Question

A horizontal pipe carries a smoothly flowing liquid of density of 1370 kg/m 3 . At Locations 1 and 2 along the pipe, the diameters are 5.75 cm and 3.45 cm, respectively.
Is the flow speed at location 2 higher or lower than at location 1?

Answer 1

Answer:

Flow speed at location 2 higher than at location 1

Explanation:

Since the pipe carries a steady flow of liquid, its volume rate can be calculated as the following

$\dot{V} = vA = v\frac{\pi d^2}{4}$

where v m/s is the flow speed, A is the cross-section area, and d is the cross-section diameter. Therefore:

$v = \frac{4\dot{V}}{d^2\pi}$

So if the volume rate stays the same, the diameter drop at location 2, then flow speed would increase at location 2.

Answer 2

The flow speed of the liquid at location 2 is higher as compared to the flow speed of the liquid at location 1.

How do you calculate the speed of the flow?

Given that density of the pipe is 1370 kg/m 3.  The volume of the pipe can be given as below.

$V = vA$

Where V is the volume of the pipe, A is the cross-sectional area of the pipe and v is the flow speed of liquid.

$V = v\times \dfrac{\pi d^2}{4}$

Where d is the cross-sectional diameter of the pipe. For location 1, d is 5.75 cm and for location 2, d is 3.45 cm.

Substituting the values in the above equation, we get the speed at both locations.

For Location 1,

$v_1 = \dfrac {4V}{\pi d_1^2}$

$v_1 = \dfrac {4\times 1370} {3.14\times (0.0575)^2}$

$v_1 = 5.3\times 10^5\;\rm m/s$

For Location 2,

$v_2 = \dfrac {4V}{\pi d_2^2}$

$v_2= \dfrac {4\times 1370} {3.14\times (0.0345)^2}$

$v_2 = 14.6 \times 10^5\;\rm m/s$

Hence we can conclude that the flow speed of the liquid at location 2 is higher as compared to the flow speed of the liquid at location 1.

To know more about the speed of liquid, follow the link given below.

https://brainly.com/question/798096.