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)²/2 - (.5307)²/2]
P1 - P2 = 140.5 Pa
P2 = 10^5 - 140.5 = 99859.5 Pa.
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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
2 answers
given;
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
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
0 answers