91Ó°ÊÓ

• The volume of 220 cans per minute is$\frac{dV}{dt}=\frac{\left(220\right)0.355L}{60s}$.
• The gauge pressure is$P_2=152kPa$.
• The distance above point 2 is$h_2=1.35m$.
• The cross-sectional area at point 2 is$A_2=8c{m}^2$.
• The cross-sectional area at point 1 is$A_1=2c{m}^2$.

In this problem, the relation between density and mass will be used to find the mass flow rate. According to the relation, density is equal to the fraction of mass and volume.

The relation of mass flow rate can be written as:

$\begin{array}{r}\mathrm{ÒÏ}=\frac{\stackrel{Ë™}{m}}{\left(\frac{dV}{dt}\right)}\\ \stackrel{Ë™}{m}=\mathrm{ÒÏ}×\left(\frac{dV}{dt}\right)\end{array}$

Here, is the density of water.

Substitute$\frac{\left(220\right)\left(0.355L\right)}{60s}$ for$\frac{dV}{dt}$ and 1 kg/L for p in the above relation.

$\begin{array}{r}\stackrel{˙}{m}=(1\mathrm{kg}/\mathrm{L})×\left[\frac{\left(220\right)\left(0.355\mathrm{L}\right)}{60\mathrm{s}}\right]\\ \stackrel{˙}{m}=1.30\mathrm{kg}/\mathrm{s}\end{array}$

Thus, the mass flow rate is$1.30kg/s$ .

The relation of volume flow rate can be written as:

$\frac{dV}{dt}=\frac{\left(220\right)\left(0.355\mathrm{L}\right)}{60\mathrm{s}}$

On further solving the above relation.

$\begin{array}{r}\frac{dV}{dt}=\frac{78.1\mathrm{L}}{60\mathrm{s}}\\ \frac{dV}{dt}=1.30\mathrm{L}/\mathrm{s}\end{array}$

Thus, the volume flow rate is 1.30 L/s .

The relation of flow speed can be written as:

$V_1=\frac{\frac{dV}{dt}}{A_1}$

Substitute 1.30 L/s for$\frac{dV}{dt}$ and$2c{m}^2$ for$A_1$ in the above relation.

$\begin{array}{r}{v}_1=\left[\frac{1.30\mathrm{L}/\mathrm{s}×\frac{1{\mathrm{m}}^3/\mathrm{s}}{1000\mathrm{L}/\mathrm{s}}}{\left(2{\mathrm{cm}}^2×\frac{1{\mathrm{m}}^2}{{10}^4{\mathrm{cm}}^2}\right)}\right]\\ v_1=6.5\mathrm{m}/\mathrm{s}\end{array}$

The relation of flow speed can be written as:

$v_2=\frac{\frac{dV}{dt}}{A_2}$

Substitute 1.30 L/s for$\frac{dV}{dt}$ and$8c{m}^2$ for$A_2$ in the above relation.

$\begin{array}{r}{v}_2=\left[\frac{1.30\mathrm{L}/\mathrm{s}×\frac{1{\mathrm{m}}^3/\mathrm{s}}{1000\mathrm{L}/\mathrm{s}}}{\left(8{\mathrm{cm}}^2×\frac{1{\mathrm{m}}^2}{{10}^4{\mathrm{cm}}^2}\right)}\right]\\ v_2=1.62\mathrm{m}/\mathrm{s}\end{array}$

Thus, the flow speed at point 1 and 2 is 6.5 m/s and 1.62 m/s respectively.

The relation from Bernoulli’s equation can be written as:

$P_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2=P_2+\frac{1}{2}\mathrm{ÒÏ}{v}_2^2+\mathrm{ÒÏ}g{h}_2âˆ\pounds$

Here, g is the gravitational acceleration and$P_2$ is the required gauge pressure.

Substitute $152kPa$for $P_2$, $1000kg/m^3$for $\mathrm{ÒÏ}$, 6.5 m/s for $v_1$, 1.62 m/s for $v_2$, $9.8m/s^2$for g and 1.35m for$h_2$ the above relation.

$\begin{array}{r}{P}_1+\frac{1}{2}\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)(6.5\mathrm{m}/\mathrm{s}{)}^2=\left(152\mathrm{kPa}×\frac{{10}^3\mathrm{Pa}}{1\mathrm{kPa}}\right)+\left[\begin{array}{c}\frac{1}{2}\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)(1.62\mathrm{m}/\mathrm{s}{)}^2+\\ \left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(1.35\mathrm{m}\right)\end{array}\right]\\ P_1=145×{10}^3\mathrm{Pa}\end{array}$

Thus, the gauge pressure is$145×{10}^{3}Pa$ .