THEORY EXAMINATION (SEM–IV) 2016-17 HYDRAULICS AND HYDRAULIC MACHINES

Course: B.Tech (Civil Engineering)
Subject Code: CE403
Subject Title: Hydraulics and Hydraulic Machines
Exam Type: Theory (Semester IV, 2016–17)
Duration: 3 Hours
Maximum Marks: 100

SECTION – A (10 × 2 = 20 Marks)

Short conceptual questions checking understanding of fundamental hydraulic principles:

SECTION – B (5 × 10 = 50 Marks)

Attempt any five. This section includes derivations, numerical analysis, and design problems.

Key Topics Covered:

(a) Flow Classification (Subcritical or Supercritical)
Given trapezoidal channel: base = 6 m, side slope 2H:1V, discharge = 17 m³/s, depth = 1.5 m.
→ Determine Froude number to identify flow type.

(b) Most Economical Rectangular Section
Derive that for maximum discharge or minimum perimeter:

b=2yandR=y2b = 2y \quad \text{and} \quad R = \frac{y}{2}b=2yandR=2y​

(c) Cavitation in Centrifugal Pump
Given:

patm=101 kPa,pv=2.34 kPa,hloss=1.55 m,H=52.5 m,σ=0.118p_{atm} = 101\,kPa, p_v = 2.34\,kPa, h_{loss} = 1.55\,m, H = 52.5\,m, \sigma = 0.118patm​=101kPa,pv​=2.34kPa,hloss​=1.55m,H=52.5m,σ=0.118.
Find maximum suction height to avoid cavitation.

(d) Inertia Head in Reciprocating Pump
Show:

hi=Lg⋅Aaω2r(1−cos⁡θ)h_i = \frac{L}{g} \cdot \frac{A}{a} \omega^2 r(1 - \cos\theta)hi​=gL​⋅aA​ω2r(1−cosθ)

where A = cylinder area, a = pipe area, L = length, r = crank radius.

(e) Chezy’s Formula
Derived as V=CRSV = C\sqrt{RS}V=CRS​ where C depends on Reynolds number and channel roughness.

(f) Most Economical Trapezoidal Section
Show that for max discharge at constant area:

m=12sin⁡(θ/2)andR=y2m = \frac{1}{2 \sin(\theta/2)} \quad \text{and} \quad R = \frac{y}{2}m=2sin(θ/2)1​andR=2y​

(g) Gradually Varied Flow Example
Rectangular channel: width = 10 m, slope changes from 0.01 to 0.0064, discharge = 125 m³/s, n=0.015n = 0.015n=0.015.
→ Determine surface profile type (M1, S2, etc.) and compute length of curve.

(h) Centrifugal Pump Construction & Working
Explain impeller, casing, suction, and delivery system with diagram; describe head, power, and efficiency relations.

SECTION – C (2 × 15 = 30 Marks)

Analytical and design-based questions on open channel flow and turbines.

Q3. Flow & Specific Energy

(a) Trapezoidal channel: bottom width = 6 m, slope 1:1, depth = 1.5 m, discharge = 15 m³/s.
→ Determine specific energy and flow type for given and critical depths.
(b) Derive hydraulic jump relation in triangular channel.

Q4. Hydraulic Jump & Pelton Wheel Design

(a) Rectangular channel: width = 4 m, discharge = 16 m³/s, initial depth = 0.5 m.
→ Determine:

If jump occurs,

Sequent depth,

Energy loss.
(b) Pelton Wheel Design:

P=1500 kW,H=160 m,N=420 rpm,η=85%P = 1500\,kW, H = 160\,m, N = 420\,rpm, \eta = 85\%P=1500kW,H=160m,N=420rpm,η=85%

Find jet diameter, wheel diameter, and bucket dimensions.

Q5. Reaction Turbine Design

Given:

Power = 300 kW, speed = 200 rpm, head = 18 m

D1=2D2,ηh=80%,ηm=95%,Q=3.6m3/sD_1 = 2D_2, \eta_h = 80\%, \eta_m = 95\%, Q = 3.6 m³/sD1​=2D2​,ηh​=80%,ηm​=95%,Q=3.6m3/s

Find:

Outer & inner diameters,

Vane inlet and exit angles,

Guide vane exit angle.
Use velocity triangles and flow relations for inward flow reaction turbine.

Key Concepts Summary

Summary

This Hydraulics and Hydraulic Machines (CE403) paper tests both conceptual and analytical skills in fluid mechanics applications.