{"id":32954,"date":"2025-06-23T04:18:40","date_gmt":"2025-06-23T04:18:40","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=32954"},"modified":"2025-06-23T04:18:42","modified_gmt":"2025-06-23T04:18:42","slug":"the-velocity-potential-for-a-certain-inviscid-flow-field-is-%cf%86-y3-3x2y-where-%cf%86-has-the-units-of-m-s-when-x-and-y-are-in-meters","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-velocity-potential-for-a-certain-inviscid-flow-field-is-%cf%86-y3-3x2y-where-%cf%86-has-the-units-of-m-s-when-x-and-y-are-in-meters\/","title":{"rendered":"The velocity potential for a certain inviscid flow field is: \u03c6 = y^3 – 3x^2y where \u03c6 has the units of m\/s when x and y are in meters."},"content":{"rendered":"\n

The velocity potential for a certain inviscid flow field is: \u03c6 = y^3 – 3x^2y where \u03c6 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.<\/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

The pressure difference between the two points is approximately 13.33 kPa<\/strong><\/p>\n\n\n\n

\n\n\n\n

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

We are asked to find the pressure difference in an inviscid, irrotational flow field where the velocity potential is:\u03d5=y3\u22123x2y\\phi = y^3 – 3x^2y\u03d5=y3\u22123x2y<\/p>\n\n\n\n

Step 1: Find the velocity components<\/strong><\/p>\n\n\n\n

The velocity components in terms of the velocity potential are: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:<\/strong>u=\u2202\u03d5\u2202x=\u2202\u2202x(y3\u22123x2y)=\u22126xyu = \\frac{\\partial \\phi}{\\partial x} = \\frac{\\partial}{\\partial x}(y^3 – 3x^2y) = -6xyu=\u2202x\u2202\u03d5\u200b=\u2202x\u2202\u200b(y3\u22123x2y)=\u22126xy<\/p>\n\n\n\n

Calculate v:<\/strong>v=\u2202\u03d5\u2202y=\u2202\u2202y(y3\u22123x2y)=3y2\u22123x2v = \\frac{\\partial \\phi}{\\partial y} = \\frac{\\partial}{\\partial y}(y^3 – 3x^2y) = 3y^2 – 3x^2v=\u2202y\u2202\u03d5\u200b=\u2202y\u2202\u200b(y3\u22123x2y)=3y2\u22123×2<\/p>\n\n\n\n

Step 2: Find velocity magnitude at both points<\/strong><\/p>\n\n\n\n

The velocity magnitude is:V=u2+v2V = \\sqrt{u^2 + v^2}V=u2+v2\u200b<\/p>\n\n\n\n

At Point 1: (x = 0.5 m, y = 0.3 m)<\/strong>u1=\u22126xy=\u22126(0.5)(0.3)=\u22120.9\u2009m\/su_1 = -6xy = -6(0.5)(0.3) = -0.9 \\, \\text{m\/s}u1\u200b=\u22126xy=\u22126(0.5)(0.3)=\u22120.9m\/sv1=3y2\u22123×2=3(0.3)2\u22123(0.5)2=0.27\u22120.75=\u22120.48\u2009m\/sv_1 = 3y^2 – 3x^2 = 3(0.3)^2 – 3(0.5)^2 = 0.27 – 0.75 = -0.48 \\, \\text{m\/s}v1\u200b=3y2\u22123×2=3(0.3)2\u22123(0.5)2=0.27\u22120.75=\u22120.48m\/sV1=(\u22120.9)2+(\u22120.48)2=0.81+0.2304=1.0404\u22481.02\u2009m\/sV_1 = \\sqrt{(-0.9)^2 + (-0.48)^2} = \\sqrt{0.81 + 0.2304} = \\sqrt{1.0404} \\approx 1.02 \\, \\text{m\/s}V1\u200b=(\u22120.9)2+(\u22120.48)2\u200b=0.81+0.2304\u200b=1.0404\u200b\u22481.02m\/s<\/p>\n\n\n\n

At Point 2: (x = 1.9 m, y = 1.6 m)<\/strong>u2=\u22126xy=\u22126(1.9)(1.6)=\u221218.24\u2009m\/su_2 = -6xy = -6(1.9)(1.6) = -18.24 \\, \\text{m\/s}u2\u200b=\u22126xy=\u22126(1.9)(1.6)=\u221218.24m\/sv2=3y2\u22123×2=3(1.6)2\u22123(1.9)2=7.68\u221210.83=\u22123.15\u2009m\/sv_2 = 3y^2 – 3x^2 = 3(1.6)^2 – 3(1.9)^2 = 7.68 – 10.83 = -3.15 \\, \\text{m\/s}v2\u200b=3y2\u22123×2=3(1.6)2\u22123(1.9)2=7.68\u221210.83=\u22123.15m\/sV2=(\u221218.24)2+(\u22123.15)2=332.86+9.92=342.78\u224818.51\u2009m\/sV_2 = \\sqrt{(-18.24)^2 + (-3.15)^2} = \\sqrt{332.86 + 9.92} = \\sqrt{342.78} \\approx 18.51 \\, \\text{m\/s}V2\u200b=(\u221218.24)2+(\u22123.15)2\u200b=332.86+9.92\u200b=342.78\u200b\u224818.51m\/s<\/p>\n\n\n\n

Step 3: Apply Bernoulli\u2019s equation<\/strong><\/p>\n\n\n\n

For steady, incompressible, inviscid flow with negligible elevation changes:P+12\u03c1V2=constantP + \\frac{1}{2} \\rho V^2 = \\text{constant}P+21\u200b\u03c1V2=constant<\/p>\n\n\n\n

Thus:P1+12\u03c1V12=P2+12\u03c1V22P_1 + \\frac{1}{2} \\rho V_1^2 = P_2 + \\frac{1}{2} \\rho V_2^2P1\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

Step 4: Calculate pressure difference<\/strong><\/p>\n\n\n\n

For water, \u03c1=1000\u2009kg\/m3\\rho = 1000 \\, \\text{kg\/m}^3\u03c1=1000kg\/m3:P1\u2212P2=12(1000)\u00d7(18.512\u22121.022)=500\u00d7(342.78\u22121.04)=500\u00d7341.74=170,870\u2009Pa=170.87\u2009kPaP_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\u200b\u2212P2\u200b=21\u200b(1000)\u00d7(18.512\u22121.022)=500\u00d7(342.78\u22121.04)=500\u00d7341.74=170,870Pa=170.87kPa<\/p>\n\n\n\n

So, the pressure at Point 1 is higher by approximately 170.87 kPa<\/strong>.<\/p>\n\n\n\n

Final Pressure Difference (P1 – P2):<\/strong>P1\u2212P2\u2248170.87\u2009kPaP_1 – P_2 \\approx 170.87 \\, \\text{kPa}P1\u200b\u2212P2\u200b\u2248170.87kPa<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

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

The velocity potential for a certain inviscid flow field is: \u03c6 = y^3 – 3x^2y where \u03c6 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 […]<\/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-32954","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32954","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=32954"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32954\/revisions"}],"predecessor-version":[{"id":32956,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/32954\/revisions\/32956"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=32954"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=32954"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=32954"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}