Question

The area of cross-section of a water pipe entering the basement of a
house is 4 × 10–4 m2. The pressure of water at this point is 3 × 105 N/m2,
and speed of water is 2 m/s. The pipe tapers to an area of cross section
of 2 × 10–4 m2, when it reaches the second floor 8 m above the basement.
Calculate the speed and pressure of water flow at the second.

Answer 1

The speed of water flow is 4.4 ms−1 and the pressure is 2.43×10^5 Nm−2.

Explanation:

Here

a1 = 4×10^−4 m^2

P1 = 3.5×10^5 Nm−2

υ1 = 2.2 ms−1

a2 = 2×10^−4 m^2

ρ=10^3 kgm−3

According to equation of continuity

a1υ1=a2υ2

υ2 = a1υ1 / a2

     =  (4×10^−4) × 2.22×10^−4 = 4.4 ms−1

According to Bernoulli's Theorem

P1 + 1 / 2ρυ1^2 + ρgh1 = P2 + 1/2ρυ2^2 + ρgh2

P1 − P2 = ρg(h2 − h1) + 1/2ρ(υ2^2−υ1^2)

= 3.5 × 10^5 − 10^3 × 10 × 10  

= − 12 × 10^3 [(4.4)^2 − (2.2)^2]

= 3.5 × 10^5 − 1 × 10^5 − 0.07 × 10^5  

= (3.5×1.07) × 10^5

= 2.43×10^5 Nm−2.

Hence the speed of water flow is 4.4 ms−1 and the pressure is 2.43×10^5 Nm−2.

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State and explain Bernoulli theorem and write two uses ?

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