Water in an irrigation ditch of width w = 3.22mand depth d= 1.04mflows with a speed of 0.207 m/s. The mass flux of the flowing water through an imaginary surface is the product of the water’s density (1000 kg/m³) and its volume flux through that surface. Find the mass flux through the following imaginary surfaces:

(a) a surface of areawd, entirely in the water, perpendicular to the flow;

(b) a surface with area 3wd / 2, of which is in the water, perpendicular to the flow;

(c) a surface of area wd / 2,, entirely in the water, perpendicular to the flow;

(d) a surface of area wd, half in the water and half out, perpendicular to the flow;

(e) a surface of area wd, entirely in the water, with its normalfrom the direction of flow.

1. Width of the ditch, w = 3.22m.
2. Depth of ditch, d = 1.04 m.
3. Speed of the water, v = 0.207 m/s.
4. The density of water, p = 1000 kg/${\mathrm{m}}^3$.

Using the concept of the mass flux, we can get the required value of the flux in different conditions considering the water flow through the given surface.

Formula:

The mass flux through a surface area, ${\varphi }_{mass}=wdpv\cos \theta$ (i)

The mass flux through the surface of the areaw d is given using equation (i) as:

$$\begin{gathered} \varphi =(3.22\mathrm{m})(1.04\mathrm{m})(1000\mathrm{kg}/{\mathrm{m}}^3)(0.207\mathrm{m}/\mathrm{s}) \\ =693\mathrm{kg}/\mathrm{s} \end{gathered}$$

Hence, the value of the mass flux is $693\mathrm{kg}/\mathrm{s}$.

Since water flows only through the area w d, the flux through the larger area is still 693 kg/s.

Now the mass flux is given using equation (i) as follows:

$$\begin{gathered} \varphi =(3.22\mathrm{m})(1.04\mathrm{m})(1000\mathrm{kg}/{\mathrm{m}}^3)(0.207\mathrm{m}/\mathrm{s})/2 \\ =(693\mathrm{kg}/\mathrm{s})/2 \\ =347\mathrm{kg}/\mathrm{s} \end{gathered}$$

Hence, the value of the flux is 347 kg/s

Since the water flows through an area (wd/2), so the mass flux is 347 kg/s.

The mass flux through the surface of the area is given using equation (i) as:

$$\begin{gathered} \varphi =(3.22\mathrm{m})(1.04\mathrm{m})(1000\mathrm{kg}/{\mathrm{m}}^3)(0.207\mathrm{m}/\mathrm{s})\cos \left({34}^{°}\right) \\ =575\mathrm{kg}/\mathrm{s} \end{gathered}$$

Hence, the value of the mass flux is 575 kg/s.