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The velocity potential for a two-dimensional flow is given by ϕ = x2-y2+3xy. The flow rate between the streamlines passing through points (1, 2) and (2, 3) is
- a)9
- b)11
- c)15
- d)21
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The velocity potential for a two-dimensional flow is given by ϕ = x2-...
ϕ = x2 - y2 + 3xy
⇒ ∂ϕ/∂x = u = 2x + 3y = ∂ψ/∂y
⇒ ψ = 2xy + 3/2y2 + f(x).... (1)
and,
- ∂ϕ/∂y = V = -2y + 3X = -∂ψ/∂x ...(ii)
And from (1)
-∂ψ/∂x = 2y-f'(x) = -2y+3x
⇒ f'(x) = -3x
⇒ f'(x) = -3/2x2
The required stream function is
ψ = 2xy- 3/2 (x2-y2)
At point (2,3);
ψ2 = 2×2×3-32(22-32)
= 12 - 3/2(-5) = 12+7.5 = 19.5 units
Flow rate between the streamlines passing through (1, 2) and (2, 3)
∆ψ = ψ2 - ψ1
= 19.5-8.5 = 11 units.
Most Upvoted Answer
The velocity potential for a two-dimensional flow is given by ϕ = x2-...
Calculation of Velocity Components:
- The velocity potential is given by ϕ = x^2 - y^2 + 3xy
- The velocity components can be calculated as follows:
- u = ∂ϕ/∂x = 2x + 3y
- v = -∂ϕ/∂y = 2y - 3x
Calculation of Flow Rate:
- The flow rate between two streamlines passing through points (1, 2) and (2, 3) can be calculated using the formula:
- Q = ∫v.dx
- To apply this formula, we need to find an equation for the streamline passing through each point:
- For point (1, 2): ϕ = 1 - 4 + 6 = 3 => 3 = x^2 - y^2 + 3xy
- For point (2, 3): ϕ = 4 - 9 + 18 = 13 => 13 = x^2 - y^2 + 3xy
- Solving these two equations simultaneously, we get the equation of the streamline passing through both points as:
- x^2 - y^2 + 3xy = 7
- We can rewrite this equation as:
- y = (x^2 + 7)/(3x)
- Now we can calculate the flow rate by integrating v.dx along this streamline:
- Q = ∫v.dx = ∫(2y - 3x)dx = ∫(2(x^2 + 7)/(3x) - 3x)dx from x = 1 to x = 2
- Q = [2/3(x^3 + 7x) - 3/2(x^2)] from x = 1 to x = 2
- Q = (16/3 - 21/2) - (2/3 + 7/3) = 11
Therefore, the flow rate between the streamlines passing through points (1, 2) and (2, 3) is 11.
- The velocity potential is given by ϕ = x^2 - y^2 + 3xy
- The velocity components can be calculated as follows:
- u = ∂ϕ/∂x = 2x + 3y
- v = -∂ϕ/∂y = 2y - 3x
Calculation of Flow Rate:
- The flow rate between two streamlines passing through points (1, 2) and (2, 3) can be calculated using the formula:
- Q = ∫v.dx
- To apply this formula, we need to find an equation for the streamline passing through each point:
- For point (1, 2): ϕ = 1 - 4 + 6 = 3 => 3 = x^2 - y^2 + 3xy
- For point (2, 3): ϕ = 4 - 9 + 18 = 13 => 13 = x^2 - y^2 + 3xy
- Solving these two equations simultaneously, we get the equation of the streamline passing through both points as:
- x^2 - y^2 + 3xy = 7
- We can rewrite this equation as:
- y = (x^2 + 7)/(3x)
- Now we can calculate the flow rate by integrating v.dx along this streamline:
- Q = ∫v.dx = ∫(2y - 3x)dx = ∫(2(x^2 + 7)/(3x) - 3x)dx from x = 1 to x = 2
- Q = [2/3(x^3 + 7x) - 3/2(x^2)] from x = 1 to x = 2
- Q = (16/3 - 21/2) - (2/3 + 7/3) = 11
Therefore, the flow rate between the streamlines passing through points (1, 2) and (2, 3) is 11.
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The velocity potential for a two-dimensional flow is given by ϕ = x2-y2+3xy. The flow rate between the streamlines passing through points (1, 2) and (2, 3) isa)9b)11c)15d)21Correct answer is option 'B'. Can you explain this answer?
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