Assuming the conditions of steady flow, the equations are as follows:

 B ds/c = Tc – W = 0, (5.3.28) B i*t = w – T = o- (5.3.29) d B dSW = /c – /t = 0- (5.3.30) Sc d Tc = Tc, ss – Tc = 0- ds (5.3.31) dB B2 (5.3.32) = (rext – /cr) = 0. ds i

From the third of the above equations, Eq. 5.3.30, in steady flow let,

3 = 0ts = /s0- (5.3.33)

The steady flow conditions are obtained from the first two of the above Eqs. 5.3.28 and 5.3.29 and are given by Wcss = Wtss; i. e.,

Wc, ss(/S0) = WC0 + f(^2 + 3 ^ _ (Ъ0 — = (/jQy)2

(5.3.34)

A parameter p, is defined as,

p =(2/H)(F/Ctyn)2, (5.3.35)

where p is the throttle non-dimensional pressure rise at minimum flow and a parameter p0,

2

P0 =- WC0 + 2 (5.3.36)

H

which is the ratio of the non-dimensional pressure rise at minimum flow to a quarter of the peak to peak variation of the pressure fluctuation at the compressor exit; then, Eq. 5.3.24 reduces to,

H H

ss(/s0) = 2 (P0 + 3x — = P 2 (1 + x)2; (5.3.37)

where the variable x is,

x =(/sc/F)- 1. (5.3.38)

If one assumes that with the minimum flow through the compressor and the throttle, the flow is always steady, then with /s0/F = 1, and one obtains from Eq. 5.3.35,

P0 = p. (5.3.39)

Assuming that the position of the throttle у is set to a nominal value у = yn when Eqs. 5.3.37-5.3.39 are satisfied, Eq. 5.3.37 may be rearranged and written

as,

H

Wc0 = – (p – 2). (5.3.40)

Eliminating Wc0, the steady flow characteristics may be defined entirely in terms of the compressor and throttle map parameters, H, F, and the product ynCt and is:

and Eq. 5.3.41 may be expressed as,

x(x2 + px + 2p — 3) = 0. (5.3.42)

From the first factor of Eq. 5.3.42, the assumed solution /s0F = 1 is recovered. Assuming x = 0 and solving for p,

P = (3 — (Us0=F — 1)2) / (Us0=F + 1). (5.3.43)

If one assumes that with the flow through the compressor and the throttle either minimum or below minimum, it is always steady, then with x = x0, when,

H H

— (p0 + 3×0 — 4) = p 2(1 + x0)2. (5.3.44)

Eliminating p0 one obtains,

(3(x — x0) — (x3 — x3)) = 2p(x — x0) + p(x2 — x0). (5.3.45)

Solving for p, one obtains,

When x0 = 0, Eq. 5.3.46 reduces to 5.3.43.

Updated: October 27, 2015 — 12:09 pm