Question

A fluid, with a density of rho = 1165 kg/m3, flows in a horizontal pipe. In one segment of the pipe the flow speed is v1 = 4.53 m/s. In a second segment the flow speed is v2 = 1.77 m/s. What is the difference between the pressure in the second segment (P2) and the pressure in the first segment (P1)?

Answer 1

Answer:

The pressure difference between two pipe is $1.01 \times 10^{4}$ Pa

Explanation:

Density $\rho = 1165$ $\frac{kg}{m^{3} }$

Speed in one pipe $v_{1} = 4.53$ $\frac{m}{s}$

Speed in second pipe $v_{2} = 1.77$ $\frac{m}{s}$

According to the bernoulli equation,

The pressure difference is given by,

     $P = \frac{1}{2} \rho v^{2}$

$P_{2} - P_{1} = \frac{1}{2} \rho (v_{1}^{2} - v_{2}^{2} )$

$P_{2} - P_{1} = \frac{1}{2} \times 1165 \times[ (4.53)^{2}- (1.77)^{2}]$

$P_{2} - P_{1} = 10128.51$

$P_{2} - P_{1} = 1.01 \times 10^{4}$ Pa

Therefore, the pressure difference between two pipe is $1.01 \times 10^{4}$ Pa