Question

A horizontal pipe has a diameter of 0.150 m at point 1 and 0.0500 m at point 2. The velocity of water at point 1 is 0.800 m/s and the pressure is 1.01 X 10^5 N/m^2. Determine the pressure at point 2.

Answer 1

principle of continuity

a1 x v1=a2 x v2

Answer 2

Answer:

Explanation:

Force/ area = pressure.

force = mass x acceleration

acceleration is change in velocity with respect to the time interval.

radius = 0.075 at point 1 = 75/1000 m

radius at point 2 = 0.025 m = 25/1000 m.          Hence 2x10^4=pressure at

velocity = 8/10 m/ s.                                             At point 2 neglecting velocity of        

acceleration = 10m/s^2 (gravity).                         Of water due to gravitational

Density of water is 1g/cc.                                    Acceleration.

pressure = effort

1.01 bar / 0.15m^2 =  X/ 0.05m^2

1.01x10^5/225/10^4= x/25/10^4

1.01x10^5/225=x/25

1/5x10^5=x

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