Water channel

The cross section of the water channel is a trapezoid. The width of the bottom is 19.7 m, the water surface width is 28.5 m, the side walls have a slope of 67°30' and 61°15'. Calculate how much water flows through the channel in 5 minutes if the water flow at rate 0.3 m/s.

Result

V =  19824.012 m3

Solution:

c=19.7 m a=28.5 m x=ac=28.519.7=445=8.8 m A=67+30/60=1352=67.5 B=61+15/60=2454=61.25 C=180(A+B)=180(67.5+61.25)=2054=51.25 S1=x2 sinA sinB/(2 sinC)=x2 sin67.5  sin67.5 /(2 sin67.5 )=8.82 sin67.5  sin67.5 /(2 sin67.5 )=x2 0.92388 0.92388/(2 0.92388)=40.21469 h=2 S1/x=2 40.2147/8.89.1397 S2=c h=19.7 9.1397180.0521 m2 S=S1+S2=40.2147+180.0521220.2668 m2 l=5 60 0.3=90 m V=l S=90 220.266819824.01219824.012 m3c=19.7 \ \text{m} \ \\ a=28.5 \ \text{m} \ \\ x=a-c=28.5-19.7=\dfrac{ 44 }{ 5 }=8.8 \ \text{m} \ \\ A=67+30/60=\dfrac{ 135 }{ 2 }=67.5 \ \\ B=61+15/60=\dfrac{ 245 }{ 4 }=61.25 \ \\ C=180 - (A+B)=180 - (67.5+61.25)=\dfrac{ 205 }{ 4 }=51.25 \ \\ S_{1}=x^2 \cdot \ \sin A ^\circ \cdot \ \sin B ^\circ / (2 \cdot \ \sin C ^\circ )=x^2 \cdot \ \sin 67.5^\circ \ \cdot \ \sin 67.5^\circ \ / (2 \cdot \ \sin 67.5^\circ \ )=8.8^2 \cdot \ \sin 67.5^\circ \ \cdot \ \sin 67.5^\circ \ / (2 \cdot \ \sin 67.5^\circ \ )=x^2 \cdot \ 0.92388 \cdot \ 0.92388/ (2 \cdot \ 0.92388)=40.21469 \ \\ h=2 \cdot \ S_{1}/x=2 \cdot \ 40.2147/8.8 \doteq 9.1397 \ \\ S_{2}=c \cdot \ h=19.7 \cdot \ 9.1397 \doteq 180.0521 \ \text{m}^2 \ \\ S=S_{1}+S_{2}=40.2147+180.0521 \doteq 220.2668 \ \text{m}^2 \ \\ l=5 \cdot \ 60 \cdot \ 0.3=90 \ \text{m} \ \\ V=l \cdot \ S=90 \cdot \ 220.2668 \doteq 19824.012 \doteq 19824.012 \ \text{m}^3

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