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Unformatted text preview: are of
flow rate)
2
ΔPhe ⎛ q ⎞
(95)
=⎜
⎟
30
⎝ 200 ⎠ ΔPv Figure 9.9 A control valve placed in series with a pump and
a heat exchanger. Pump discharge pressure is constant.
di ⎛q⎞
ΔPs = ΔPhe = 30 ⎜
⎟
⎝ 200 ⎠ 2 (96) Because the pump head is constant at 40 psi, the pressure drop
available for the valve is
⎛q⎞
ΔPv = 40 − ΔPhe = 40 − 30 ⎜
⎟
⎝ 200 ⎠ 2 (97) Figure 9.11 illustrates these relations. Note that in all four design
cases ΔPv / ΔPs = 10 / 30 = 33% at qd.
18 19 a) First calculate the rated Cv from q = Cv f for f = 0.5, gs = 1.0. Chapter 9 Chapter 9 Cv = Figure 9.11 Pump characteristic and system pressure drop
for Example 9.2. As valve is opened to allow more flow, the
pressure drop across the valve drops and you get less flow
than if ΔPv had stayed constant. () 200 gpm
= 126.5
0.5 (10 psig/1.0) ΔPv
gs (92) (98) We will “pick” Cv = 125. For a linear characteristic valve, use
the relation between f and q from Eq. 92 with f = :
q
=
where ΔPv depends on q (99)
Cv ΔPv / g s
Using Eq. 99 and values of ΔPv ( q ) from Eq. 97 for a set of
values of flow q, the installed valve characteristic curve
(q vs. ) can be plotted from values of for a set of q values....
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 Winter '08
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