Water at 70^◦ F flows in a prototype rectangular open channel with a width of 10 ft, and a sluice gate with a free outflow is inserted in the channel, as illustrated in Figure EP 11.16, in order to measure the flowrate. The opening of the gate is set at 3.28 ft, the depth of flow upstream of the

Question

Water at [latex] 70^{\circ } F [/latex] flows in a prototype rectangular open channel with a width of 10 ft, and a sluice gate with a free outflow is inserted in the channel, as illustrated in Figure EP 11.16, in order to measure the flowrate. The opening of the gate is set at 3.28 ft, the depth of flow upstream of the sluice gate is 6.56 ft, and the depth of flow downstream of the sluice gate is 3.11 ft. A smaller model of the larger prototype is designed in order to study the flow characteristics of open channel flow in the flow-measuring device. The model fluid is also water at [latex] 70^{\circ } F [/latex] F, the ideal downstream velocity (see Equation 9.285 [latex] v_{2i} = \frac{\sqrt{2g(y_{1}- y_{2})} }{\sqrt{1- \left(\frac{A_{2}}{A_{1}} \right)^{2} } } = \frac{\sqrt{2g(y_{1}- y_{2})} }{\sqrt{1- \left(\frac{by_{2}}{by_{1}} \right)^{2} } } = \frac{\sqrt{2g(y_{1}- y_{2})} }{\sqrt{1- \left(\frac{y_{2}}{y_{1}} \right)^{2} } } [/latex]) of flow of the water, [latex] v_{2i} [/latex] in the smaller model open channel fitted with a sluice gate is 8.46 ft/sec, and the model scale, λ is 0.25. (a) Determine the actual discharge in the model sluice gate. (b) Determine the ideal downstream velocity of flow of the water in the prototype sluice gate in order to achieve dynamic similarity between the model and the prototype. (c) Determine the actual discharge in the prototype sluice gate in order to achieve dynamic similarity between the model and the prototype.

Accepted Answer

(a) In order to determine the actual discharge in the model sluice gate, the actual discharge equation, Equation 11.158 [latex] Q_{a} = \underbrace{\left[\underbrace{\left(\frac{v_{a}}{\sqrt{2g\Delta y } } ight) }_{C_{v}} \underbrace{\left(\frac{A_{a}}{A_{i}} ight) }_{C_{c}} ight] }_{C_{d}} Q_{i} [/latex], is applied as follows:

[latex] Q_{a} = \underbrace{\left[\underbrace{\left(\frac{v_{a}}{\sqrt{2g\Delta y } } ight) }_{C_{v}} \underbrace{\left(\frac{A_{a}}{A_{i}} ight) }_{C_{c}} ight] }_{C_{d}} Q_{i} [/latex]

where the [latex] C_{D} [/latex], or in the case of a sluice gate, the discharge coefficient, [latex] C_{d} [/latex], is used to model the flow resistance, and is a function of the geometry of the flow-measuring device, [latex] L_{1}/L =y_{1}/a [/latex], as illustrated in Figure 9.37. Furthermore, in order to determine the geometry [latex] y_{1} [/latex], [latex] y_{2} [/latex], and a of the model channel and sluice gate, the model scale, λ (inverse of the length ratio) is applied as follows:

[latex] b_{p}: = 10 ft [/latex] [latex] y_{1p}: = 6.56 ft [/latex] [latex] y_{2p}: = 3.11 ft [/latex] [latex] a_{p}: = 3.28 ft [/latex] [latex] \lambda : = 0.25 [/latex]

Guess value: [latex] b_{m}: = 1 ft [/latex] [latex] y_{1m}: = 1 ft [/latex] [latex] y_{2m}: = 0.5 ft [/latex] [latex] a_{m}: = 0.5 ft [/latex]

Given

[latex] \lambda = \frac{b_{m}}{b_{p}} [/latex] [latex] \lambda = \frac{y_{1m}}{y_{1p}} [/latex] [latex] \lambda = \frac{y_{2m}}{y_{2p}} [/latex] [latex] \lambda = \frac{a_{m}}{a_{p}} [/latex]
[latex] \left ( \begin{matrix} b_{m} \ y_{1m} \ y_{2m} \ a_{m} \end{matrix} ight ) : = Find (b_{m}, y_{1m},y_{2m} , a_{m}) = \left ( \begin{matrix} 2.5 \ 1.64 \ 0.778 \ 0.82 \end{matrix} ight ) ft [/latex]

[latex] slug: = 1 lb \frac{sec^{2}}{ft} [/latex] [latex] ho _{m} : = 1.936 \frac{slug}{ft^{3}} [/latex] [latex] \mu _{m} : = 20.5 imes 10^{-6} lb \frac{sec}{ft^{2}} [/latex]

[latex] V_{2im}: = 8.46 \frac{ft}{sec} [/latex] [latex] R_{m}: = \frac{ ho _{m} .V_{2im} .a_{m}}{\mu _{m}} = 6.551 imes 10^{5} [/latex]

[latex] A_{2im}: = b_{m} .a_{m} = 2.05 ft^{2} [/latex] [latex] Q_{im}: = V_{2im}. A_{2im} = 17.343 \frac{ft^{3}}{sec} [/latex]

[latex] \frac{y_{1m}}{a_{m}} =2 [/latex] [latex] C_{dm}: = 0.51 [/latex] [latex] Q_{am}: =C_{dm}. Q_{im} = 8.845 \frac{ft^{3}}{sec} [/latex]

(b)–(c) To determine the ideal downstream velocity of flow of the water in the prototype sluice gate in order to achieve dynamic similarity between the model and the prototype, and to determine the actual discharge in the prototype sluice gate in order to achieve dynamic similarity between the model and the prototype, the geometry [latex] L_{i}/L [/latex] must remain a constant between the model and prototype as follows:

[latex] \left(\frac{L_{i}}{L} ight)_{p} = \left(\frac{L_{i}}{L} ight)_{m} [/latex]

where the geometry is modeled as follows:

[latex] \frac{y_{1p}}{a_{p}} = 2 [/latex] [latex] \frac{y_{1m}}{a_{m}} = 2 [/latex] [latex] \frac{y_{2p}}{a_{p}} = 0.948 [/latex] [latex] \frac{y_{2m}}{a_{m}} = 0.948 [/latex]

However, because the discharge coefficient, [latex] C_{d} [/latex] is independent of R for a sluice gate, R does not need to remain a constant between the model and the prototype.
(b)–(c) To determine the ideal downstream velocity of flow of the water in the prototype sluice gate in order to achieve dynamic similarity between the model and the prototype, and to determine the actual discharge in the prototype sluice gate in order to achieve dynamic similarity between the model and the prototype, the discharge coefficient, [latex] C_{d} [/latex] must remain a constant between the model and the proto type (which is a direct result of maintaining a constant [latex] L_{i}/L [/latex] between the model and the prototype and applying the “gravity model” similitude scale ratio; specifically the velocity ratio, [latex] v_{r} [/latex] given in Table 11.2) as follows:

[latex] \underbrace{\left[\frac{Q_{a}}{\sqrt{2g\Delta y} (A_{i}) } ight]_{p} }_{C_{D_{p}} = C_{d_{p}}} = \underbrace{\left[\frac{Q_{a}}{\sqrt{2g\Delta y} (A_{i}) } ight]_{m} }_{C_{D_{m}} = C_{d_{m}}} [/latex]
[latex] v_{r} = \frac{v_{p}}{v_{m}} = \frac{\left(\sqrt{gL } ight)_{p} }{\left(\sqrt{ gL} ight)_{m} } = L_{r}^{\frac{1}{2} } [/latex]

[latex] ho _{p} : = 1.936 \frac{slug}{ft^{3}} [/latex] [latex] \mu _{p} : = 20.5 imes 10^{-6} lb \frac{sec}{ft^{2}} [/latex] [latex] g: = 32.174 \frac{ft}{sec^{2}} [/latex]

[latex] A_{2ip}: = b_{p} .a_{p} = 32.8 ft^{2} [/latex]

Guess value: [latex] V_{2ip}: = 1 \frac{ft}{sec} [/latex] [latex]Q_{ap} : = 1 \frac{ft^{3}}{sec} [/latex] [latex]Q_{ip} : = 1 \frac{ft^{3}}{sec} [/latex] [latex]C_{dp} : = 0.1 [/latex]

Given

[latex]C_{dp} = \frac{Q_{ap}}{Q_{ip}} [/latex] [latex] \frac{V_{2ip}}{\sqrt{g.a_{p}}} = \frac{V_{2im}}{\sqrt{g.a_{m}}}[/latex]

[latex]C_{dp} = C_{dm} [/latex] [latex]Q_{ip} = V_{2ip}. A_{2ip} [/latex]
[latex] \left ( \begin{matrix} V_{2ip} \ Q_{ap} \ Q_{ip} \ C_{dp} \end{matrix} ight ) : = Find (V_{2ip}, Q_{ap},Q_{ip} ,C_{dp}) [/latex]

[latex] V_{2ip} = 16.92 \frac{ft}{s} [/latex] [latex]Q_{ap} = 283.038 \frac{ft^{3}}{sec} [/latex] [latex]Q_{ip} = 554.976 \frac{ft^{3}}{sec} [/latex] [latex]C_{dp} = 0.51 [/latex]

Furthermore, the Froude number, F remains a constant between the model and the prototype as follows:

[latex] F_{m}: = \frac{V_{2im}}{\sqrt{g.a_{m}}} = 1.647 [/latex] [latex] F_{p}: = \frac{V_{2ip}}{\sqrt{g.a_{p}}} = 1.647 [/latex]

Therefore, although the similarity requirements regarding the independent π term, [latex] L_{i}/L (y_{1p}/a_{p}=y_{1m}/a_{m} = 2 and y_{2p}/a_{p}=y_{2m}/a_{m} = 0.948)[/latex] and the dependent π term, F (“gravity model”) ([latex] F_{p} = F_{m} = 1.647 [/latex]) are theoretically satisfied, the dependent π term (i.e., the discharge coefficient, [latex] C_{d} [/latex]) will actually/practically remain a constant between the model and its prototype ( [latex] C_{dp} [/latex] = [latex] C_{dm} [/latex] = 0.51) only if it is practical to maintain/attain the model velocity, flow depth, fluid, scale, and cost. Furthermore, because the discharge coefficient, [latex] C_{d} [/latex] is independent of R for a sluice gate, R does not need to remain a constant between the model and the prototype as follows:

[latex]R_{m} = 6.551 imes 10^{5} [/latex] [latex] R_{p} : = \frac{ ho _{p} . V_{2ip}. a_{p}}{\mu _{p}} =5.241 imes 10^{6} [/latex]

Table 11.2
Similitude Scale Ratios for Physical Quantities for a Gravity Model
Physical Quantity	FLT System	MLT System	Primary Scale Ratios	Secondary/Similitude Scale Ratios for a Pressure Model
			[latex] F_{r} = \frac{F_{G_{p}}}{F_{G_{m}}} = \frac{F_{I_{p}}}{F_{I_{m}}} = constant [/latex]	[latex] \underbrace{\left[\left(\frac{ v}{\sqrt{gL} } ight)_{p} ight] }_{F_{p}} = \underbrace{\left[\left(\frac { v}{\sqrt{ gL} } ight)_{m} ight] }_{F_{m}} [/latex]
Geometrics Length, L	L	L	[latex] L_{r} = \frac{L_{p}}{L_{m}} [/latex]	[latex] L_{r} = \frac{L_{p}}{L_{m}} [/latex]
Area, A	[latex] L^{2} [/latex]	[latex] L^{2} [/latex]	[latex] L_{r}^{2} = \frac{L_{p}^{2}}{L_{m}^{2}} [/latex]	[latex] L_{r}^{2} = \frac{L_{p}^{2}}{L_{m}^{2}} [/latex]
Volume, V	[latex] L^{3} [/latex]	[latex] L^{3} [/latex]	[latex] L_{r}^{3} = \frac{L_{p}^{3}}{L_{m}^{3}} [/latex]	[latex] L_{r}^{3} = \frac{L_{p}^{3}}{L_{m}^{3}} [/latex]
Kinematics Time, T	T	T	[latex] T_{r} = \frac{L_{r}}{v_{r}} [/latex]	[latex] T_{r} = \frac{L_{r}}{v_{r}} = L_{r}^{1/2} [/latex]
Velocity, v	[latex] LT^{-1}[/latex]	[latex] LT^{-1}[/latex]	[latex] v_{r} = \frac{v_{p}}{v_{m}} [/latex]	[latex] v_{r} = \frac{v_{p}}{v_{m}} = \frac{\left(\sqrt{gL} ight) _{p} }{\left(\sqrt{gL} ight) _{m}} = L_{r}^{1/2} [/latex]
Acceleration, a	[latex] LT^{-2}[/latex]	[latex] LT^{-2}[/latex]	[latex] a_{r} = \frac{L_{r}}{T_{r}^{2}} = \frac{v_{r}^{2}}{L_{r}} [/latex]	[latex] a_{r} = \frac{v_{r}^{2}}{L_{r}} =1[/latex]
Discharge, Q	[latex] L^{3}T^{-1}[/latex]	[latex] L^{3}T^{-1}[/latex]		[latex] Q_{r} = v_{r}. L_{r}^{2} = L_{r}^{5/2} [/latex]
Dynamics Mass, M	[latex] FL^{-1}T^{2}[/latex]	M		[latex] M_{r} = F_{r}a_{r}^{-1} = ho _{r} L_{r}^{3} [/latex]
Force, F	F	[latex] MLT^{-2}[/latex]	[latex] F_{r} = \frac{F_{G_{p}}}{F_{G_{m}}} = \frac{F_{I_{p}}}{F_{I_{m}}} [/latex]	[latex] F_{r} = ho_{r} L_{r}^{3} g_{r} = ho _{r} v^{2}_{r} L_{r}^{2} [/latex]
Pressure, p	[latex] FL^{-2} [/latex]	[latex] ML^{-1}T^{-2}[/latex]		[latex] p_{r} = F_{r} L_{r}^{-2} = ho _{r} L_{r} [/latex]
Momentum, Mv or Impulse, FT	FT	[latex] MLT^{-1}[/latex]		[latex] F_{r} T_{r} = ho _{r} L_{r}^{7/2} [/latex]
Energy, E or Work, W	FL	[latex] ML^{2}T^{-2}[/latex]		[latex] W_{r} = F_{r} L_{r}= ho _{r} L_{r}^{4} [/latex]
Power, P	[latex] FLT^{-1} [/latex]	[latex] ML^{2}T^{-3}[/latex]		[latex] p_{r} = W_{r} T_{r}^{-1} = ho _{r} L^{7/2}_{r} [/latex]

Question 11.16: Water at 70^◦ F flows in a prototype rectangular open channe...

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