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 sin(Arad=A π180 rad=40.2146853917 rad) sin(Brad=B π180 rad=40.2146853917 rad)/(2 sin(Crad=C π180 rad=40.2146853917 rad)) 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=90 220.266819824.012=19824.012 m3c = 19.7 \ m \ \\ a = 28.5 \ m \ \\ x = a-c = 28.5-19.7 = \frac{ 44 }{ 5 } = 8.8 \ m \ \\ A = 67+30/60 = \frac{ 135 }{ 2 } = 67.5 \ \\ B = 61+15/60 = \frac{ 245 }{ 4 } = 61.25 \ \\ C = 180 - (A+B) = 180 - (67.5+61.25) = \frac{ 205 }{ 4 } = 51.25 \ \\ S_{ 1 } = x^2 \cdot \ \sin( A \rightarrow rad = A \cdot \ \frac{ \pi }{ 180 } \ rad = 40.2146853917 \ rad) \cdot \ \sin( B \rightarrow rad = B \cdot \ \frac{ \pi }{ 180 } \ rad = 40.2146853917 \ rad)/ (2 \cdot \ \sin( C \rightarrow rad = C \cdot \ \frac{ \pi }{ 180 } \ rad = 40.2146853917 \ rad)) \ \\ 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 \ m^2 \ \\ S = S_{ 1 }+S_{ 2 } = 40.2147+180.0521 \doteq 220.2668 \ m^2 \ \\ l = 5 \cdot \ 60 \cdot \ 0.3 = 90 \ m \ \\ V = l \cdot \ S = 90 \cdot \ 220.2668 \doteq 19824.012 = 19824.012 \ m^3

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