Flynnwood
Established Member
The question at the end is a yes/no
We've had a large gas fired heater fitted at work. Four large air inlets (one each side), with one large heat outlet - with heat being pumped out by a large fan. It's a big heater.
This started blowing a 10 amp fuse inside the heater.
On the 2nd visit by the engineer, the manufacturer told him to cover up 75% of the air inlet, to reduce the amps down to 7.8 (which it did). I saw it with my own eyes on his meter.
This whole approach seemed odd to me though (and the engineer), so my research led to this:
"This equation, i=v/r, tells us that the current (i), flowing through a circuit is directly proportional to the voltage (v), and inversely proportional to the resistance (r). So if the voltage is increased, the current will increase." ( I get that bit)
"But if the resistance is increased, the current will decrease."
My question: Why would/how covering up 75% of the air inlets actually reduce the resistance in the 'actual fan', or is it some by-product? (I don't get that bit, despite the measured reduction).
Why a manufacturer would make such a heater this way (with 4 large intakes that could cause too many amps to blow internal fuses) is beyond me. I won't mention the manufacturer, as I would like purely impartial electrical advice if possible.
Thank you.
We've had a large gas fired heater fitted at work. Four large air inlets (one each side), with one large heat outlet - with heat being pumped out by a large fan. It's a big heater.
This started blowing a 10 amp fuse inside the heater.
On the 2nd visit by the engineer, the manufacturer told him to cover up 75% of the air inlet, to reduce the amps down to 7.8 (which it did). I saw it with my own eyes on his meter.
This whole approach seemed odd to me though (and the engineer), so my research led to this:
"This equation, i=v/r, tells us that the current (i), flowing through a circuit is directly proportional to the voltage (v), and inversely proportional to the resistance (r). So if the voltage is increased, the current will increase." ( I get that bit)
"But if the resistance is increased, the current will decrease."
My question: Why would/how covering up 75% of the air inlets actually reduce the resistance in the 'actual fan', or is it some by-product? (I don't get that bit, despite the measured reduction).
Why a manufacturer would make such a heater this way (with 4 large intakes that could cause too many amps to blow internal fuses) is beyond me. I won't mention the manufacturer, as I would like purely impartial electrical advice if possible.
Thank you.