91Ó°ÊÓ

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

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

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

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

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

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

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 $\mathrm{ÒÏ}$ is density.

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

$V=\sqrt{\frac{2a^2\mathrm{Δ}p}{\mathrm{ÒÏ}\left(a^2-A^2\right)}}$

By using the given values in the above equation,

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

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

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

The volume flow rate $Q$is,

$Q=VA$

$Q=4.0825×1.90×{10}^2$

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

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

The volume flow rate $Q$is,

$Q=\frac{m}{\mathrm{ÒÏ}}$

$\mathrm{ÒÏ}Q$$=900\mathrm{kg}/{\mathrm{m}}^3×0.0775675{\mathrm{m}}^3/\mathrm{s}$

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

$≈69.8\mathrm{kg}/\mathrm{s}$

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