what is the answer and could yo u go into detail on how you worked the problem?

Question

what is the answer and could yo

u go into detail on how you worked the problem?

Sapling Learning A liquid of density 1.37 x103 kg/m3 flows steadily through a pipe of varying diameter and height. At location 1 along the pipe the flow speed is 9.51 m/s and the pipe diameter 12.5 cm is 12.5 cm. At location 2 the pipe diameter is 15.3 cm. At location 1 the pipe is 8.13 m higher than it is at location 2. lgnoring viscosity, calculate the difference between the fluid pressure at location 2 and the fluid pressure at location 1. Number 8.13 m 15.3 cm

Explanation / Answer

v1 = 9.51 m/s

use below formula to find v2
A1*v1 = A2*v2
d1^2*v1 = d2^2*v2
12.5^2 * 9.51 = 15.3^2*v2
v2 = 6.35 m/s

you need to use Bernoulli's pronciple:
P1 + 0.5*rho*v1^2 + rho*g*h1 = P2 + 0.5*rho*v2^2 + rho*g*h2
P2-P1 = 0.5*rho*(v1^2 - v2^2) + rho*g*(h1-h2)
= 0.5*1.37*10^3*(9.51^2 - 6.35^2) + 1.37*10^3*9.8*8.13
= 34330.6 + 109153.4
= 143484 Pa

Answer: 143484 Pa

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