Question

A liquid of density $\mathbf{900}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$ flows through a horizontal pipe that has a cross-sectional area of $\mathbf{1}\mathbf{.}\mathbf{90}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}{\mathbf{}\mathbf{m}}^{\mathbf{2}}$ in region $A$ and a cross-sectional area of $\mathbf{9}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{.}\mathbf{2}}{\mathbf{}\mathbf{m}}^{\mathbf{2}}$ in region $\mathit{B}$. The pressure difference between the two regions is $\mathbf{7}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{Pa}$.

(a) What is the volume flow rate?

(b) What is the mass flow rate?

Short answer

(a) The volume flow rate$Q$ is $0.0776{\mathrm{m}}^3/\mathrm{s}$.

(b) The mass flow rate $m$ is $m=69.8\mathrm{kg}/\mathrm{s}$.

Step-by-step solution

Given information

$\rho$ is the density of the liquid, $\rho =900\mathrm{kg}/{\mathrm{m}}^3$

The cross-sectional area at region $A,A=1.90\times {10}^{-2}{\mathrm{m}}^2$

The cross-sectional area at region $B,a=9.50\times {10}^{-2}{\mathrm{m}}^2$

The pressure difference, $\Delta P=7.20\times {10}^3\mathrm{Pa}$.

Understanding the concept of Bernoulli’s equation

By using Bernoulli's equation and the equation of continuity, find the speed of a given liquid $\mathit{V}$. Then by using the calculated value of the speed of a given liquid$\mathit{V}$, find the value of the volume flow rate $\mathit{Q}$. Then, using the value of volume flow rate $\mathit{Q}$, find the mass flow rate $\mathit{m}$of the liquid. According to Bernoulli's equation, the speed of a moving fluid increases, and the pressure within the fluid decreases.

Formulae are as follows:

i) According to Bernoulli's equation and the equation of continuity, the speed of a given liquid $V$at regions $A$and $B$is

$V=\sqrt{\frac{2a^2\Delta p}{\rho \left(a^2-A^2\right)}}\text{.}$

ii) Rate of flow of water $Q,Q=VA$.

iii) The mass flow rate $m,p$↑

Where, $p$ is pressure, $V$ is velocity, $h$ is height, $g$ is an acceleration due to gravity, $h$ is height, $A,a$ are areas, $Q$ is the rate of flow, $m$ is mass flow rate, and $\rho$ is density.

(a) Determining the volume flow rate Q

According to Bernoulli's equation and the equation of continuity, the speed of a given liquid $V$at regions $A$and$B$is,

$V=\sqrt{\frac{2a^2\Delta p}{\rho \left(a^2-A^2\right)}}$

By using the given values in the above equation,

$V=\sqrt{\frac{2{\left(9.50\times {10}^{-2}{\mathrm{m}}^2\right)}^2\times \left(7.20\times {10}^3\mathrm{Pa}\right)}{900\mathrm{kg}/{\mathrm{m}}^3\left({\left(9.50\times {10}^2{\mathrm{m}}^2\right)}^2-{\left(1.90\times {10}^{-2}{\mathrm{m}}^2\right)}^2\right)}}$

$=\sqrt{\frac{1299600}{77976}}$

$=4.0825\mathrm{m}/\mathrm{s}$

The volume flow rate $Q$is,

$Q=VA$

$Q=4.0825\times 1.90\times {10}^2$

$Q=0.0775675{\mathrm{m}}^3/\mathrm{s}$

$\approx 0.0776{\mathrm{m}}^3/\mathrm{s}$

Hence, the volume flow rate $Q$ is $0.0776{\mathrm{m}}^3/\mathrm{s}$.

(b) Determining the mass flow rate m

The volume flow rate $Q$is,

$Q=\frac{m}{\rho }$

$\rho Q$$=900\mathrm{kg}/{\mathrm{m}}^3\times 0.0775675{\mathrm{m}}^3/\mathrm{s}$

=$69.8108\mathrm{kg}/\mathrm{s}$

$\approx 69.8\mathrm{kg}/\mathrm{s}$

Hence, the mass flow rate m is $69.8\mathrm{kg}/\mathrm{s}$.