Question

A horizontal tube consists of a 7.0-cm-diameter pipe that narrows to a 2.0-cm-diameter throat. In the pipe, the water pressure is twice atmospheric pressure and the water flows with a speed of 0.40 m/s. What is the pressure in the throat, assuming that the water behaves like an ideal fluid? The density of water is 1000 kg/m3 , and atmospheric pressure is 1.01 × 105 Pa.

Answer 1

Answer:

$1.9\times10^{5} Pa$

Explanation:

$d_{p}$ = diameter of the pipe = 7 cm

$v_{p}$ = speed of water in the pipe = 0.40 m/s

$A_{p}$ = Area of cross-section of pipe = $(0.25)\pi d_{p}^{2}$

$d_{t}$ = diameter of the throat = 2 cm

$v_{t}$ = speed of water in the throat

$A_{t}$ = Area of cross-section of throat = $(0.25)\pi d_{t}^{2}$

Using equation of continuity

$A_{p} v_{p} = A_{t} v_{t} \\(0.25)\pi d_{p}^{2} v_{p} = (0.25)\pi d_{t}^{2} v_{t} \\(7)^{2} (0.40) = (2)^{2} v_{t}\\v_{t} = 4.9 ms^{-1}$

$P_{o}$ = atmospheric pressure = 1.01 x 10⁵ Pa

$P_{p}$ = Pressure in the pipe = $2 P_{o}$ = 2.02 x 10⁵ Pa

$P_{t}$ = Pressure in the throat

Using Bernoulli's theorem

$P_{t} + (0.5)\rho v_{t}^{2} = P_{p} + (0.5)\rho v_{p}^{2}\\P_{t} + (0.5)(1000) (4.9)^{2} = 2.02\times10^{5} + (0.5)(1000) (0.40)^{2}\\P_{t} + 12005 = 202080\\P_{t} = 190075 Pa\\P_{t} = 1.9\times10^{5} Pa$