Solution :-

P = 1750 kW

H = 100 m

Unit power of the Pelton wheel (Pu),

= P/H^(3/2)

= 1750/(100)^(3/2)

= 1750/1000

= 1.75 kW

Solution :-

U = xyÎ + 3yzĵ - (2yz + z²)k + 3t

Let, u, v, and w are velocity components along x, y and z-direction respectively.

u = xy

v = 3yz

w = - (2yz + z²)

For z -direction, apply the derivative to the z-component (w) of velocity.

∂w/∂t = 0

∂w/∂x = 0

∂w/∂y = -2z

∂w/∂z = -2y - 2z

The acceleration in z direction a(z),

= ∂w/∂t + u(∂w/∂x) + v(∂w/∂y) + w(∂w/∂z)

= 0 + xy(0) + 3yz(-2z) + (-2yz - z²)(-2y - 2z)

= - 6yz² + 4zy² + 4yz² + 2yz² + 2z³

= - 6yz² + 4zy² + 6yz² + 2z³

= 4zy² + 2z³

= (4 × 4 × 2²) + (2 × 4³)     [∴ at point (1, 2, 4)]

= 64 + 128

= 192 unit

Explanation :-

(ii) has a viscosity that always decrease with the increase in temperature (False)

• For liquid viscosity decrease with the increase in temperature
• But, for gases viscosity increase with the increase in temperature

(iii) is essentially incompressible (False)

• Liquids are nearly incompressible, but gases are highly compressible.

Explanation :-

(i) Darcy-Weisbach equation

• Used to calculate the head loss due to friction along each pipe branch.

(ii) Bernoulli’s equation.

• Describes the relationship between pressure, velocity, and elevation in a fluid flow, and it's used to analyze energy conservation along a streamline.
• In branched pipes, it helps determine how energy is distributed among the different branches.

(iii) Momentum equation

• Used to analyze the forces acting on a fluid and how those forces affect its momentum.

(iv) Cauchy’s distribution equation

• Related to the distribution of stress in a fluid.

Only (i) and (ii) Equations are required to solve branching of pipe problems.

Solution :-

R = D/2 = 20/2 = 10 cm

V(max) = 2 m/s

r = 5 cm

Velocity at a radial distance r,

= V(max) × [1 - r²/R²]

= 2 × [1 - 5²/10²]

= 2 × [1 - 25/100]

= 2 × (1 - 0.25)

= 2 × 0.75

= 1.5 m/s

Explanation :-

The unit of dynamic viscosity = N⋅s/m2

The dimension of dynamic viscosity,

= MLT⁻² ⋅ T / L²

= ML⁻¹T⁻¹

Explanation :-

Venturiflume,

Solution :-

D1 = 6 cm

D2 = 12 cm

Continuity equation,

A1⋅V1 = A2⋅V2

(π/4)⋅D1²⋅V1 = (π/4)⋅D2²⋅V2

V2 = (D1/D2)² × V1

V2 = (6/12)² × V1

V2 = (1/4) × V1

Head loss due to sudden expansion (hL),

= (V1 - V2)²/2g

= [V1 - (1/4) × V1]²/2g

= [(3/4) × V1]²/2g

= (9/16)(V1²/2g)

Explanation :-

(i) The rise of water level which takes place due to the transformation of the sub-critical flow to the super-critical flow (False)

• A hydraulic jump occurs when flow transitions is from super-critical (high velocity, low depth) to sub-critical (low velocity, high depth).

Solution :-

y1 = pre jump depth = 1 m

y2 = post jump depth = 2 m

Head loss due to hydraulic jump (hL),

= (y2 - y1)³/(4⋅y1⋅y2)

= (2 - 1)³/(4 × 1 × 2)

= 1/8

= 0.125 m