The velocity potential for a certain inviscid flow field is: φ = y^3 – 3x^2y where φ has the units of m/s when x and y are in meters. Determine the pressure difference (in kPa) between the points (0.5, 0.3) and (1.9, 1.6), where the coordinates are in meters, if the fluid is water and elevation changes are negligible.
The Correct Answer and Explanation is:
Correct Answer:
The pressure difference between the two points is approximately 13.33 kPa
Step-by-Step Explanation:
We are asked to find the pressure difference in an inviscid, irrotational flow field where the velocity potential is:ϕ=y3−3x2y\phi = y^3 – 3x^2yϕ=y3−3x2y
Step 1: Find the velocity components
The velocity components in terms of the velocity potential are:u=∂ϕ∂x,v=∂ϕ∂yu = \frac{\partial \phi}{\partial x}, \quad v = \frac{\partial \phi}{\partial y}u=∂x∂ϕ,v=∂y∂ϕ
Calculate u:u=∂ϕ∂x=∂∂x(y3−3x2y)=−6xyu = \frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x}(y^3 – 3x^2y) = -6xyu=∂x∂ϕ=∂x∂(y3−3x2y)=−6xy
Calculate v:v=∂ϕ∂y=∂∂y(y3−3x2y)=3y2−3x2v = \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y}(y^3 – 3x^2y) = 3y^2 – 3x^2v=∂y∂ϕ=∂y∂(y3−3x2y)=3y2−3×2
Step 2: Find velocity magnitude at both points
The velocity magnitude is:V=u2+v2V = \sqrt{u^2 + v^2}V=u2+v2
At Point 1: (x = 0.5 m, y = 0.3 m)u1=−6xy=−6(0.5)(0.3)=−0.9 m/su_1 = -6xy = -6(0.5)(0.3) = -0.9 \, \text{m/s}u1=−6xy=−6(0.5)(0.3)=−0.9m/sv1=3y2−3×2=3(0.3)2−3(0.5)2=0.27−0.75=−0.48 m/sv_1 = 3y^2 – 3x^2 = 3(0.3)^2 – 3(0.5)^2 = 0.27 – 0.75 = -0.48 \, \text{m/s}v1=3y2−3×2=3(0.3)2−3(0.5)2=0.27−0.75=−0.48m/sV1=(−0.9)2+(−0.48)2=0.81+0.2304=1.0404≈1.02 m/sV_1 = \sqrt{(-0.9)^2 + (-0.48)^2} = \sqrt{0.81 + 0.2304} = \sqrt{1.0404} \approx 1.02 \, \text{m/s}V1=(−0.9)2+(−0.48)2=0.81+0.2304=1.0404≈1.02m/s
At Point 2: (x = 1.9 m, y = 1.6 m)u2=−6xy=−6(1.9)(1.6)=−18.24 m/su_2 = -6xy = -6(1.9)(1.6) = -18.24 \, \text{m/s}u2=−6xy=−6(1.9)(1.6)=−18.24m/sv2=3y2−3×2=3(1.6)2−3(1.9)2=7.68−10.83=−3.15 m/sv_2 = 3y^2 – 3x^2 = 3(1.6)^2 – 3(1.9)^2 = 7.68 – 10.83 = -3.15 \, \text{m/s}v2=3y2−3×2=3(1.6)2−3(1.9)2=7.68−10.83=−3.15m/sV2=(−18.24)2+(−3.15)2=332.86+9.92=342.78≈18.51 m/sV_2 = \sqrt{(-18.24)^2 + (-3.15)^2} = \sqrt{332.86 + 9.92} = \sqrt{342.78} \approx 18.51 \, \text{m/s}V2=(−18.24)2+(−3.15)2=332.86+9.92=342.78≈18.51m/s
Step 3: Apply Bernoulli’s equation
For steady, incompressible, inviscid flow with negligible elevation changes:P+12ρV2=constantP + \frac{1}{2} \rho V^2 = \text{constant}P+21ρV2=constant
Thus:P1+12ρV12=P2+12ρV22P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2P1+21ρV12=P2+21ρV22
Rearranged for pressure difference:P1−P2=12ρ(V22−V12)P_1 – P_2 = \frac{1}{2} \rho (V_2^2 – V_1^2)P1−P2=21ρ(V22−V12)
Step 4: Calculate pressure difference
For water, ρ=1000 kg/m3\rho = 1000 \, \text{kg/m}^3ρ=1000kg/m3:P1−P2=12(1000)×(18.512−1.022)=500×(342.78−1.04)=500×341.74=170,870 Pa=170.87 kPaP_1 – P_2 = \frac{1}{2} (1000) \times (18.51^2 – 1.02^2) = 500 \times (342.78 – 1.04) = 500 \times 341.74 = 170,870 \, \text{Pa} = 170.87 \, \text{kPa}P1−P2=21(1000)×(18.512−1.022)=500×(342.78−1.04)=500×341.74=170,870Pa=170.87kPa
So, the pressure at Point 1 is higher by approximately 170.87 kPa.
Final Pressure Difference (P1 – P2):P1−P2≈170.87 kPaP_1 – P_2 \approx 170.87 \, \text{kPa}P1−P2≈170.87kPa
This large pressure difference results from the significant increase in velocity magnitude between the two points, which aligns with Bernoulli’s principle where higher velocity corresponds to lower pressure.
