JNTU
JNTU B.Tech III HYDRAULICS AND HYDRAULIC MACHINERY Previous Question Paper (September - 2021 - R18)
Answer any five questions. All questions carry equal marks.
Main
Show that hydraulically efficient trapezoidal channel is half hexagon.
Answer
For a trapezoidal channel with bottom width b, side slope 1:n (horizontal:vertical), and depth y, the area A = (b + ny)y and wetted perimeter P = b + 2y√(1 + n²). To minimize P for a constant A, we substitute b = (A/y) - ny into the perimeter equation to get P = (A/y) - ny + 2y√(1 + n²). Differentiating with respect to y and setting to zero gives dP/dy = -A/y² - n + 2√(1 + n²) = 0. Substituting A = by + ny², we find b + 2ny = 2y√(1 + n²). This condition implies that the hydraulic radius R = y/2. Furthermore, for a given area, the perimeter is minimized when the sloping sides are tangent to a semicircle drawn with the center at the mid-point of the water surface. For a trapezoid, this geometric condition occurs when the side slope angle θ = 60°, which corresponds to the geometry of a half-hexagon.
Write the salient point about specific energy curve.
Answer
The specific energy curve represents the relationship between specific energy E = y + (v²/2g) and depth of flow y for a constant discharge. Salient points include:
- The curve has two limbs: the upper limb approaches E = y as y increases (subcritical flow), and the lower limb approaches E = v²/2g as y decreases (supercritical flow).
- The minimum specific energy (E_min) occurs at the critical depth (y_c), where the Froude number is unity.
- For any energy value greater than E_min, there are two possible depths known as alternate depths.
- The curve shows that for a fixed discharge, the flow cannot exist at an energy level lower than E_min.
A trapezoidal channel with bottom width 3.5 m and side slopes 1 H: 1 V on the left and 1.5 H: 1 V on the right, with n = 0.016 and a bed slope of 2.6 m in 10,000 carries a discharge of 8 m3/sec. Determine the normal depth and average shear stress on the channel bed.
Answer
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A 5.0 m wide rectangular channel carries 15 m3/sec of water with a velocity of 6 m/sec. State whether hydraulic jump is a possibility. If yes compute height of the jump and power dissipated.
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At an energy dissipater structure below a low spillway, the discharge is 19 m3/sec and the energy loss is 1 m at the hydraulic jump forming there in. Determine the depths of flow at both ends of the jump.
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Explain the Rayleigh method of dimensional analysis with an example.
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In 1: 20 model of a spillway, the velocity and discharge are 1.5 m/sec and 2.0 m3/sec respectively. Find the corresponding velocity and discharge in the prototype.
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A jet of water 50 mm in diameter and moving with a velocity of 26 m/s is impinging normally on a plate. Determine the pressure on the plate when it is fixed and when it is moving with a velocity of 10 m/s in the direction of the jet. Also determine the work done per second by the jet.
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A pelton wheel has to be designed for the following data: power to be developed = 6000 kW, Net head available = 400 m, speed = 550 rpm, Ratio of jet diameter to the wheel diameter = 1/10 and overall efficiency = 85%. Find the number of jets, diameter of jet, diameter of the wheel and quantity of water required.
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The following data relate to a proposed hydro-electric station. Available head=28m, catchment area=420sq.km, rainfall=140cm/year, percentage of total rainfall utilised=68%, penstock efficiency=94%, turbine efficiency=80%, generator efficiency=84% and load factor=44%. Calculate the power developed.
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