Hydraulics

Question 1 A long culvert has the cross-section of an equilateral triangle with side 0.7 m. The surfaces of the culvert have a Manning's roughness n = 0.012 m-¹/³s and the streamwise slope is 0.01. The culvert carries a flow of 0.35 m³ s. Assuming that the culvert does not run full, find the normal depth (measured from the lowest point) and the Froude number if the apex of the triangle is: (a) at the bottom; (b) at the top.

Q2. A rectangular channel b= 2.0 m, Q= 2 m3/s, the depth of uniform flow before the hump is 0.8 m. What should be the height of the hump (Az) to have critical flow over it (no head loss).

An existing trapezoidal aqueduct (b= 15 ft, S = 0.01%, n = 0.025) with a 34:1 side slope was designed to flow at D/b = 0.30. The channel is sides and bottom are modified to increase side slope to 2½:1 and decrease roughness to 0.021. Calculate percent change in flow capacity assuming base width, slope and design D/b remain constant.

Question [2] A steady, two-dimensional velocity field is given by: \vec{V}=(u, v)=(-0.781-4.67 x) \vec{\imath}+(-2.7-2.1 y) \vec{\jmath} Calculate the location of the stagnation point.

Water @ 80°F flows from a storage tank through a horizontal, 80-ft long, 4-in diameter galvanized iron pipe that includes a flow control valve. a) Ignore minor losses and calculate depth of water in the tank above the discharge pipe to convey 2 cfs. Use Darcy equation for friction head loss. b) Repeat calculation using Hazen-Williams equation for friction head loss c) List 2 reasons that water hammers could develop in this pipe and list 2 mitigations to either eliminate or reduce the potential for water hammers.

Q1. Calculate capillary rise/fall in a glass tube 2 mm diameter when immersed in (a) water (b) mercury.Both the liquids are at 20°C and the surface tension values at this temperature for water and mercury are 0.072 N/m and 0.052 N/m respectively. The specific gravity of mercury is 13.6. The contact angle of water and mercury are 0° and 130° respectively.

Water flows smoothly and falls on top of a plat surface (anupside-down baking tray in the picture). Two interestingphenomena are involved here: A) vena contracta; and B)hydraulic jump. Google these terms and choose one to study inyour video: A) Explain phenomenon A in the picture. Study water column diameter versus z position based on mass conservation. Make it quantitative. Compare your theory with your own experimental measurements. B) Explain phenomenon B in the picture. Study the diameter of the circle on the flat surface versus flow rate based on mass conservation. Make it quantitative Compare your theory with your own experimental measurements.

Two tanks A and B are connected by a 6,000-meter long, 250-mm diameter pipe steel pipe (f= 0.020).The pipe is tapped at its midpoint to draw out 0.04 m³/sec (Q3). Ignore minor losses. a) Determine flow rate out of Tank A b) Determine flow rate into Tank B

Water @ 50°F flows between an upper reservoir and a lower reservoir through a 2-ft diameter,15,000-ft long cast-iron pipe. The difference in water levels between the two reservoirs is 100 ft.There is a hill between the reservoirs with a summit 3 ft above the upper reservoir. The pipeline must cross the hill where its length is 3,000-ft. Ignore minor losses and assume f = 0.016. a) Determine minimum depth below summit of hill that pipeline can be constructed if pressure at the top of the siphon cannot fall below atmospheric b) Briefly explain why siphon pressure is typically designed to fall to no more than 60-70% of negative atmospheric pressure at any point c) If a third reservoir is added to this system and all reservoirs meet at a common junction, briefly explain why junction pressure is constant at steady state

Q2. A 75 m long cast iron pipe, 15 cm in diameter, connects two tanks (open to the atmosphere) that have a water surface elevation difference of 5 m. The entrance of the pipe from the supply tank is square edged, and the pipe contains a 90° bend with a sharp, 15 cm radius. Determine the flow rate in the \text { pipe. Use } f=\frac{0.25}{\left[\log \left(\frac{e / D}{3.7}+\frac{6.71}{N \cdot 9.9}\right)\right]^{2}}