## Fluid Mechanics and Hydraulic Machinery Miscellaneous

#### Fluid Mechanics and Hydraulic Machinery

1. A two-dimensional flow field has velocities along the x and y directions given by u = x²t and v = –2xyt respectively, where t is time. The equation of streamline is

1. Given, u = x²t and v = – 2xyt

= v = -2xyt..........(i)
 We know, δψ δy

= -u = x²t..........(ii)
 δψ δy

Integrating equation (i), we get
ψ = –x²yt + f(y) ...(iii)
Differentiating equation (iii) with respect to y, we get
= -x²t + f(y)........(v)
 δψ δy

Equating the value of δψ/δy from equations (ii) and (iv), we get
–x²t =–x²t + f'(y)
Since, f'(y) = 0, thus f(y) = C
(where ‘C’ is constant of integration)
ψ = –x²yt + C
C is a numerical constant so it can be taken as zero
ψ = –x²yt
For equation of stream lines,
ψ = constant
–x²yt = constant
For a particular instance,
x²y = constant

##### Correct Option: A

Given, u = x²t and v = – 2xyt

= v = -2xyt..........(i)
 We know, δψ δy

= -u = x²t..........(ii)
 δψ δy

Integrating equation (i), we get
ψ = –x²yt + f(y) ...(iii)
Differentiating equation (iii) with respect to y, we get
= -x²t + f(y)........(v)
 δψ δy

Equating the value of δψ/δy from equations (ii) and (iv), we get
–x²t =–x²t + f'(y)
Since, f'(y) = 0, thus f(y) = C
(where ‘C’ is constant of integration)
ψ = –x²yt + C
C is a numerical constant so it can be taken as zero
ψ = –x²yt
For equation of stream lines,
ψ = constant
–x²yt = constant
For a particular instance,
x²y = constant

1. In a flow field the stream lines and equipotential lines

1. NA

##### Correct Option: D

NA

1. A fluid flow is represented by the velocity field = axî + ayĵ, where a is a constant. The equation of stream line passing through a point (1, 2) is

1. Given: ux = ax and uy = ay
Equation of steam line is,

 du = dy ux uy

 ⇒ du = dy ux uy

Integrating both sides, we have
log(ax) = log(ay) + log c
or ax = c'ay
or x = cy
Since the steam line is passed through point (1, 2), therefore
1 = 2c
 ⇒ c = 1 2

 ∴ x = y 2

Hence equation of steam line is
2x – y = 0.

##### Correct Option: C

Given: ux = ax and uy = ay
Equation of steam line is,

 du = dy ux uy

 ⇒ du = dy ux uy

Integrating both sides, we have
log(ax) = log(ay) + log c
or ax = c'ay
or x = cy
Since the steam line is passed through point (1, 2), therefore
1 = 2c
 ⇒ c = 1 2

 ∴ x = y 2

Hence equation of steam line is
2x – y = 0.

1. A streamlined body is defined as a body about which

1. NA

##### Correct Option: C

NA

1. Figure shows the schematic for the measurement of velocity of air (density = 1.2 kg/m³) through a constant-area duct using a pitot tube and a water tube manometer. The differential head of water (density = 1000 kg/ m³) in the two columns of the manometer is 10 mm. Take acceleration due to gravity as 9.8 m/s². The velocity of air in m/s is 10 m m

1. From Bernoulli's equation

 v²1 - v²2 = p2 - p1 2g 2g

 ⇒ v1 = √ (p2 - p1) pa

But p2 - p1 = (ρgh)water
= 9810 × 10 × 10– 3
= 98.1 N/m²
 ⇒ v1 = √ 2 × 98.1 = 12.8 m/s 1.2

##### Correct Option: C

From Bernoulli's equation

 v²1 - v²2 = p2 - p1 2g 2g

 ⇒ v1 = √ (p2 - p1) pa

But p2 - p1 = (ρgh)water
= 9810 × 10 × 10– 3
= 98.1 N/m²
 ⇒ v1 = √ 2 × 98.1 = 12.8 m/s 1.2