Fluid Mechanics and Hydraulic Machinery Miscellaneous
 A twodimensional 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

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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 = constantCorrect 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
 In a flow field the stream lines and equipotential lines

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NA
Correct Option: D
NA
 A fluid flow is represented by the velocity field = ax_{î} + ay_{ĵ}, where a is a constant. The equation of stream line passing through a point (1, 2) is

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Given: u_{x} = ax and u_{y} = ay
Equation of steam line is,du = dy u_{x} u_{y} ⇒ 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: u_{x} = ax and u_{y} = ay
Equation of steam line is,du = dy u_{x} u_{y} ⇒ 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.
 A streamlined body is defined as a body about which

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NA
Correct Option: C
NA
 Figure shows the schematic for the measurement of velocity of air (density = 1.2 kg/m³) through a constantarea 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

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From Bernoulli's equation
v²_{1}  v²_{2} = p_{2}  p_{1} 2g 2g ⇒ v_{1} = √ (p_{2}  p_{1}) p_{a}
But p_{2}  p_{1} = (ρgh)_{water}
= 9810 × 10 × 10^{– 3}
= 98.1 N/m²⇒ v_{1} = √ 2 × 98.1 = 12.8 m/s 1.2 Correct Option: C
From Bernoulli's equation
v²_{1}  v²_{2} = p_{2}  p_{1} 2g 2g ⇒ v_{1} = √ (p_{2}  p_{1}) p_{a}
But p_{2}  p_{1} = (ρgh)_{water}
= 9810 × 10 × 10^{– 3}
= 98.1 N/m²⇒ v_{1} = √ 2 × 98.1 = 12.8 m/s 1.2