Chapter 16, Problem 19P

The flow $Q\left(m^3/s\right)$ in an open channel can be predicted with the Manning equation

$Q=\frac{1}{n}{A}_c{R}^{2/3}{S}^{1/2}$

where $n=$ Manning roughness coefficient (a dimensionless number used to parameterize the channel friction), $A_c=$ cross-sectional area of the channel $\left(m^2\right),S=$ channel slope (dimensionless, meters drop per meter length), and $R=$ hydraulic radius (m), which is related to more fundamental parameters by $R=A_c/P,\text{ where }P=$ wetted perimeter (m). As the name implies, the wetted perimeter is the length of the channel sides and bottom that is under water. For example, for a rectangular channel, it is defined as $P=B+2H,\text{ where H=}$ depth (m). Suppose that you are using this formula to design a lined canal (note that farmers line canals to minimize leak-age losses).

(a) Given the parameters n $n=0.035,S=0.003,S\text{ and }Q=1m^3/s$, determine the values of B and H that minimize the wetted perimeter. Note that such a calculation would minimize cost if lining costs were much larger than excavation costs.

(b) Repeat part (a), but include the cost of excavation. To do this minimize the following cost function,

$C=c_1{A}_c+c_{2}P$

where $c_1$ is a cost factor for excavation $=\$100/m^2\text{ and }{c}_2$ is a cost factor for lining $50/m.

(c) Discuss the implications of your results.

To determine

To calculate: The values of that will minimize the wetted perimeter P if the Manning equation of flow Q $\left({\text{m}}^{\text{3}}/\text{s}\right)$

can be written as $Q=\frac{1}{n}{A}_c{R}^{2/3}{S}^{1/2}$.

Answer to Problem 19P

Solution:

The values of $B\text{and}H$

that will minimize the wetted perimeter P is $1.547046\text{m}\text{and}\text{0}\text{.77328m}$

respectively.

Explanation of Solution

Given Information:

The Manning equation of flow Q $\left({\text{m}}^{\text{3}}/\text{s}\right)$

can be written as $Q=\frac{1}{n}{A}_c{R}^{2/3}{S}^{1/2}$

where n is Manning roughness coefficient, $A_c$

is cross-sectional are of channel $\left({\text{m}}^{\text{2}}\right)$, S is channel slope and R is hydraulic radius $\left(\text{m}\right)$

is given by $R=A_c/P$

where P is wetted perimeter and it is defined by $P=B+2H$

where H is depth $\left(\text{m}\right)$

and the parameters value are given as $n=0.035,S=0.003\text{and}Q=1{\text{m}}^{\text{3}}/\text{s}$.

Calculation:

Consider the equation,

$P=B+2H$

The constrained can be written as,

$Q=\frac{1}{n}{A}_c{R}^{2/3}{S}^{1/2}$

This problem can be solved out by Linear programming formulation.

To minimize the wetted perimeter P function with the given constraint, the excel solver can be used.

The excel solver steps are,

Step 1. Initiate quantity $B\text{and}H=1$

and then write the parameter as shown below,

Step 2. Apply the formula in $A_c$

as shown below,

Step 3. Apply the formula in P as shown below,

Step 4. Apply the formula in R as shown below,

Step 5. Apply the formula in Qmanning as shown below,

Step 6. Go to DATA and then click on Solver. This dialog box will appear.

Step 7. Select the set objective, min, changing variable and then add,

Step 8. Click OK then this dialog box appears.

Step 9. Click on Solve and then OK.

Hence, the values of $B\text{and}H$

that will minimize the wetted perimeter P is $1.547046\text{m}\text{and}\text{0}\text{.77328m}$

respectively.

To determine

To calculate: The values of $B\text{and}H$

that will minimize the excavation cost function $C=c_1{A}_c+c_{2}P$ where cost factor for excavation $c_1$

is $\$100/{\text{m}}^{\text{2}}$

and cost factor for lining $c_2$ is $\text{\$50}/\text{m}$

if the Manning equation of flow Q $\left({\text{m}}^{\text{3}}/\text{s}\right)$

can be written as $Q=\frac{1}{n}{A}_c{R}^{2/3}{S}^{1/2}$.

Answer to Problem 19P

Solution:

The values of $B\text{and}H$

that will minimize the cost function P is $1.547046\text{m}\text{and}\text{0}\text{.77328m}$

respectively.

Explanation of Solution

Given Information:

The Manning equation of flow Q $\left({\text{m}}^{\text{3}}/\text{s}\right)$

can be written as $Q=\frac{1}{n}{A}_c{R}^{2/3}{S}^{1/2}$

where n is Manning roughness coefficient, $A_c$

is cross-sectional are of channel $\left({\text{m}}^{\text{2}}\right)$, S is channel slope and R is hydraulic radius $\left(\text{m}\right)$

is given by $R=A_c/P$

where P is wetted perimeter and it is defined by $P=B+2H$

where H is depth $\left(\text{m}\right)$

and the parameters value are given as $n=0.035,S=0.003\text{and}Q=1{\text{m}}^{\text{3}}/\text{s}$.

Calculation:

Consider the function,

$C=c_1{A}_c+c_{2}P$

The constrained can be written as,

$Q=\frac{1}{n}{A}_c{R}^{2/3}{S}^{1/2}$

This problem can be solved out by Linear programming formulation.

To minimize the cost function with the given constraint, the excel solver can be used.

The excel solver steps are,

Step 1. Initiate quantity $B\text{and}H=1$

and then write the parameter as shown below,

Step 2. Apply the formula in $A_c$

as shown below,

Step 3. Apply the formula in P as shown below,

Step 4. Apply the formula in R as shown below,

Step 5. Apply the formula in Qmanning as shown below,

Step 6. Apply the formula in Cost as shown below,

Step 7. Go to DATA and then click on Solver. This dialog box will appear.

Step 8. Select the set objective, min, changing variable and then add,

Step 9. Click OK then this dialog box appears.

Step 10. Click on Solve and then OK.

Hence, the values of $B\text{and}H$

that will minimize the cost function is $1.547046\text{m}\text{and}\text{0}\text{.77328m}$

respectively.

To determine

The implication of the result obtained in part (a) and part (b).

Answer to Problem 19P

Solution:

This can be interpreted that both the excavation and limiting cost can be minimized simultaneously by considering the bottom width B that is twice the length of each vertical side H.

Explanation of Solution

To interpret the result, first consider the constraint equation,

$\frac{1}{n}BH{\left(\frac{BH}{B+2H}\right)}^{2/3}{S}^{1/2}=Q$

And,

$\begin{array}{c}BH\left(\frac{BH}{B+2H}\right)={\left(\frac{nQ}{S^{1/2}}\right)}^{3/2}\\ \frac{{\left(BH\right)}^{5/2}}{\left(B+2H\right)}={\left(\frac{nQ}{S^{1/2}}\right)}^{3/2}\\ =\text{Constant}\end{array}$

On further simplification,

$BH=\text{Constant}\times {\left(B+2H\right)}^{2/5}$

As $A_c\text{and}P$

are dependent on B and H, thus both have minimized.

Thus, the excavation cost function $C=c_1{A}_c+c_{2}P$

is minimized and as excavation cost C is directly proportional to cross-sectional area.

Hence, both the excavation and limiting cost can be minimized simultaneously by considering the bottom width B that is twice the length of each vertical side H as obtained in part (a) and (b) as $1.547046\text{m}\text{and}\text{0}\text{.77328m}$.