QUESTION 30 In the figure shown, sub-critical uniform flow occurs in a 1.2 m width rectangular channel at the depth of 0.6 m. A hump of height 10 cm is placed in the bottom of the channel. Flow rate of the channel is 0.8 m/s. Assuming no energy losses, 1. Show that the flow depth on the step is 0.453 m(Y2) 2. If the flow depth on the step were to be critical depth (Y2) calculate the minimum height of the step (AH) Y - 0.6 m Y: Q-0.8 m'/s DAH = 0.1 m

3. Water is moving between two reservoirs at a flow rate of 0.08 m³/s via the pipes shown in the diagram (D₁ = 10 cm, D₂ = 20 cm). Assume both pipes have a roughness ks = 0.08 mm. If the difference in water surface elevation between the two reservoirs is 70 m and L₁ = 50 m, what is the length of the larger pipe (L2)? Further, what is the entrance length of pipe 1? For what fraction of the length of pipe 1 is flow fully developed? Include form losses due to pipe components (entrance, expansion, and exit) in your calculations. For more info on component losses, use Table 10.5 and 0 = 180°. Water T = 20°C D₁ D₂ 1₂2.

Water flows over the spillway shown in the figure below. If the velocity is uniform at sections (1) and (2) and viscous effects are negligible, determine the flowrate per unit width of the spillway. 4.2 m Q= (1) 2.8 m m³/s 0.5 m 1 (2)

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In the parallel water distribution network below, what percentage of the flow rate (Q) passes through each branch? Assume constant friction factors in the system, and ignore minor losses.CHOOSE ONE: 40% of the flow passes through Branch#1 and 60% of the flow passes through Branch#2 20% of the flow passes through Branch#1 and 80% of the flow passes through Branch#2 60% of the flow passes through Branch#1 and 40% of the flow passes through Branch#2 80% of the flow passes through Branch#1 and 20% of the flow passes through Branch#2

In the model test of a spillway the discharge and velocity of flow over the model were 2 m/s and 1.5 m/s, respectively. Calculate the velocity and discharge over the prototype which is 36 times the model size. Note:Use Froud Model Law to solve the question. Velocity Ratio: V = -= JT, %3D V. L T, LN, - = JL, T LV L, Time Ratio: Tr=P3= V,/T, Vr a V IT Tr Acceleration Ratio: a, Discharge Ratio: Q, AV Force Ratio: Fr=m,a, = p_Q}, = pFT =pEF =p1L = pL, Power Ratio: Pr=Fr. Vr=p LV V = p_LV? =pI(L) =p,L

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Calculate the daily flow in (m³/d) using Darcy's law for a confined aquifer with a hydraulic conductivity of 1.9 (m/d), a thickness of 50 (m) which is 5 (km) wide. Two observation wells are located 1.8 (km) apart in the direction of flow and the head in well 1 is 105 (m) and in well 2 is 95 (m). (Please enter your answer with one decimal)

The top width and the depth of flow in a triangular channel were measured as 5 m and 1 m respectively. The measured velocity on the centre line of flow area at the water surface, 0.2 m and 0.8 m below the surface are 0.8 m/s, 0.6 m/s and 0.5 m/s respectively. Using two point method of velocity measurement, the discharge (in m3/s) in the channel is

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