Question

Water flows through a horizontal pipe of which the cross - section is not constant. The
pressure is 1cm of mercury where the velocity is 0.35m/s. Find the pressure at a point where
the velocity is 0.65m/s.
​

Answer 1

$v_1=0.35\ m\ s^{-1}\quad\quad P_1=1\ cm\ Hg=1333\ Pa\\\\v_2=0.65\ m\ s^{-1}\quad\quad P_2=?$

By Bernoulli's Theorem,

$P_1+\dfrac {1}{2}\rho(v_1)^2=P_2+\dfrac {1}{2}\rho(v_2)^2$

Potential energy is not considered since water flows through a horizontal pipe.

$1333+\dfrac {1}{2}\times 1000(0.35)^2=P_2+\dfrac {1}{2}\times1000(0.65)^2\\\\\\\underline {\underline {P_2=1183\ Pa}}$

Or,

$\underline {\underline {P_2=0.887\ cm\ Hg}}$

Answer 2

Explanation:

\begin{gathered}v_1=0.35\ m\ s^{-1}\quad\quad P_1=1\ cm\ Hg=1333\ Pa\\\\v_2=0.65\ m\ s^{-1}\quad\quad P_2=?\end{gathered}

v

1

=0.35 m s

−1

P

1

=1 cm Hg=1333 Pa

v

2

=0.65 m s

−1

P

2

=?

By Bernoulli's Theorem,

P_1+\dfrac {1}{2}\rho(v_1)^2=P_2+\dfrac {1}{2}\rho(v_2)^2P

1

+

2

1

ρ(v

1

)

2

=P

2

+

2

1

ρ(v

2

)

2

Potential energy is not considered since water flows through a horizontal pipe.

\begin{gathered}1333+\dfrac {1}{2}\times 1000(0.35)^2=P_2+\dfrac {1}{2}\times1000(0.65)^2\\\\\\\underline {\underline {P_2=1183\ Pa}}\end{gathered}

1333+

2

1

×1000(0.35)

2

=P

2

+

2

1

×1000(0.65)

2

P

2

=1183 Pa

Or,

\underline {\underline {P_2=0.887\ cm\ Hg}}

P

2

=0.887 cm Hg