Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is 2 cm, and the areas of cross section at A and B are 4 cm2 and 2 cm2 respectively, find the rate of flow of water across any section.
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student-name Athan Tay asked in Physics
Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes is 2 cm, and the areas of cross section at A and B are 4 cm2 and 2 cm2 respectively, find the rate of flow of water across any section.
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student-name Global Expert answered this
in Physics, Class
Given:
Difference in the heights of water columns in vertical tubes = 2 cm
Area of cross section at A, aA = 4 cm2
Area of cross section at B, aB = 2 cm2
Now, let vA and vB be the speeds of water at A and B, respectively.
From the equation of continuity, we have:
vAaA=vB×aB⇒ vA×4=vB×2⇒ vB=2vA ...(i)
From Bernoulli's equation, we have:
12ρv2A+ρghA+pA=12ρv2B+ρghB+pB⇒12ρv2A+pA=12ρv2B+pB⇒pA−pB=12ρ(v2B−v2A)
Here,
pA and pB are the pressures at A and B, respectively.
hA and hB are the heights of water columns at point A and B, respectively.
ρ is the density of the liquid.
Thus, we have:
12×1×(4v2A−v2A)⇒2×1×1000=12×1×3v2A[pA−pB=2 cm=2×1×1000 dyne/cm2 (water column)]⇒ v2A=40003−−−−√=36.51 cm/s∴Rate of flow=vAaA=36.51×4=146 cm3/s
Hence, the required rate of flow of water across any section is 146 cm3/s.