Title (eBook) Transient Pipe Flow in Pipelines and Networks - The Newest Simulation Methods 6.3 MB 369
##### Document Text Contents
Page 184

168 Simplified liquid flow solution

Pipe Flow 1: Single-phase Flow Assurance For non-commercial use only

enough for the pressure loss across the valve to be negligible or as low as the minimum

Kf for the actual valve.

7.3.3 Valve located any other place than inlet or outlet

In the previous example, we saw if the valve is located at the outlet of a pipe, simple

algebra could deliver a simple equation for computing an imaginary KB at the

boundary. If the valve was located at the inlet, KA could be determined in a similar way.

But what happens if the valve is located somewhere along a pipe? We will see that

„cleanest‟ formulation of the problem leads us to a relatively complicated set of

equations, but those equations may be solved relatively easily with Newton-iteration.

Figure 7.3.2. Inline valve.

The valve is again described by equation 7.3.3, but the

valve equation is now relevant at both sides of the valve.

The pipe where the valve is located is computationally

regarded as two pipes connected via the valve.

At the valve boundary of pipe 1, equations 7.2.13 and 7.2.14 allow us to express pc as a

function of the known quantity KA1 and vc:

𝑝𝐶1 = 𝐾𝐴1 − 𝜌𝑎𝑠1𝑣𝐶1 (7.3.6)

The density is not indexed, meaning we assume it is the same everywhere. Speed of

sound, on the other hand, may differ in the different pipes, so we need to put an index

on it when more than one pipe is involved.

Similarly, at pipe 2‟s inlet, we express:

𝑝𝐶2 = 𝐾𝐵2 + 𝜌𝑎𝑠2𝑣𝐶2 (7.3.7)

Continuity means 𝑣𝐶1 = 𝑣𝐶2, and 𝑝𝐶1 − 𝑝𝐶2 must equal the pressure loss in the valve

according to equation 7.3.3:

1 2

V-1