Question

Consider a fluid of density 3.43 g⋅cm−3 flowing through a pipe of varying cross-section. The diameter of the pipe in one section is 9.1 cm, while the diameter in a second section is 12.6 cm. When the diameter of the pipe is 9.1 cm, the flow speed of the fluid is 339 cm⋅s−1 and the pressure is 2.93 × 105 Pa.A)Calculate the flow speed (in m⋅s−1) of the fluid when the diameter of the pipe is 12.6 cm. B)Calculate the pressure (in × 105 Pa) when the pipe has a diameter of 12.6 cm

Answer 1

Given that the pipe has varying cross-sections.

The diameter of one section is d1 = 9.1 cm and the diameter of second section is d2 = 12.6 cm.

Also, the fluid has the density,

`\rho=3.43gcm^(-3)`

The area of the cross-section for the first section is

`\begin{gathered} A_1=(\pi(d1)^2)/(4) \\ =(\pi(9.1)^2)/(4)cm^2 \end{gathered}`

The area of the cross-section for the second section is

`\begin{gathered} A_2=(\pi(d2)^2)/(4) \\ =(\pi(12.6)^2)/(4)cm^2 \end{gathered}`

The flow speed for the first section is v1 = 339 cm s^-1

The flow speed for the second section will be v2.

(a) The flow speed for the second section can be calculated as

`\begin{gathered} A_1v1=A_2_{}v2 \\ v2=(A_1v1)/(A_2) \\ =(\pi(9.1)^2*339*4)/(4*\pi*(12.6)^2) \\ =\text{ 176.82 cm/s} \\ =1.7682\text{ m/s} \end{gathered}`

(b) The pressure for first section is p1 = 2.93 x 10^5 Pa

The pressure for the second section will be p2.

The pressure for the second section can be calculated by the formula,

`\begin{gathered} p2=p1+(1)/(2)\rho\mleft\lbrace(v1)^2-(v2\mright)^2\} \\ =2.93*10^5+(1)/(2)*3.43\mleft\lbrace(339)^2-(176.82)^2\mright\rbrace \\ =4.36\text{ }*10^5\text{ Pa} \end{gathered}`