Question

A pipe contains a gradually tapering section where the diameter decreases from 400 mm to 250 mm. The pipe contains an incompressible fluid of density 1000 kgm−3 and runs full. If the flow velocity is 2 ms−1 in the smaller diameter, determine:
a) the velocity in the larger diameter
b) the volume flow rate
c) the mass flow rate.

Answer 1

Answer on Question 59308, Physics, Mechanics, Relativity

Question:

A pipe contains a gradually tapering section where the diameter decreases from $400 \, \text{mm}$ to $250 \, \text{mm}$ . The pipe contains an incompressible fluid of density $1000 \, \text{kgm}^{-3}$ and runs full. If the flow velocity is $2 \, \text{ms}^{-1}$ in the smaller diameter, determine:

a) the velocity in the larger diameter

b) the volume flow rate

c) the mass flow rate

Solution:

Here's the sketch of our task:

a) We can find the velocity of the fluid from the Law of Continuity:

$$A _ {l} v _ {l} = A _ {s} v _ {s},$$

here, $A_{l} = \pi r_{l}^{2}$ , $A_{s} = \pi r_{s}^{2}$ are the large and small cross-sectional areas of the pipe, respectively; $r_{l}, r_{s}$ are the large and small radii of the pipe, respectively; $v_{l}$ is the velocity of the fluid in the larger diameter, $v_{s}$ is the velocity of the fluid in the smaller diameter.

Then, from this formula we can calculate the velocity of the fluid in the larger diameter:

$$v _ {l} = \frac {A _ {s} v _ {s}}{A _ {l}} = v _ {s} \cdot \frac {\pi r _ {s} ^ {2}}{\pi r _ {l} ^ {2}} = 2 m s ^ {- 1} \cdot \frac {\pi \cdot (1 2 5 \cdot 1 0 ^ {- 3} m) ^ {2}}{\pi \cdot (2 0 0 \cdot 1 0 ^ {- 3} m) ^ {2}} = 0. 7 8 m s ^ {- 1}.$$

b) Let us determine the volume flow rate – the rate of flow through the volume $V$ per unit time $t$ :

$$V = v t A,$$ $$\frac {\Delta V}{\Delta t} = A _ {l} v _ {l} = \frac {\pi D _ {l} ^ {2}}{4} v _ {l} = \frac {3.14}{4} (400 \cdot 10^{-3} \, \text{m}) ^ {2} \cdot 0.78 \, \frac {\text{m}}{\text{s}} = 0.098 \, \frac {\text{m} ^ {3}}{\text{s}}.$$

c) As we know the volume flow rate, we can calculate the mass flow rate:

$$\frac {m}{\rho} = v t s,$$ $$\frac {\Delta m}{\Delta t} = \rho \frac {\Delta V}{\Delta t} = \rho v _ {l} A _ {l} = 1000 \, \frac {\text{kg}}{\text{m} ^ {3}} \cdot 0.098 \, \frac {\text{m} ^ {3}}{\text{s}} = 98 \, \frac {\text{kg}}{\text{s}}.$$

Answer:

a) $v_{l} = 0.78 \, \text{ms}^{-1}$.

b) $\frac{\Delta V}{\Delta t} = 0.098 \, \frac{\text{m}^3}{\text{s}}$.

c) $\frac{\Delta m}{\Delta t} = 98 \, \frac{\text{kg}}{\text{s}}$.

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