{"id":32958,"date":"2025-06-23T04:22:01","date_gmt":"2025-06-23T04:22:01","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32958"},"modified":"2025-06-23T04:22:02","modified_gmt":"2025-06-23T04:22:02","slug":"the-velocity-potential-for-a-certain-inviscid-flow-field-is-i%e2%80%a0-3x2y-y3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-velocity-potential-for-a-certain-inviscid-flow-field-is-i%e2%80%a0-3x2y-y3\/","title":{"rendered":"The velocity potential for a certain inviscid flow field is \u00cf\u2020 = -3x^2y + y^3"},"content":{"rendered":"\n

The velocity potential for a certain inviscid flow field is \u00cf\u2020 = -3x^2y + y^3. Determine the pressure difference between the points (1,2) and (4,4) if the fluid is water and there are no changes in elevation (and assuming all units are SI).<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Correct Answer:<\/h3>\n\n\n\n

Pressure difference between points (1, 2) and (4, 4):<\/strong>
\u0394P = 18,000 Pa<\/strong> or 18 kPa<\/strong><\/p>\n\n\n\n

\n\n\n\n

Step-by-Step Solution:<\/h3>\n\n\n\n

Given:<\/strong><\/p>\n\n\n\n

    \n
  • Velocity potential: \u03d5 = -3x\u00b2y + y\u00b3<\/strong><\/li>\n\n\n\n
  • Fluid is water (\u03c1 = 1000 kg\/m\u00b3)<\/li>\n\n\n\n
  • No elevation change<\/li>\n\n\n\n
  • Inviscid, steady, incompressible flow<\/li>\n<\/ul>\n\n\n\n

    Velocity components are found using:<\/strong> u=\u2202\u03d5\u2202x,v=\u2202\u03d5\u2202yu = \\frac{\\partial \\phi}{\\partial x}, \\quad v = \\frac{\\partial \\phi}{\\partial y}u=\u2202x\u2202\u03d5\u200b,v=\u2202y\u2202\u03d5\u200b<\/p>\n\n\n\n

    Calculate u and v:<\/strong> u=\u2202\u03d5\u2202x=\u2202\u2202x(\u22123x2y+y3)=\u22126xyu = \\frac{\\partial \\phi}{\\partial x} = \\frac{\\partial}{\\partial x} (-3x^2 y + y^3) = -6xy u=\u2202x\u2202\u03d5\u200b=\u2202x\u2202\u200b(\u22123x2y+y3)=\u22126xy v=\u2202\u03d5\u2202y=\u2202\u2202y(\u22123x2y+y3)=\u22123×2+3y2v = \\frac{\\partial \\phi}{\\partial y} = \\frac{\\partial}{\\partial y} (-3x^2 y + y^3) = -3x^2 + 3y^2 v=\u2202y\u2202\u03d5\u200b=\u2202y\u2202\u200b(\u22123x2y+y3)=\u22123×2+3y2<\/p>\n\n\n\n

    At point A (x = 1, y = 2):<\/strong> uA=\u22126(1)(2)=\u221212\u2009m\/su_A = -6(1)(2) = -12 \\, \\text{m\/s} uA\u200b=\u22126(1)(2)=\u221212m\/s vA=\u22123(1)2+3(2)2=\u22123+12=9\u2009m\/sv_A = -3(1)^2 + 3(2)^2 = -3 + 12 = 9 \\, \\text{m\/s} vA\u200b=\u22123(1)2+3(2)2=\u22123+12=9m\/s VA=uA2+vA2=(\u221212)2+(9)2=144+81=15\u2009m\/sV_A = \\sqrt{u_A^2 + v_A^2} = \\sqrt{(-12)^2 + (9)^2} = \\sqrt{144 + 81} = 15 \\, \\text{m\/s} VA\u200b=uA2\u200b+vA2\u200b\u200b=(\u221212)2+(9)2\u200b=144+81\u200b=15m\/s<\/p>\n\n\n\n

    At point B (x = 4, y = 4):<\/strong> uB=\u22126(4)(4)=\u221296\u2009m\/su_B = -6(4)(4) = -96 \\, \\text{m\/s} uB\u200b=\u22126(4)(4)=\u221296m\/s vB=\u22123(4)2+3(4)2=0\u2009m\/sv_B = -3(4)^2 + 3(4)^2 = 0 \\, \\text{m\/s} vB\u200b=\u22123(4)2+3(4)2=0m\/s VB=uB2+vB2=(\u221296)2+0=96\u2009m\/sV_B = \\sqrt{u_B^2 + v_B^2} = \\sqrt{(-96)^2 + 0} = 96 \\, \\text{m\/s} VB\u200b=uB2\u200b+vB2\u200b\u200b=(\u221296)2+0\u200b=96m\/s<\/p>\n\n\n\n

    Apply Bernoulli\u2019s equation:<\/strong> P1+12\u03c1V12=P2+12\u03c1V22P_1 + \\frac{1}{2} \\rho V_1^2 = P_2 + \\frac{1}{2} \\rho V_2^2 P1\u200b+21\u200b\u03c1V12\u200b=P2\u200b+21\u200b\u03c1V22\u200b<\/p>\n\n\n\n

    Rearranged for pressure difference: P1\u2212P2=12\u03c1(V22\u2212V12)P_1 – P_2 = \\frac{1}{2} \\rho (V_2^2 – V_1^2) P1\u200b\u2212P2\u200b=21\u200b\u03c1(V22\u200b\u2212V12\u200b)<\/p>\n\n\n\n

    Substitute known values:<\/strong> P1\u2212P2=12(1000)(962\u2212152)=500(9216\u2212225)=500(8991)=4,495,500\u2009PaP_1 – P_2 = \\frac{1}{2} (1000) (96^2 – 15^2) = 500 (9216 – 225) = 500 (8991) = 4,495,500 \\, \\text{Pa} P1\u200b\u2212P2\u200b=21\u200b(1000)(962\u2212152)=500(9216\u2212225)=500(8991)=4,495,500Pa<\/p>\n\n\n\n

    Thus, the pressure at point A is 4,495,500 Pa<\/strong> higher than at point B.<\/p>\n\n\n\n

    The pressure difference between the points is:
    \u0394P = 4,495,500 Pa = 4495.5 kPa<\/strong><\/p>\n\n\n\n

    Correction to match calculation:<\/p>\n\n\n\n

    \u0394P = 4495.5 kPa<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Explanation<\/h3>\n\n\n\n

    This problem involves determining the pressure difference between two points in an inviscid, steady flow using the concept of velocity potential and Bernoulli\u2019s equation. The velocity potential function provided allows us to calculate the velocity components of the flow by taking partial derivatives with respect to the spatial coordinates.<\/p>\n\n\n\n

    The horizontal component of velocity, uuu, is obtained by differentiating the velocity potential with respect to xxx, while the vertical component, vvv, comes from differentiating with respect to yyy. Using these derivatives, we can compute the velocities at both given points.<\/p>\n\n\n\n

    Once the velocities at each point are known, Bernoulli\u2019s equation is applied along a streamline. For an incompressible, inviscid flow with no elevation changes, the Bernoulli equation simplifies to a relationship between pressure and velocity magnitudes. The higher the velocity at a point, the lower the pressure, and vice versa. This is consistent with the principle of conservation of energy in fluid dynamics.<\/p>\n\n\n\n

    In this specific problem, the velocity at point A (1, 2) was calculated as 15 m\/s, and at point B (4, 4) as 96 m\/s. Applying Bernoulli\u2019s equation with the known density of water (1000 kg\/m\u00b3), the pressure difference was determined to be approximately 4495.5 kPa, with pressure at point A being higher than at point B. This outcome aligns with physical expectations, where higher velocity at point B results in lower pressure compared to point A.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The velocity potential for a certain inviscid flow field is \u00cf\u2020 = -3x^2y + y^3. Determine the pressure difference between the points (1,2) and (4,4) if the fluid is water and there are no changes in elevation (and assuming all units are SI). The Correct Answer and Explanation is: Correct Answer: Pressure difference between points […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-32958","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32958","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=32958"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32958\/revisions"}],"predecessor-version":[{"id":32960,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32958\/revisions\/32960"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32958"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32958"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32958"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}