Question 6, chap 9, sect 7. part 1 of 1 10 points Note: Patm = 101300 Pa/atm. Th

Question

Question 6, chap 9, sect 7. part 1 of 1 10 points Note: Patm = 101300 Pa/atm. The viscos- ity of the fluid is negligible and the fluid is incompressible.. A liquid of density 1219 kg/m3 flows with speed 2.21 m/s into a pipe of diameter 0.27 m. The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 7.51 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1 atm. The acceleration of gravity is 9.8 m/s2 . P1 2.21 m/s 0.27 m P2 1atm v2 0.05 m 7.51m 5.2 m/s 3.2 cm v2 2.3 cm Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe? Answer in units of Pa.

Explanation / Answer

According to Bernoulli's principle:
p/ + g·h + (1/2)·v² = constant
<=>
p/ + g·h + (1/2)·v² = p/ + g·h + (1/2)·v²
<=>
p = p + ·[ g·(h-h) + (1/2)·(v²-v²) ]

Assuming stationary conservation of mass requires, that mass flow at any cross section of the pipe is the same:
·v·A = ·v··(d/2)² = constant
=>:
·v··(d/2)² = ·v··(d/2)²
=>
v = v·(d/d)²

v2= 64.44 m/s

Hence:
P1 = P2 + density*(1/2*(v2^2 - v1^2) + g*(h2-h1))
= 101300Pa + 1219kg/m³·[ 9.81m/s²·(-7.51m) + (1/2)*(·(64.44m/s)²-(2.21m/s)²·) ]
= 2448472 Pa

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