[amp_mcq option1=”$${\\text{D}}_1^2 – {\\text{D}}_2^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_1} – {{\\text{V}}_2}} \\right)$$” option2=”$${\\text{D}}_2^2 – {\\text{D}}_1^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_1} – {{\\text{V}}_2}} \\right)$$” option3=”$${\\text{D}}_1^2 – {\\text{D}}_2^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_2} – {{\\text{V}}_1}} \\right)$$” option4=”$${\\text{D}}_1^2 + {\\text{D}}_2^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_2} – {{\\text{V}}_1}} \\right)$$” correct=”option1″]<\/p>\n
The correct answer is: $${\\text{D}}_1^2 – {\\text{D}}_2^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_1} – {{\\text{V}}_2}} \\right)$$<\/p>\n
The fundamental momentum equation for a hydraulic jump is a relationship between the depth of water upstream and downstream of the jump, the velocity of water upstream and downstream of the jump, and the discharge of water over the jump. The equation can be derived from the principle of conservation of momentum.<\/p>\n
The equation states that the change in the kinetic energy of the water is equal to the work done by the pressure forces on the water. The change in the kinetic energy of the water is given by:<\/p>\n
$$\\Delta K = \\frac{1}{2} \\left( {\\text{D}}_1^2{{\\text{V}}_1^2 – {\\text{D}}_2^2{{\\text{V}}_2^2} \\right)$$<\/p>\n
The work done by the pressure forces on the water is given by:<\/p>\n
$$W = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_1} – {{\\text{V}}_2}} \\right)$$<\/p>\n
where ${\\text{D}}_1$ and ${\\text{D}}_2$ are the depths of water upstream and downstream of the jump, respectively, ${\\text{V}}_1$ and ${\\text{V}}_2$ are the velocities of water upstream and downstream of the jump, respectively, and ${\\text{q}}$ is the discharge of water over the jump.<\/p>\n
Substituting the expression for the work done by the pressure forces on the water into the expression for the change in the kinetic energy of the water, we get:<\/p>\n
$$\\frac{1}{2} \\left( {\\text{D}}_1^2{{\\text{V}}_1^2 – {\\text{D}}_2^2{{\\text{V}}_2^2} \\right) = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_1} – {{\\text{V}}_2}} \\right)$$<\/p>\n
Simplifying the equation, we get:<\/p>\n
$${\\text{D}}_1^2 – {\\text{D}}_2^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_1} – {{\\text{V}}_2}} \\right)$$<\/p>\n
This is the fundamental momentum equation for a hydraulic jump.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$${\\text{D}}_1^2 – {\\text{D}}_2^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_1} – {{\\text{V}}_2}} \\right)$$” option2=”$${\\text{D}}_2^2 – {\\text{D}}_1^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_1} – {{\\text{V}}_2}} \\right)$$” option3=”$${\\text{D}}_1^2 – {\\text{D}}_2^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_2} – {{\\text{V}}_1}} \\right)$$” option4=”$${\\text{D}}_1^2 + {\\text{D}}_2^2 = \\frac{{2{\\text{q}}}}{{\\text{g}}}\\left( {{{\\text{V}}_2} – {{\\text{V}}_1}} \\right)$$” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[649],"tags":[],"class_list":["post-7559","post","type-post","status-publish","format-standard","hentry","category-irrigation-engineering","no-featured-image-padding"],"yoast_head":"\n