{"id":6269,"date":"2021-10-01T20:19:30","date_gmt":"2021-10-01T19:19:30","guid":{"rendered":"https:\/\/www.nickzom.org\/blog\/?p=6269"},"modified":"2024-05-16T09:14:19","modified_gmt":"2024-05-16T08:14:19","slug":"how-to-calculate-and-solve-for-velocity-of-newtonian-fluid-transport-phenomena","status":"publish","type":"post","link":"https:\/\/www.nickzom.org\/blog\/2021\/10\/01\/how-to-calculate-and-solve-for-velocity-of-newtonian-fluid-transport-phenomena\/","title":{"rendered":"How to Calculate and Solve for Velocity of Newtonian Fluid | Transport Phenomena"},"content":{"rendered":"

\"\"<\/p>\n

The image above represents the velocity of Newtonian fluid. To calculate the velocity of Newtonian fluid, five essential parameters are needed, and these parameters are Change in Pressure (\u0394P), Distance between Fluid Flow (L), Diameter of Pipe (y), Separation Distance of Atoms (\u03b4) and Viscosity (\u03b7).<\/p>\n

The formula for calculating the velocity of Newtonian fluid:<\/strong><\/h3>\n

v = (\u03b4\u00b2 – y\u00b2)\u0394P<\/sup> \/ 2\u03b7.L<\/sub><\/p>\n

Where:<\/p>\n

v = Velocity of Newtonian Fluid
\n\u0394P = Change in Pressure
\nL = Distance between Fluid Flow
\ny = Diameter of Pipe
\n\u03b4 = Separation Distance of Atoms
\n\u03b7 = Viscosity<\/p>\n

Let’s solve an example;
\nFind the velocity of newtonian fluid when the change in pressure is 22, the distance between fluid flow is 13, the diameter of pipe is 7, the separation distance of atoms is 11 and the viscosity is 5.<\/p>\n

This implies that;<\/p>\n

\u0394P = Change in Pressure = 22
\nL = Distance between Fluid Flow = 13
\ny = Diameter of Pipe = 7
\n\u03b4 = Separation Distance of Atoms = 11
\n\u03b7 = Viscosity = 5<\/p>\n

v = (\u03b4\u00b2 – y\u00b2)\u0394P<\/sup> \/ 2\u03b7.L<\/sub>
\nThat is, v = (11\u00b2 – 7\u00b2)22<\/sup> \/ 2(5).(13)<\/sub>
\nv = (121 – 49)22<\/sup> \/ (10).(13)<\/sub>
\nv = (72)22<\/sup> \/ 130<\/sub>
\nSo, v = 1584<\/sup> \/ 130<\/sub>
\nv = 12.18<\/p>\n

Therefore, the velocity of newtonian fluid <\/strong>is 12.18 m\/s.<\/strong><\/p>\n

Calculating the Change in Pressure when the Velocity of Newtonian Fluid, the Distance between Fluid Flow, the Diameter of Pipe, the Separation Distance of Atoms and the Viscosity are Given<\/strong><\/h3>\n

\u0394P = v x 2\u03b7.L<\/sup> \/ (\u03b4\u00b2 – y\u00b2)<\/sub><\/p>\n

Where:<\/p>\n

\u0394P = Change in Pressure
\nv = Velocity of Newtonian Fluid
\nL = Distance between Fluid Flow
\ny = Diameter of Pipe
\n\u03b4 = Separation Distance of Atoms
\n\u03b7 = Viscosity<\/p>\n

Let’s solve an example;
\nFind the change in pressure when the velocity of newtonian fluid is 4, the distance between fluid flow is 2, the diameter of pipe is 3, the separation distance of atoms is 8 and the viscosity is 6.<\/p>\n

This implies that;<\/p>\n

v = Velocity of Newtonian Fluid = 4
\nL = Distance between Fluid Flow = 2
\ny = Diameter of Pipe = 3
\n\u03b4 = Separation Distance of Atoms = 8
\n\u03b7 = Viscosity = 6<\/p>\n

\u0394P = v x 2\u03b7.L<\/sup> \/ (\u03b4\u00b2 – y\u00b2)<\/sub>
\n\u0394P = 4 x 2(6).(2)<\/sup> \/ (8\u00b2 – 3\u00b2)<\/sub>
\nSo, \u0394P = 4 x 24<\/sup> \/ (64 – 9)<\/sub>
\n\u0394P = 96<\/sup> \/ 55<\/sub>
\n\u0394P = 1.745<\/p>\n

Therefore, the change in pressure <\/strong>is 1.745.<\/strong><\/p>\n

Read more:\u00a0How to Calculate and Solve for Viscosity of Newtonian Fluids | Rheology<\/a><\/p>\n

Calculating the Distance between Fluid Flow when the Velocity of Newtonian Fluid, the Change in Pressure, the Diameter of Pipe, the Separation Distance of Atoms and the Viscosity are Given<\/strong><\/h3>\n

L = (\u03b4\u00b2 – y\u00b2)\u0394P<\/sup> \/ v x 2\u03b7<\/sub><\/p>\n

Where:<\/p>\n

L = Distance between Fluid Flow
\nv = Velocity of Newtonian Fluid
\n\u0394P = Change in Pressure
\ny = Diameter of Pipe
\n\u03b4 = Separation Distance of Atoms
\n\u03b7 = Viscosity<\/p>\n

Let’s solve an example;
\nFind the distance between fluid flow when the velocity of newtonian fluid is 10, the change in pressure is 4, the diameter of pipe is 1, the separation distance of atoms is 2 and the viscosity is 7.<\/p>\n

This implies that;<\/p>\n

v = Velocity of Newtonian Fluid = 10
\n\u0394P = Change in Pressure = 4
\ny = Diameter of Pipe = 1
\n\u03b4 = Separation Distance of Atoms = 2
\n\u03b7 = Viscosity = 7<\/p>\n

L = (\u03b4\u00b2 – y\u00b2)\u0394P<\/sup> \/ v x 2\u03b7<\/sub>
\nL = (2\u00b2 – 1\u00b2)4<\/sup> \/ 10 x 2(7)<\/sub>
\nThat is, L = (4 – 1)4<\/sup> \/ 10 x 14<\/sub>
\nL = 12<\/sup> \/ 140<\/sub>
\nL = 0.0857<\/p>\n

Therefore, the distance between fluid flow <\/strong>is 0.0857 m.<\/strong><\/p>\n

Calculating the Diameter of Pipe when the Velocity of Newtonian Fluid, the Change in Pressure, the Distance between Fluid Flow, the Separation Distance of Atoms and the Viscosity are Given<\/strong><\/h3>\n

y = \u221a\u03b4\u00b2 – (v x 2\u03b7.L<\/sup> \/ \u0394P<\/sub>)<\/p>\n

Where:<\/p>\n

y = Diameter of Pipe
\nv = Velocity of Newtonian Fluid
\n\u0394P = Change in Pressure
\nL = Distance between Fluid Flow
\n\u03b4 = Separation Distance of Atoms
\n\u03b7 = Viscosity<\/p>\n

Let’s solve an example;
\nFind the diameter of pipe when the velocity of newtonian fluid is 3, the change in pressure is 5, the distance between fluid flow is 7, the separation distance of atoms is 24 and the viscosity is 11.<\/p>\n

This implies that;<\/p>\n

v = Velocity of Newtonian Fluid = 3
\n\u0394P = Change in Pressure = 5
\nL = Distance between Fluid Flow = 7
\n\u03b4 = Separation Distance of Atoms = 24
\n\u03b7 = Viscosity = 11<\/p>\n

y = \u221a\u03b4\u00b2 – (v x 2\u03b7.L<\/sup> \/ \u0394P<\/sub>)
\nThat is, y = \u221a24\u00b2 – (3 x 2(11).(7)<\/sup> \/ 5<\/sub>)
\ny = \u221a576 – (462<\/sup> \/ 5<\/sub>)
\ny = \u221a576 – 92.4
\nSo, y = \u221a483.6
\ny = 21.99<\/p>\n

Therefore, the diameter of pipe <\/strong>is 21.99 m.<\/strong><\/p>\n

Read more:\u00a0How to Calculate and Solve for Non-Newtonian Fluid Shear Stress | Transport Phenomena<\/a><\/p>\n

Calculating the Separation Distance of Atoms when the Velocity of Newtonian Fluid, the Change in Pressure, the Distance between Fluid Flow, the Diameter of Pipe and the Viscosity are Given<\/strong><\/h4>\n

\u03b4 = \u221a(v x 2\u03b7.L<\/sup> \/ \u0394P<\/sub>) + y\u00b2<\/p>\n

Where:<\/p>\n

\u03b4 = Separation Distance of Atoms
\nv = Velocity of Newtonian Fluid
\n\u0394P = Change in Pressure
\nL = Distance between Fluid Flow
\ny = Diameter of Pipe
\n\u03b7 = Viscosity<\/p>\n

Let’s solve an example;
\nFind the separation distance of atoms when the velocity of newtonian fluid is 12, the change in pressure is 10, the distance between fluid flow is 14, the diameter of pipe is 6 and the viscosity is 8.<\/p>\n

v = Velocity of Newtonian Fluid = 12
\n\u0394P = Change in Pressure = 10
\nL = Distance between Fluid Flow = 14
\ny = Diameter of Pipe = 6
\n\u03b7 = Viscosity = 8<\/p>\n

\u03b4 = \u221a(v x 2\u03b7.L<\/sup> \/ \u0394P<\/sub>) + y\u00b2
\nThen, \u03b4 = \u221a(12 x 2(8).(14)<\/sup> \/ 10<\/sub>) + 6\u00b2
\n\u03b4 = \u221a(2688<\/sup> \/ 10<\/sub>) + 36
\n\u03b4 = \u221a268.8 + 36
\nThat is, \u03b4 = \u221a304.8
\n\u03b4 = 17.45<\/p>\n

Therefore, the separation distance atoms <\/strong>is 17.45.<\/strong><\/p>\n

Calculating the Viscosity when the Velocity of Newtonian Fluid, the Change in Pressure, the Distance between Fluid Flow, the Diameter of Pipe and the Separation Distance of Atoms are Given<\/strong><\/h3>\n

\u03b7 = (\u03b4\u00b2 – y\u00b2)\u0394P<\/sup> \/ v x 2\u03b7.L<\/sub><\/p>\n

Where:<\/p>\n

\u03b7 = Viscosity
\nv = Velocity of Newtonian Fluid
\n\u0394P = Change in Pressure
\nL = Distance between Fluid Flow
\ny = Diameter of Pipe
\n\u03b4 = Separation Distance of Atoms<\/p>\n

Let’s solve an example;
\nFind the viscosity when the velocity of newtonian fluid is 22, the change in pressure is 6, the distance between fluid flow is 10, the diameter of pipe is 3 and the separation distance of atoms is 4.<\/p>\n

This implies that;<\/p>\n

v = Velocity of Newtonian Fluid = 22
\n\u0394P = Change in Pressure = 6
\nL = Distance between Fluid Flow = 10
\ny = Diameter of Pipe = 3
\n\u03b4 = Separation Distance of Atoms = 4<\/p>\n

\u03b7 = (\u03b4\u00b2 – y\u00b2)\u0394P<\/sup> \/ v x 2.L<\/sub>
\nThat is, \u03b7 = (4\u00b2 – 3\u00b2)6<\/sup> \/ 22 x 2.(10)<\/sub>
\n\u03b7 = (16 – 9)6<\/sup> \/ 440<\/sub>
\n\u03b7 = (7)6<\/sup> \/ 440<\/sub>
\nSo, \u03b7 = 42<\/sup> \/ 440<\/sub>
\n\u03b7 = 0.095<\/p>\n

Therefore, the viscosity <\/strong>is 0.095.<\/strong><\/p>\n

Read more:\u00a0How to Calculate and Solve for Viscosity | Non Polar Gases | Transport Phenomena<\/a><\/p>\n

How to Calculate the Velocity of Newtonian Fluid With Nickzom Calculator<\/strong><\/h3>\n

Nickzom Calculator \u2013\u00a0The Calculator Encyclopedia<\/strong> is capable of calculating the velocity of newtonian fluid.<\/p>\n

\n

To get the answer and workings of the velocity of newtonian fluid using the Nickzom Calculator \u2013 The Calculator Encyclopedia.\u00a0<\/strong>First, you need to obtain the app.<\/p>\n

You can get this app via any of these means:<\/p>\n

Web<\/strong>\u00a0\u2013\u00a0https:\/\/www.nickzom.org\/calculator-plus<\/a><\/p>\n

To get access to the\u00a0professional\u00a0<\/strong>version via web, you need to\u00a0register<\/strong>\u00a0and\u00a0subscribe <\/strong>to have utter access to all functionalities.
\nYou can also try the\u00a0demo\u00a0<\/strong>version via\u00a0https:\/\/www.nickzom.org\/calculator<\/a><\/p>\n

Android (Paid)<\/strong>\u00a0\u2013\u00a0https:\/\/play.google.com\/store\/apps\/details?id=org.nickzom.nickzomcalculator<\/a>
\nAndroid (Free)<\/strong>\u00a0\u2013\u00a0https:\/\/play.google.com\/store\/apps\/details?id=com.nickzom.nickzomcalculator<\/a><\/p>\n

Apple (Paid)<\/strong>\u00a0\u2013\u00a0https:\/\/itunes.apple.com\/us\/app\/nickzom-calculator\/id1331162702?mt=8<\/a>
\nOnce, you have obtained the calculator encyclopedia app, proceed to the\u00a0Calculator Map,\u00a0<\/strong>then click on\u00a0Materials and Metallurgical\u00a0<\/b>under\u00a0Engineering<\/strong>.<\/strong><\/p>\n<\/div>\n

<\/p>\n

Now, Click on Transport Phenomena <\/b>under\u00a0Materials and Metallurgical<\/strong><\/p>\n

\"\"<\/p>\n

Now, Click on Velocity of Newtonian Fluid <\/strong>under Transport Phenomena
\n<\/strong><\/p>\n

\"\"<\/p>\n

The screenshot below displays the page or activity to enter your values, to get the answer for the velocity of newtonian fluid according to the respective parameter which is the Change in Pressure (\u0394P), Distance between Fluid Flow (L), Diameter of Pipe (y), Separation Distance of Atoms (\u03b4) and Viscosity (\u03b7).<\/strong><\/p>\n

\"How<\/p>\n

Now, enter the values appropriately and accordingly for the parameters as required by the\u00a0Change in Pressure (\u0394P)<\/strong> is 22<\/strong>, Distance between Fluid Flow (L)<\/strong> is 13<\/strong>, Diameter of Pipe (y)<\/strong> is 7<\/strong>, Separation Distance of Atoms (\u03b4)\u00a0<\/strong>is\u00a011<\/strong> and Viscosity (\u03b7)<\/strong> is 5<\/strong>.<\/p>\n

\"\"<\/p>\n

Finally, Click on Calculate<\/p>\n

\"How
\n\"How<\/p>\n

As you can see from\u00a0the screenshot above,\u00a0Nickzom Calculator<\/strong>\u2013 The Calculator Encyclopedia solves for the velocity of newtonian fluid and presents the formula, workings and steps too.<\/p>\n","protected":false},"excerpt":{"rendered":"The image above represents the velocity of Newtonian fluid. To calculate the velocity of Newtonian fluid, five essential…","protected":false},"author":5,"featured_media":15237,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_opengraph-title":"","_yoast_wpseo_opengraph-description":"","_yoast_wpseo_twitter-title":"","_yoast_wpseo_twitter-description":"","_lmt_disableupdate":"","_lmt_disable":"","_sitemap_exclude":false,"_sitemap_priority":"","_sitemap_frequency":"","_yoast_wpseo_focuskw_text_input":"","csco_display_header_overlay":false,"csco_singular_sidebar":"","csco_page_header_type":"","footnotes":"","_members_access_role":[],"_members_access_error":""},"categories":[47],"tags":[972,1796,1798,60,543,1788,1775,1799,118],"class_list":["post-6269","post","type-post","status-publish","format-standard","has-post-thumbnail","category-engineering","tag-change-in-pressure","tag-diameter-of-pipe","tag-distance-between-fluid-flow","tag-engineering","tag-materials-and-metallurgical","tag-separation-distance-of-atoms","tag-transport-phenomena","tag-velocity-newtonian-fluid","tag-viscosity","cs-entry"],"yoast_head":"\nHow to Calculate and Solve for Velocity of Newtonian Fluid | Transport Phenomena<\/title>\n<meta name=\"description\" content=\"Master all the formulas and steps on How to Calculate Velocity of Newtonian Fluid. 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