## Fluid Mechanics and Hydraulic Machinery Miscellaneous

#### Fluid Mechanics and Hydraulic Machinery

1. If P is the gauge pressure within a spherical droplet, then gauge pressure within a bubble of the same fluid and of same size will be

1. P = Gauge pressure within a spherical droplet

 Pdroplet = 4σ for droplet d

 Pbubble = 8σ for bubble d

Pbubble = 2 Pdroplet

##### Correct Option: D

P = Gauge pressure within a spherical droplet

 Pdroplet = 4σ for droplet d

 Pbubble = 8σ for bubble d

Pbubble = 2 Pdroplet

1. The dimension of surface tension is

1. Surface tension (σ)

 σ = F = N = kg − m = MT−2 L m s2 × m

##### Correct Option: D

Surface tension (σ)

 σ = F = N = kg − m = MT−2 L m s2 × m

1. In a Pelton wheel, the bucket peripheral speed is 10 m/s, the water jet velocity is 25 m/s and volumetric flow rate of the jet is 0.1 m3/s. If the jet deflection angle is 120° and the flow is ideal, the power developed is

1. u = 10 m/s
Vi = 25 m/s
Q = 0.1 m3 /s
Jet deflection angle = 120°
∴ β = 180° – 120° = 60°
P = ρQ (Vi – u)(1 + cosβ)u
P = 1000 × 0.1(25 – 10)(1 + cos60°) × 10 P = 22500 W = 22.5 kW

##### Correct Option: C

u = 10 m/s
Vi = 25 m/s
Q = 0.1 m3 /s
Jet deflection angle = 120°
∴ β = 180° – 120° = 60°
P = ρQ (Vi – u)(1 + cosβ)u
P = 1000 × 0.1(25 – 10)(1 + cos60°) × 10 P = 22500 W = 22.5 kW

1. At a hydro electric power plant site, available head and flow rate are 24.5 m and 10.1 m3/s respectively. If the turbine to be installed is required to run at 4.0 revolution per second (rps) with an overall efficiency of 90%, the suitable type of turbine for this site is

1. Given: H = 24.5 m, Q = 10.1 m3 /s
N = 4 rev/s = 4 × 60 = 240 r.p.m.
η0 = 0.90
∴ Power generated P = ρgQH × 0.9
P = 1000 × 9.81 × 10.1 × 24.5 × 0.9
P = 2184.7 kW
Again ,

 Ns = N√P H5 / 4

 Ns = 240√2184.7 = 205.80 (24.5)5 / 4

51 < Ns < 255, hence turbine in Francis.

##### Correct Option: A

Given: H = 24.5 m, Q = 10.1 m3 /s
N = 4 rev/s = 4 × 60 = 240 r.p.m.
η0 = 0.90
∴ Power generated P = ρgQH × 0.9
P = 1000 × 9.81 × 10.1 × 24.5 × 0.9
P = 2184.7 kW
Again ,

 Ns = N√P H5 / 4

 Ns = 240√2184.7 = 205.80 (24.5)5 / 4

51 < Ns < 255, hence turbine in Francis.

1. Match List-I with List-II and select the correct answer using the codes given below the lists:
 List-I List-II P. Curtis 1. Reaction steam turbine Q. Rateau 2. Gas turbine R. Kaplan 3. Velocity compounding S. Francis 4. Pressure compounding 5. Impulse water turbine 6. Axial turbine 7. Mixed flow turbine 8. Centrifugal pump

1. NA

NA