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Engineering

Valve-controlled actuator

Move the command of a four-way spool and watch the slide follow: with a closed-centre spool nothing happens inside the lap, then the speed lands on the quasi-static balance of the two metering orifices and friction, and at null the load is held tight. Switch to an open-centre spool and the valve leaks 2 Cd w u sqrt(ps/rho) at null while both chambers sit at half the supply, but the smallest command tilts the pressures steeply towards the load. Push against the slide and read the pressure difference it needs. The supply’s pQ is booked against the throttling, the load’s Fv, the friction and the oil’s compression.

By Mykhailo Makarov

Valve-controlled actuator — Actual application image
Actual application image

What happens

A four-way spool valve meters 100 bar oil to a symmetric cylinder sliding a 100 kg block on rails. The spool follows the command through a 5 ms lag; its four slots open by w·(x − u) for a closed-centre (overlapped) spool or w·(x + u) for an open-centre (underlapped) one, so the first has a deadband of the lap and the second a null leakage and a steep pressure gain. Turbulent orifice flows feed two compressible chambers, the piston carries the mass against Coulomb and viscous friction, an external force and elastic end stops. RK4 at 20 µs with the chamber pressures held above the tank’s; the supply’s pQ meets the throttling, the load’s Fv, the friction and the compression to rounding.

Try this

Give a small command and nothing moves inside the lap; a larger one and the slide runs at the orifice balance. Switch the centre open and read the null leakage and the pressure gain with the slide held.

The model and its limits

Rectangular slots without radial clearance or flow forces, a first-order spool, constant bulk modulus, friction as tanh. Gains and leakages are those of this idealized spool.

Equations used in the module
Equations
A_PA = w·(x − u) (closed centre),  A_PA = w·(x + u) (open centre)
Q = C_d·A·√(2Δp/ρ),  dp_A/dt = (β/V_A)·(Q_PA − Q_AT − A·v)
m·dv/dt = (p_A − p_B)·A − F_fr(v) − F_ext + F_stop
Q_null = 2C_d·w·u·√(p_s/ρ),  p_L = p_s·((u+x)² − (u−x)²)/((u+x)² + (u−x)²)

Related experiments

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Air compressor — Actual application image

Engineering

Air compressor

Start a single-cylinder compressor on a ten-litre receiver and watch the gauge climb: each stroke draws air through the intake plate, squeezes it hot and pushes it past the delivery plate once the cylinder beats the tank. The mass delivered per cycle shrinks as the tank fills, close to the clearance-volume closed form 1 - c ((p/p0)^(1/n) - 1); a larger clearance delivers less. At the set pressure the switch stops the motor, the unloader vents the cylinder and the machine coasts to rest; open the consumer and the tank settles where delivery meets the bleed. Motor work and intake enthalpy are booked against the gas’s internal energy, the heat to wall and room and the air sent on.

Steam engine mechanism — Actual application image

Engineering

Steam engine mechanism

Open the throttle on a horizontal engine fed with 6 bar air and watch the slide valve lead the piston: the eccentric’s lap and advance are solved so the steam port cracks open by the lead at dead centre and shuts at the chosen cutoff, exactly acos(1 - 2c) of crank angle. Shorten the cutoff and the air expands in the cylinder, less air per joule of work but less work per cycle; load the brake until the mean effective pressure can no longer carry it and the engine stalls; park the crank at dead centre and it will not start. The admitted enthalpy is booked against the exhaust, the wall, the gas’s energy and the indicated work, and the indicated work against the flywheel, the friction and the brake.

Stirling mechanism — Actual application image

Engineering

Stirling mechanism

Spin a beta-type Stirling engine whose hot cap stands at 650 K and whose cold space is kept at 300 K: the displacer moves the air between them, the power piston a quarter turn behind takes the work, and the pressure follows the Schmidt law M R / (Vh/Th + Vc/Tc + Vr/Tr). Read the cycle work against the Schmidt closed form and the heat in and out against Carnot’s efficiency; load the brake until the engine stalls; cool the hot end and watch the work shrink; set both ends to one temperature and the flywheel coasts to a stop. Change the phase angle and find the ninety-odd degrees that give the most work. The gas work is booked against the flywheel, the friction and the brake.