As discussed, blowdown in relief valves arises from a geometric change, as seen by the fluid. It’s induced by the moving part(s) internal to the valve, and/or changes in pressure distribution on these moving internal parts during the relief-flow phase.

Digging deeper into these causes, consider the simplest form of relief valve typical of fluid systems — a valve utilizing a spring-loaded ball, Figure 1. When the valve is closed, the poppet (ball) experiences a net seating force, where p1A1 < F, Figure 1a.

Here, the system pressure, p1, acting over the seat area, A1, is less than the spring force, F. When p1 exceeds the set value, Figure 1b, the force balance shifts and p1A1 exceeds F. This lifts the ball off the seat, permitting flow to the reservoir, which is at p0. Notice that the ball sees a totally different pressure distribution due to the flow as it passes around it: some pressure is applied locally around the seat area; some pressure (p2) is applied over a greater area (A2) around the sides of the ball near the bottom; and some pressure (p3) is applied to an area A3 in the spring zone at the top of the ball. Although the relief flow ultimately occurs between p1 and p0, intermediate paths sustain pressure drops so that p1 > p2 > p3 > p0. This variation is due to intermediate valve geometry, essentially a series of orifices that the flow must pass through.  The area of these orifices is related to the geometry, Figure 2. In this example, O1 is the equivalent orifice at the inlet seat; O2 is the equivalent orifice for flow past the bottom of the ball; O3 is the equivalent orifice for some flow past the top of the ball; and O4 is the equivalent orifice for the final relief flow into the sump.

The exact modeling of these orifices can be quite a difficult task, because the valve’s dynamics may vary considerably. Therefore, the above orifice concept must only be regarded qualitatively as a method to help explain blowdown.

Using the orifice concept to determine the pressure distributions on areas A1, A2, and A3 (Figure 1b, again) is a complex task beyond the scope of this article. However, for simplicity, the following general relations hold:

When p1 < pset, then F >p1A1 with the valve closed.

When p1 = pset, then F = p1 A1 with the valve about to open.

After p1 > pset, flow commences, and pressures in the valve adjust so that:

F + p3 A3 = p1 A1 + p2 A2

Note that as the spring lifts, spring force rises slightly due to the spring rate (F2 > F1). The discrete pressures and areas were arbitrarily assigned to simplify explanation of the action of flow on the valve’s internal components. In actuality, discrete pressures at specific locations as shown might apply differently.

For example, pressures might vary from one vertical location on the ball to another as a function of the ball’s vertical movement. p1 and A1 are the pressures and areas that interact at the various vertical ball positions to define the force balance during flows. Using these notations, the generalized force balance along the poppet motion direction becomes the vectoral summation of all the individual pressures and areas. Note that pressures act normal to the areas. Actual performance of this task is complex and beyond the scope of this discussion.

### Single-Stage Valve Example

Consider Figure 3, a simple direct operating single-stage valve. The system pressure is seen as p1 applied over the port seat diameter, d1. The poppet is loaded by spring force, F. The valve will begin to open when F1 = p1A1. Flow then passes through the long relief pipe to the reservoir at pressure p0. The pipe’s length exhibits a resistance to flow, so a pressure drop occurs along this path. This is represented schematically as an orifice. Thus, it’s assumed that the poppet is completely open (and shows little resistance to flow) during relief flow, and the entire internal portion of the valve between the seat and the orifice is at system pressure.

Note that the upper body of the poppet is larger than the seat diameter, d2, which sees a pressure of p1. The upward force balance on the poppet is p1 A2. So, for this example:

A2 = π(d22)/4.

An additional lift force now exists on the poppet, and it will not reseat until p1 drops below the original set pressure. Assuming that the spring force remains constant during the flow period, the reseat pressure is:

p1 (reseat) = p1 (set) × (d1/d2)2.

Because d1/d2 is carried as a square ratio, the reseat pressure can be quite low with a significant loss of system pressure. For example, if the poppet diameter, d2, is 0.5 in., and the seat diameter is 0.25 in. — a ratio of 2:1 — blowdown is 75%, which is excessive. In this simple case, two solutions are possible:  shorten the relief pipe so that the valve internals see p0 during flow; or remove the poppet’s O-ring so that the additional area due to d2 will be partially balanced in the spring zone. The latter solution may result in limit-cycle chatter.