明渠水理計算_曼寧公式
https://play.google.com/store/apps/details?id=com.yuchihung.manning
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明渠水理計算小工具是依曼寧公式所製作，可以提供設計者在已知流量Q條件下計算出水深D，並依此來設計明渠之渠牆高度，在現場也可以根據水深D計算出流量Q，以了解該渠道之通水能力。
程式對於左右坡度不對稱之渠道，也能通過左右m值來描述其坡度，預設之m=0為垂直狀態，圖示中1:m代表垂直為1，水平為m。
水理計算範例(適用:矩形渠道或梯形渠道)
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明渠水理計算
一、已知條件:
Q=5.0cms B=3.0m S=0.001 n=0.016
mL=0(左岸邊坡比) mR=0(右岸邊坡比) 註:矩形渠道邊坡比為0
二、由曼寧公式及通水面積以試誤法求解
曼寧公式:Q=1/n*R^(2/3)*S^(1/2)*A......................(1)
水力半徑:R=A/P ......................................................(2)
(A:通水面積 P:濕周)
令 T=Q*n/[S^(1/2)]...................................................(3)
則(1)式可簡化為 T=R^(2/3)*A
(2)式帶入上式
T=(A/P)^(2/3)*A=A^(5/3)/[P^(2/3)]
由上式可導出 A^(5/3)=T*P^(2/3)
=> A=T^(3/5)*P^(2/5).........................................(4)
濕周P=B+x*SQRT(1+mL^2)+x*SQRT(1+mR^2)......(5) 其中x為假設水深
通水面積:A=x*[0.5(2B+mL*x+mR*x)]......................(6)
(5)式及(6)式代入(4)式
=> x*[0.5(2B+mL*x+mR*x)]=T^(3/5)*[B+x*SQRT(1+mL^2)+x*SQRT(1+mR^2)]^(2/5)
整理上式
x={T^(3/5)*[B+x*SQRT(1+mL^2)+x*SQRT(1+mR^2)]^(2/5)}/[0.5(2B+mL*x+mR*x)]
令上式左側之x為欲求解水深，以d取代
d={T^(3/5)*[B+x*SQRT(1+mL^2)+x*SQRT(1+mR^2)]^(2/5)}/[0.5(2B+mL*x+mR*x)]....(7)
試誤法求解
1.假設水深為x=5m，代入(7)式等號右側，並計算出d值
2.比較 |d-x| 之差值，若 |d-x|>允許誤差值(例如允許誤差值=0.00001)
則令計算出之d為新的x，代入(7)式計算新的d
3.若 |d-x|<允許誤差值，則d為近似解，否則重複執行步驟2.
三、計算結果
d=1.130m A=3.390m^2
P=5.260m R=0.645m
V=1.475m/s
其中d=水深 A=通水面積 P=濕周 R=水力半徑 V=Q/A=流速
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註:上述紅字部分，讀者可依不同水理條件輸入
APP計算後，將計算結果貼上置換。
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圖例區及計算結果區可上下滑動觀看後續資料。
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Example of hydraulic calculation (applicable: rectangular channel or trapezoidal channel)
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OPEN CHANNEL HYDRAULIC CALCULATION
1. Known conditions:
Q= 5.0 cms B= 3.0 m S= 0.001 n= 0.016
mL= 0 (left bank slope ratio) mR= 0 (right bank slope ratio)
Note: the slope ratio of a rectangular channel is 0
2. Solve by trial and error method by Manning's formula and water flow area
Manning's formula: Q=1/n*R^(2/3)*S^(1/2)*A..........( 1)
Hydraulic Radius: R=A/P ................................................ ............(2)
(A: water flow area P: wet area)
Let T=Q*n/[S^(1/2)].......................... ..............(3)
Then formula (1) can be simplified as T=R^(2/3)*A
Bring (2) into the above formula
T=(A/P)^(2/3)*A=A^(5/3)/[P^(2/3)]
A^(5/3)=T*P^(2/3) can be derived from the above formula
=> A=T^(3/5)*P^(2/5).......................... .........(4)
Wet area P=B+x*SQRT(1+mL^2)+x*SQRT(1+mR^2)......(5)
where x is the assumed water depth
Water flow area: A=x*[0.5(2B+mL*x+mR*x)]..........(6)
Substitute (5) and (6) into (4)
=> x*[0.5(2B+mL*x+mR*x)]=T^(3/5)*[B+x*SQRT(1+mL^2)+x*SQRT(1+mR^2 )]^(2/5)
Organize the above formula
x={T^(3/5)*[B+x*SQRT(1+mL^2)+x*SQRT(1+mR^2)]^(2/5)}/[0.5(2B+mL *x+mR*x)]
Let x on the left side of the above formula be the water depth to be solved, and replace it with d
d={T^(3/5)*[B+x*SQRT(1+mL^2)+x*SQRT(1+mR^2)]^(2/5)}/[0.5(2B+mL *x+mR*x)]....(7)
Trial and error solution
1. Assuming that the water depth is x=5m, substitute it into the right side of the equation (7) and calculate the value of d
2. Compare the difference between |d-x|, if |d-x|> allowable error value (for example, allowable error value=0.00001)
Then let the calculated d be the new x, and substitute it into (7) to calculate the new d
3. If |d-x|<allowable error value, then d is an approximate solution, otherwise repeat step 2.
3. Calculation results
d= 1.130 m A= 3.390 m^2
P= 5.260 m R= 0.645 m
V= 1.475 m/s
Among them, d=water depth, A=water flow area, P=wet area, R=hydraulic radius,
V=Q/A=flow velocity
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Note: For the part in red above, readers can input according to different water conditions
Open channel hydraulic calculation_Manning formula
After the APP calculates, paste the calculation result as a replacement.
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