HOUSING-TURBINE 7E2746 - Caterpillar

Illustration 2 g00952767
When the engine is first started and the hydraulic oil is cold, the oil from the piston motor can not easily flow through the hydraulic oil cooler. The oil pressure will increase in the hydraulic oil cooler. The check valve for the hydraulic oil cooler bypass will open. The check valve limits the maximum oil pressure in the oil cooler to 450 55 kPa (65 8 psi). The oil from the case drain is filtered through a screen in the hydraulic oil tank.As the hydraulic oil temperature increases, the pressure of the oil through the oil cooler will decrease. The force of the spring in the check valve (hydraulic oil cooler bypass) is greater than the force of the oil pressure. Then, the check valve will close. The hydraulic oil will flow through the oil cooler into the hydraulic tank.When the engine is started, all temperatures for the four sensors are below the key target temperatures. The engine ECM sends the maximum current to the solenoid valve. Signal oil to the flow control spool is open to the hydraulic tank through the solenoid valve. Supply oil is directed to the actuator piston in order to destroke the pump. The angle of the pump swashplate is at a minimum. The pump will produce a minimum flow. As one of the temperatures of the four sensors increases above the key target temperature, the engine ECM sends a proportional reduction in current to the solenoid valve. The solenoid valve will start to shift. This will allow some of the supply oil to flow to the flow control spool. The flow control spool starts to shift to the left. A proportional amount of oil that is on the right side of the actuator piston will flow back to the hydraulic tank. As the pressure behind the actuator piston begins to decrease, the actuator spring will increase the swashplate angle. The pump output flow will increase. The fan speed will increase.As the temperatures of the machine continue to increase, the engine ECM will continue to reduce the current that is sent to the solenoid valve. The solenoid valve will continue to shiftin the downward direction. This will increase the hydraulic signal to the flow control spool. The control spool will shift more to the left in order to continue to drain oil that is behind the actuator piston. The swashplate moves more toward a maximum angle and the pump flow continues to increase. The fan speed continues to increase.
Illustration 3 g00952768
If the current from the engine ECM to the solenoid is at a minimum, the solenoid valve will send the maximum hydraulic signal to the flow control. Pump output flow will increase due to the swashplate that is moving toward the maximum angle.As the rpm of the hydraulic fan motor approaches the maximum speed,the pressure of the pump output also increases. The increase in pressure of the pump supply oil will work on the left side of both the flow