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Derive the diagram to this equation 2.2-15

Derive the diagram to this equation 2.2-15

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Derive the diagram to this equation 2.2-15
own in Fig. 2.2-4a, is being used to measure the head or pres- How meter. The heavier fluid is mercury, with a density of op fluid is water, with a density of 1.00 g/cm³. The reading on = 32.7 cm. Calculate the pressure difference in N/m² using mg R to m, R = -fluid B PB - fluid A PA 32.7 100 M = 0.327 m PB- tay R. Pb (b) to measure pressure differences: (a) U tube; (b) two-fluid PA
Transcribed Image Text:own in Fig. 2.2-4a, is being used to measure the head or pres- How meter. The heavier fluid is mercury, with a density of op fluid is water, with a density of 1.00 g/cm³. The reading on = 32.7 cm. Calculate the pressure difference in N/m² using mg R to m, R = -fluid B PB - fluid A PA 32.7 100 M = 0.327 m PB- tay R. Pb (b) to measure pressure differences: (a) U tube; (b) two-fluid PA
2. Two-fluid U tube. In Fig. 2.2-4b a two-fluid U tube is shown, which is a sensitive device for measuring small heads or pressure differences. Let A m² be the cross-sectional area of each of the large reservoirs and a m² be the cross-sectional area of each of the tubes forming the U. Proceeding and making a pressure balance as for the U tube, a a R - RO) (P₁ - PB + PB - PC) 8 Pc PaPb (R- RO) PA (2.2-15) where Ro is the reading when p. = Pb, R is the actual reading, PA is the density of the heavier fluid, and PB is the density of the lighter fluid. Usually, a/A is made sufficiently small as to be negligible, and also Ro is often adjusted to zero; then Pa- Pb = R(PA-PB)8 (SI) (22-10
Transcribed Image Text:2. Two-fluid U tube. In Fig. 2.2-4b a two-fluid U tube is shown, which is a sensitive device for measuring small heads or pressure differences. Let A m² be the cross-sectional area of each of the large reservoirs and a m² be the cross-sectional area of each of the tubes forming the U. Proceeding and making a pressure balance as for the U tube, a a R - RO) (P₁ - PB + PB - PC) 8 Pc PaPb (R- RO) PA (2.2-15) where Ro is the reading when p. = Pb, R is the actual reading, PA is the density of the heavier fluid, and PB is the density of the lighter fluid. Usually, a/A is made sufficiently small as to be negligible, and also Ro is often adjusted to zero; then Pa- Pb = R(PA-PB)8 (SI) (22-10
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