where is heatercore located on a 1999 chevyprism
where is heatercore in a 1995 Chevy cavilier
you have to remove your steering wheel and dash completely out of the car then you can access the heatercore
in the engine compartment, passenger side fire wall by the heatercore hose connections. probably black, but could also be grey.
connect a piece of heater hose from the thermostate housing (where one of the hoses connected to from the heater core already is) to the connector for the coolant at the bottom of the enigine (where the second line should be connected).
Tits
There will be an input of some kind
You are going to want a service manual for this, because it involves removing a sizable portion of the dash board including the passenger airbag. You will likely need some (inexpensive) specialized tools to manage the hose connections as well.
If its the little "elbow" that comes out of the fire wall on the engine side then it means the heatercore is leaking. This is a drain hose for the ac and if the heater core leaks it drains thru this tube
First, with engine off make sure the heater hose will come off easy. With engine at operating temp. and heater inside car on high, slowly open the heater hose to let out any air that is trapped. Yes, fluid will come out also. Do this a few times and check inside temp.
Disconnect both heater hoses from engine side, fabricate a fitting to attach a garden hose to one of the hoses coming from the heater, direct the other hose into a container, slowly turn on the water and flush until it runs clear, then reverse hoses and flush again until clear, reconnect hoses to engine and refill reservoir.
The force at the input of a sprinkler can be found from F = PA; where P is the pressure of the water at the spigot or source of the water to the sprinkler and A is the x-sectional area of the input hose to the sprinkler. EX: A typical house-hold water pressure is about P = 60 psi in the US. And a typical hose is a 1/2 inch diameter; so A = pi (D/2)^2 = pi (.5/2)^2 in^2 is the x-sectional area. Which means the force at the input would be F = PA = 60pi()(.5/2)^2 = 11.78097245 ~ 12 pounds of input force. ANS.