12) Water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm with a flow rate is 1.00 L
/s. At a valve, the effective cross-sectional area of the pipe is reduced to 5.00 cm2. What is the pressure at the valve? The density of water is 1000 kg/m3, and treat it as an ideal incompressible fluid
1 answer:
Answer:
The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]
Explanation:
We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.
For this case, we don't use the elevation difference and therefore those terms can be cancelled.
When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.
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Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be