El Presidente Posted October 3, 2012 Share Posted October 3, 2012 Swapped a 93' EJ22 into my 87 GL wagon last Feb., but ever since this last august its been getting on the warm side in +85* weather. Before august, I didn't have any issues. When I have turned the heat on to boost the cooling system, it gets warmer instead(the motor). The air coming out is still nice and toasty, same as always, but recently I've noticed it gets warmer in cooler weather too when the heat is on. Any Ideas? I've never encountered this.. Josh Link to comment Share on other sites More sharing options...
Gloyale Posted October 3, 2012 Share Posted October 3, 2012 If your heater core is working really well the water returning from it could be signifigantly cooler than water in coming from the rad. 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. Link to comment Share on other sites More sharing options...
kanurys Posted October 3, 2012 Share Posted October 3, 2012 (edited) I was going to try to answer this but Gloyale pretty much sums it up. The other question I have is where are you getting your temp reading from? The modified cluster gauge or a scan tool and the ecu? If the EA cluster gauge is hooked up wrong, it might read backwards. Edited October 3, 2012 by kanurys Link to comment Share on other sites More sharing options...
El Presidente Posted October 3, 2012 Author Share Posted October 3, 2012 Thanks for the info guys! Sounds like I need an OEM t-stat! 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 Link to comment Share on other sites More sharing options...
kanurys Posted October 3, 2012 Share Posted October 3, 2012 I'd put the OEM stat in anyways. Stant makes a direct replacement and a cheapo. Search around. Hook the computer up to a scan tool and read the temp. My point is it might be cooling, not getting hotter. Link to comment Share on other sites More sharing options...
El Presidente Posted October 4, 2012 Author Share Posted October 4, 2012 Thanks again for the advice. 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 Link to comment Share on other sites More sharing options...
kanurys Posted October 4, 2012 Share Posted October 4, 2012 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. Link to comment Share on other sites More sharing options...
Gloyale Posted October 5, 2012 Share Posted October 5, 2012 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. Link to comment Share on other sites More sharing options...
kanurys Posted October 5, 2012 Share Posted October 5, 2012 Opps, my bad. It is a pull-down resistor. 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. Link to comment Share on other sites More sharing options...
Gloyale Posted October 6, 2012 Share Posted October 6, 2012 "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 Link to comment Share on other sites More sharing options...
kanurys Posted October 6, 2012 Share Posted October 6, 2012 Well, not exactly halves the voltage, but makes a current divider, proportionally changing the current going through the sensor, instead of all the current going through the sensor. If the sensor and the resistor were at exactly 275 ohms it would divide the current evenly. 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.org/wiki/Current_divider Link to comment Share on other sites More sharing options...
Gloyale Posted October 9, 2012 Share Posted October 9, 2012 (edited) 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 Edited October 9, 2012 by Gloyale Link to comment Share on other sites More sharing options...
kanurys Posted October 9, 2012 Share Posted October 9, 2012 (edited) 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 October 9, 2012 by kanurys math error Link to comment Share on other sites More sharing options...
kanurys Posted October 9, 2012 Share Posted October 9, 2012 And c is just another resistor (gauge). 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? Link to comment Share on other sites More sharing options...
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