak?
$endgroup$
– user77232
9 hours ago
2
$begingroup$
@user77232 Indeed.
$endgroup$
– DKNguyen
9 hours ago
add a comment |
$begingroup$
When you say 120V @ 50Hz AC you are implicitly saying 120Vrms.
The RMS voltage is qualitatively defined as the voltage which will give the same resistive heating (averaged out over time) as a DC voltage of the same number. Therefore, by the definition of RMS, the heating will be the same because the RMS voltages are the same.
If you said 120Vpeak or something different, then things would be different.
answered 9 hours ago
DKNguyenDKNguyen
5,4431 gold badge6 silver badges25 bronze badges
$endgroup$
1
$begingroup$
So the utility supply is rated at 120Vrms which is actually 170V peak?
$endgroup$
– user77232
9 hours ago
2
$begingroup$
@user77232 Indeed.
$endgroup$
– DKNguyen
9 hours ago
add a comment |
$begingroup$
When you say 120V @ 50Hz AC you are implicitly saying 120Vrms.
The RMS voltage is qualitatively defined as the voltage which will give the same resistive heating (averaged out over time) as a DC voltage of the same number. Therefore, by the definition of RMS, the heating will be the same because the RMS voltages are the same.
If you said 120Vpeak or something different, then things would be different.
answered 9 hours ago
DKNguyenDKNguyen
5,4431 gold badge6 silver badges25 bronze badges
$endgroup$
When you say 120V @ 50Hz AC you are implicitly saying 120Vrms.
The RMS voltage is qualitatively defined as the voltage which will give the same resistive heating (averaged out over time) as a DC voltage of the same number. Therefore, by the definition of RMS, the heating will be the same because the RMS voltages are the same.
If you said 120Vpeak or something different, then things would be different.
answered 9 hours ago
DKNguyenDKNguyen
5,4431 gold badge6 silver badges25 bronze badges
edited 9 hours ago
answered 9 hours ago
DKNguyenDKNguyen
5,4431 gold badge6 silver badges25 bronze badges
answered 9 hours ago
DKNguyenDKNguyen
5,4431 gold badge6 silver badges25 bronze badges
answered 9 hours ago
DKNguyenDKNguyen
5,4431 gold badge6 silver badges25 bronze badges
5,4431 gold badge6 silver badges25 bronze badges
1
$begingroup$
So the utility supply is rated at 120Vrms which is actually 170V peak?
$endgroup$
– user77232
9 hours ago
2
$begingroup$
@user77232 Indeed.
$endgroup$
– DKNguyen
9 hours ago
add a comment |
1
$begingroup$
So the utility supply is rated at 120Vrms which is actually 170V peak?
$endgroup$
– user77232
9 hours ago
2
$begingroup$
@user77232 Indeed.
$endgroup$
– DKNguyen
9 hours ago
1
1
$begingroup$
So the utility supply is rated at 120Vrms which is actually 170V peak?
$endgroup$
– user77232
9 hours ago
$begingroup$
So the utility supply is rated at 120Vrms which is actually 170V peak?
$endgroup$
– user77232
9 hours ago
2
2
$begingroup$
@user77232 Indeed.
$endgroup$
– DKNguyen
9 hours ago
$begingroup$
@user77232 Indeed.
$endgroup$
– DKNguyen
9 hours ago
add a comment |
$begingroup$
No, because 120V RMS is the AC voltage that produces the same heat as 120 VDC. Theoretically it should be no difference at all.
answered 9 hours ago
Marko BuršičMarko Buršič
11.2k2 gold badges9 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
No, because 120V RMS is the AC voltage that produces the same heat as 120 VDC. Theoretically it should be no difference at all.
answered 9 hours ago
Marko BuršičMarko Buršič
11.2k2 gold badges9 silver badges12 bronze badges
$endgroup$
add a comment |
$begingroup$
No, because 120V RMS is the AC voltage that produces the same heat as 120 VDC. Theoretically it should be no difference at all.
answered 9 hours ago
Marko BuršičMarko Buršič
11.2k2 gold badges9 silver badges12 bronze badges
$endgroup$
No, because 120V RMS is the AC voltage that produces the same heat as 120 VDC. Theoretically it should be no difference at all.
answered 9 hours ago
Marko BuršičMarko Buršič
11.2k2 gold badges9 silver badges12 bronze badges
answered 9 hours ago
Marko BuršičMarko Buršič
11.2k2 gold badges9 silver badges12 bronze badges
answered 9 hours ago
Marko BuršičMarko Buršič
11.2k2 gold badges9 silver badges12 bronze badges
answered 9 hours ago
Marko BuršičMarko Buršič
11.2k2 gold badges9 silver badges12 bronze badges
11.2k2 gold badges9 silver badges12 bronze badges
add a comment |
add a comment |
$begingroup$
For heating, the only thing that matters is active power. If the load is a resistor, active power is $ V^2/R $ with V being the RMS voltage.
The RMS value of 120V DC is 120V RMS, which will give the same active power in a resistor than 120V AC RMS.
AC power will be pulsed, but a 750W resistor will have enough thermal mass to smooth it out, so there is no difference.
answered 9 hours ago
peufeupeufeu
26.4k2 gold badges39 silver badges78 bronze badges
$endgroup$
add a comment |
$begingroup$
For heating, the only thing that matters is active power. If the load is a resistor, active power is $ V^2/R $ with V being the RMS voltage.
The RMS value of 120V DC is 120V RMS, which will give the same active power in a resistor than 120V AC RMS.
AC power will be pulsed, but a 750W resistor will have enough thermal mass to smooth it out, so there is no difference.
answered 9 hours ago
peufeupeufeu
26.4k2 gold badges39 silver badges78 bronze badges
$endgroup$
add a comment |
$begingroup$
For heating, the only thing that matters is active power. If the load is a resistor, active power is $ V^2/R $ with V being the RMS voltage.
The RMS value of 120V DC is 120V RMS, which will give the same active power in a resistor than 120V AC RMS.
AC power will be pulsed, but a 750W resistor will have enough thermal mass to smooth it out, so there is no difference.
answered 9 hours ago
peufeupeufeu
26.4k2 gold badges39 silver badges78 bronze badges
$endgroup$
For heating, the only thing that matters is active power. If the load is a resistor, active power is $ V^2/R $ with V being the RMS voltage.
The RMS value of 120V DC is 120V RMS, which will give the same active power in a resistor than 120V AC RMS.
AC power will be pulsed, but a 750W resistor will have enough thermal mass to smooth it out, so there is no difference.
answered 9 hours ago
peufeupeufeu
26.4k2 gold badges39 silver badges78 bronze badges
answered 9 hours ago
peufeupeufeu
26.4k2 gold badges39 silver badges78 bronze badges
answered 9 hours ago
peufeupeufeu
26.4k2 gold badges39 silver badges78 bronze badges
answered 9 hours ago
peufeupeufeu
26.4k2 gold badges39 silver badges78 bronze badges
26.4k2 gold badges39 silver badges78 bronze badges
add a comment |
add a comment |
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$begingroup$
Since the duty cycle of power is 50% and assuming that the heating element is somthing like nichrome, the inertia of this element can not respond that fast. Using proper adjustable PWM is different. In any case you can calculate the thermal time constant of the material
$endgroup$
– GR Tech
6 hours ago
$begingroup$
Note that while the heating element will probably not care, the fan motor might not work with DC. In any case, it's probably not a good idea to run these things out of spec due to the fire hazard…
$endgroup$
– wrtlprnft
54 mins ago