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September 20, 2014

September 20, 2014

Posted by **Chris** on Tuesday, April 27, 2010 at 7:21pm.

A tube of radius 5 cm is connected to tube of radius 1 cm as shown above. Water is forced through the tube at a rate of 10 liters/min. The pressure in the 5 cm tube is 1×105 Pa. The density of water is 1000 kg/m3. Assume that the water is nonviscous and uncompressible.

(a) What is the velocity of the water in the 5 cm radius tube?

v1 = m/s

(b) What is the velocity of the water in the 1 cm radius tube?

v2 = m/s

(c) What is the water pressure in the 1 cm radius tube?

P2 = N/m2

- Physics -
**nick**, Wednesday, April 28, 2010 at 11:54pmA) V1= 10L/((pie)(.05^2)(1000)(60))=.0212m/s

B) V2= 10L/((pie)(.01^2)(1000)(60)=.5307m/s

C)P=P1 - P2 = 1000*[(.0212)²/2 - (.5307)²/2]

P1 - P2 = 140.5 Pa

P2 = 10^5 - 140.5 = 99859.5 Pa.

- Physics -
**nick**, Wednesday, April 28, 2010 at 11:58pmgiven;

r2 = 0.05 m

r1 = 0.01 m

Q = 10 liters/min = 10 E-3/60 = 1.67 E-4 m³/s

P2 = 10^5 Pa

ρ = 1000 kg/m³

no. # (a)

Q = (A2)(v2)

Q = π(r2)²)(v2)

1.67 E-4 = π(0.05)²)(v2)

V2 = 0.0212 m/s

no. # (b)

V1 = (r2/r1)² (V2) = (5/1)² (0.0212)

V1 = 0.5316 m/s

no. # (c)

using Bernoulli's equation in dynamics fluid

P1 + ½ ρ V1² + ρ g h = P2 + ½ ρ V2² + ρ g h

I assume there's no level differences and no losses of energy.

P1 + ½ ρ V1² = P2 + ½ ρ V2²

P1 + ½ (1000) (0.5316)² = 10^5 + ½ (1000) (0.0212)²

P1 = 99859 Pa

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