Question 23.79: Water in an irrigation ditch of width w = 3.22 m and depth d......
Water in an irrigation ditch of width w = 3.22 m and depth d = 1.04 m flows 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 area wd, entirely in the water, perpendicular to the flow; (b) a surface with area 3wd/2, of which wd 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 normal 34.0° from the direction of flow.
(a) The mass flux is w d \rho ν=(3.22 \,m )(1.04 \,m )\left(1000\, kg / m ^3\right)(0.207 \,m / s )=693 \,kg / s .
(b) Since water flows only through area wd, the flux through the larger area is still 693 kg/s.
(c) Now the mass flux is (wd/2)ρν = (693 kg/s)/2 = 347 kg/s.
(d) Since the water flows through an area (wd/2), the flux is 347 kg/s.
(e) Now the flux is (wd cosθ )ρν = (693kg /s)(cos34°) = 575 kg /s .