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EJ22 gets warmer with the heat on?
#1
Posted 03 October 2012 - 01:29 AM
Any Ideas? I've never encountered this..
Josh
#2
Posted 03 October 2012 - 10:19 AM
Heater core returns to right before the Thermostat(heater core is also the warm up bypass)
Here's my theory:
Cool water returns from heater core, somewhat closes the stat. Flow through the radiator is reduced. Engine runs a bit hotter, till the Stat opens up again....the whole dance balances back out, but at a slightly higher temp.
Heater off = Thermostat opens at normal engine temp.
Heater on = Thermostat opens at higher engine temp, because bypass water is cooling off stat.....overall temp has to be higher to open to radiator.
I've seen this happen before.....usually means that the T-stat is not a Subaru OEM.
All other things functioning well, this shouldn't make more than a few degrees difference.
If the car is overheating, there is another issue.
#3
Posted 03 October 2012 - 11:13 AM
If the EA cluster gauge is hooked up wrong, it might read backwards.
Edited by kanurys, 03 October 2012 - 11:17 AM.
#4
Posted 03 October 2012 - 04:33 PM
I'm getting my readings from the stock ea82 cluster gauge without a resistor spliced in, so normal for me is at about 1/8" above the last cold line. It seems to work fine, starts at cold in the morning and moves up to it usual spot when going down the road. It doesn't seem like its overheating, but the needle is moving enough off its normal spot to let me know somethings going on and makes me feel uneasy using the heat.
Josh
#5
Posted 03 October 2012 - 06:34 PM
Hook the computer up to a scan tool and read the temp. My point is it might be cooling, not getting hotter.
#6
Posted 04 October 2012 - 12:28 PM
My '93 EJ uses an OBD1 ECU, so scan tools are hard if not impossible to find and I haven't adapted the SSM port to connect to my PC yet. The reading in ohms, at normal operating temperture, at the sensor should be around 200-500 ohms right?
Josh
#7
Posted 04 October 2012 - 12:54 PM
#8
Posted 05 October 2012 - 10:37 AM
You could put in the pull-up resistor and see what the gauge does. The 270 ohm 1/4w resistor worked great on my EJ swap. Make sure you connect it to 12v, not ground.
Uh....
Resitor hooked to 12v? That's not right.
Resistor should be parallel to the sender, to ground. Works best if done near the engine.
Easy way is to simply strip some insulation back on the Temp sender wire, solder on another pigtail. Solder the resistor inline between that pigtail and ground to intake.
#9
Posted 05 October 2012 - 11:49 AM
from numbchux ea2ea: "Solution 4: In theory, you could wire a 275 Ohm resistor from the wire between the sender and the gauge to ground,..."
That's what I did and it worked great.
#10
Posted 06 October 2012 - 11:16 AM
"Solution 4: In theory, you could wire a 275 Ohm resistor from the wire between the sender and the gauge to ground,..."
.
this is correct.
No mention of adding 12v.
add a wire between the sender and gauge (think "T"), then that wire goes to ground.
This means there are 2 paths to ground for the gauge. This halves the total resistance
#11
Posted 06 October 2012 - 12:41 PM
A smaller resistor would mean that the sensor would have less effect on the gauge reading (more correction from stock). A larger resistor would mean the sensor would have more effect on the gauge reading (less correction from stock).
http://en.wikipedia....Current_divider
#12
Posted 08 October 2012 - 08:38 PM
Halves the resitance of each item in parallel.
A= gauge resistance
B= correction resitance
C= resitance at gauge/display value
(A/2)+(B/2)=C
Edited by Gloyale, 08 October 2012 - 08:41 PM.
#13
Posted 09 October 2012 - 12:46 AM
my understanding is this
Halves the resitance of each item in parallel.
A= gauge resistance
B= correction resitance
C= resitance at gauge/display value
(A/2)+(B/2)=C
Close ... actually it's C+ 1/((1/A) + (1/B) ) = total resistance
Edited by kanurys, 09 October 2012 - 01:03 AM.
math error
#14
Posted 09 October 2012 - 12:57 AM
So really, we're talking
volts = current*total resistance
Not sure how much current flows through the gauge in a stock setup, but the equation would look like
13.5=(plug measured current in here in amps)*(C+(1/((1/A)+(1/B))))
And solve for B
I'm sure there are different voltages or other changes because of the signal coming from the ecu.
How goes the cooling system woes?
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