A funnel of water is connected to pipes as shown in the figure. The top of the f
ID: 1520975 • Letter: A
Question
A funnel of water is connected to pipes as shown in the figure. The top of the funnel is sufficiently large that the speed downwards of the water (? = 1000 kg/m3) at the top of the funnel is nearly zero. The water is observed to be moving at 2.0 m/s through the pipe at point B and the water is observed to exit pipe C at a speed of 2.8 m/s. Note that the pressure at point C is the atmospheric pressure.
1. Compare the pressure at point A in the funnel with the pressure at point C in the horizontal pipe. Assume that the speed at point A is 0.5 m/s.
(a) PA > PC
(b) PA = PC
(c) PA < PC
2. Calculate the difference between the pressure at point B, 2 m below the surface of water, and atmospheric pressure.
(a) PB - PAtm = 1.6 × 104 N/m2
(b) PB - PAtm = 1.8 × 104 N/m2
(c) PB - PAtm = 2.0 × 104 N/m2
(d) PB - PAtm = 2.2 × 104 N/m2
(e) PB - PAtm = 2.4 × 104 N/m2
3. The radius of the pipe at point B is 0.15 m. What is the radius of the pipe at point C?
(a) rc = 0.110 m (b) rc = 0.127 m (c) rc = 0.178 m
Explanation / Answer
from the Bernoulis equation
pA-pC = 1/2 rho ( vC^2 - vA^2
pA = pC + 1/2 rho ( vC^2 - vA^2
= 101325 pa + 1/2 * 1000 ( 2.8^2 - 0.5^2 )
= 105120 Pa
sp pA> pC
(2)
PB- PA = rho gh = 1000 ( 9.8) ( 2 m)= 2 * 10^4 N/m^2
(3)
AB vB = Ac vC
pi rB^2 vB = pi rC^2 vC
rC =sqrt rB^2 vB /vC
= sqrt (0.15)^2 ( 2)/2.8
=0.127 m
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