Question

The pressure of water inside the closed
pipe is 3 x 105 N/m². This pressure
reduces to 2 x 105 N/m² on opening the value of the pipe. Calculate the speed of water flowing through the pipe. Density of water =1000kg/s
​

Answer 1

ANSWER:

• The speed of water flowing through the pipe = 17.3 m/s.

GIVEN:

• The pressure of water inside the closed pipe is 3 x 10⁵ N/m².

• This pressure reduces to 2 x 10⁵ N/m² on opening the value of the pipe.

• Density of water = 1000 kg/m³.

TO FIND:

• The speed of water flowing through the pipe.

EXPLANATION:

By using Bernoulli's theorem,

$\boxed{\bold{\large{\gray{\dfrac{{P}_{1}}{\rho}+ \dfrac{({v}_{1})^{2}}{ 2}=\dfrac{{P}_{2}}{\rho}+ \dfrac{({v}_{2})^{2}}{2}}}}}$

$\sf P_{1}=3.5 \times 10^{5} \ N {m}^{ - 2}$

$\sf v_1 = 0$

$\sf P_2 =2 \times 10^{5}\ N {m}^{ - 2}$

$\sf\rho=1000 \ kgm^{-3}$

$\sf\dfrac{{P}_{1}}{\rho}+ \dfrac{({v}_{1})^{2}}{ 2}=\dfrac{3.5 \times 10^{5}}{1000}+ 0$

$\sf\dfrac{{P}_{1}}{\rho}+ \dfrac{({v}_{1})^{2}}{ 2}=3.5 \times 10^{2}$

$\sf\dfrac{{P}_{2}}{ \rho}+ \dfrac{({v}_{2})^{2}}{ 2}=\dfrac{2\times 10^{5}}{1000}+ \dfrac{({v}_{2})^{2}}{ 2}$

$\sf\dfrac{{P}_{2}}{ \rho}+ \dfrac{({v}_{2})^{2}}{ 2}=2 \times 10^{2}+ \dfrac{({v}_{2})^{2}}{ 2}$

$\sf3.5 \times 10^{2}=2 \times 10^{3}+ \dfrac{({v}_{2})^{2}}{ 2}$

$\sf3.5 \times 10^{2} - 2 \times 10^{2}=\dfrac{({v}_{2})^{2}}{ 2}$

$\sf1.5 \times 10^{2}=\dfrac{({v}_{2})^{2}}{ 2}$

$\sf 3\times 10^{2}=({v}_{2})^{2}$

$\sf v_2= 10\sqrt{3}$

$\sf{v}_{2} = 17.3\ m {s}^{ - 1}$

Hence the speed of water flowing through the pipe = 17.3 m/s.