AT&T LTE 9:19 PM 24%) x CO4_ F15_Q01_Stati... 1. Consider the below sketch of a

Question

AT&T LTE 9:19 PM 24%) x CO4_ F15_Q01_Stati... 1. Consider the below sketch of a differential mercury manometer with absolute pressure at Point A-14.7 psi absolate pressurek Point B-9.0 psi (gage pecssarc), 13.0I-11.0L a) Determine AMR in inches b) How moch change in head in feet (Ah) should be expocted between points A and B when AMR - 10 inches? Left AMR Right Flow Assumptions: Valves A, F, are open, while valves B, C. D and F are closed -Left side was peoperly bled Flow coateol valve Flow 1. Identify the valves that should be open and closed required to bleed the right side of the OPEN CLOSED 2. What was the last step used in the lab to verify that the manometer was calibrated?

Explanation / Answer

Prob.1

Apply the static pressure equilibrium at the given section passing through right and left limbs.

PA + (ZA-ZB)*unit weight of H2O + dMR*unit weight of Hg  = PB+dMR*unit weight of H2O

(a)

PA = 14.7 psi = 14.7*144 lbs/ft2 (Absolute) = (0 gage), and PB = 9 psie (gage) = 9*144 lbs/ft2,

unit weight of water = 62.4 pcf, and unit weight of Hg = 13.6*62.4 pcf

or 0+ (13-11)*62.4 + dMR*13.6*62.4 = (9)*144 + dMR*62.4

so, DMR = 1.489 feet = 17.875 inches

(b) put dMR = 10 inches = 0.8333 feet

0+ (dh)*62.4 + 0.8333*13.6*62.4 = (9)*144 + 0.833*62.4

=> dh= 10.269 feet

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