Homework Help: Physics

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

it didnt let me put the link to image..

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 11:54pm
A) 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 - (.5307)˛/2]
P1 - P2 = 140.5 Pa
P2 = 10^5 - 140.5 = 99859.5 Pa.
Physics - nick, Wednesday, April 28, 2010 at 11:58pm
given;
r2 = 0.05 m
r1 = 0.01 m
Q = 10 liters/min = 10 E-3/60 = 1.67 E-4 mł/s
P2 = 10^5 Pa
ρ = 1000 kg/mł

no. # (a)

Q = (A2)(v2)

Q = π(r2)˛)(v2)

1.67 E-4 = π(0.05)˛)(v2)

V2 = 0.0212 m/s

no. # (b)

V1 = (r2/r1)˛ (V2) = (5/1)˛ (0.0212)

V1 = 0.5316 m/s

no. # (c)

using Bernoulli's equation in dynamics fluid

P1 + ˝ ρ V1˛ + ρ g h = P2 + ˝ ρ V2˛ + ρ g h

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

P1 + ˝ ρ V1˛ = P2 + ˝ ρ V2˛

P1 + ˝ (1000) (0.5316)˛ = 10^5 + ˝ (1000) (0.0212)˛

P1 = 99859 Pa

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