12_Control_Valves_Chap9S10_4

# 9 a control valve placed in series with a pump and a

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: are of flow rate) 2 ΔPhe ⎛ q ⎞ (9-5) =⎜ ⎟ 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 (9-6) 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 (9-7) 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 (9-2) (9-8) We will “pick” Cv = 125. For a linear characteristic valve, use the relation between f and q from Eq. 9-2 with f = : q = where ΔPv depends on q (9-9) Cv ΔPv / g s Using Eq. 9-9 and values of ΔPv ( q ) from Eq. 9-7 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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online