Lecture 5

# Lecture 5 - ater Water esources Resources ngineering...

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Unformatted text preview: ater Water esources Resources ngineering Engineering ecture 5- Lecture 5 Prepared by T. Wagener & P. Reed enn State University 1 Penn State University ontent Content omentum on t try this) 1. Momentum (Don,t try this) 2. Pressure and Pressure Forces in Static luids Fluids • Buoyancy • Hydrostatic Forces 2 http://focus.msn.de/sport/leichtathletik/goeteborg_aid_23462.html?interface=galerie&drucken=1 omentum Momentum order to derive the general momentum In order to derive the general momentum equation for fluid flow in a hydrologic or hydraulic system, we use the control volume pproach long with ewton’s second law 3 approach along with Newton s second law . omentum Momentum What does the following derivative represent? ( ) momentum ∂ t ∂ Soln: 1 1 1 * * * * * * L L t F t F 2 * * * * * * * m m t F t F orce t t t t t = = = = 4 xtensive Property = Momentum = m Extensive Property Momentum m v When applying the control volume approach to fluid momentum, the extensive property is momentum, B = m v (mass times velocity) and the intensive property is e omentum per unit mass = m so that the momentum per unit mass, ß = d ( m v )/ dm , so that B m o m e n t u m d m v ( ) ( ) sys dB d m o m e n t u m d m v dt dt dt = = 5 ewton’s 2 nd aw Newton s 2 Law states that the time rate of change of momentum is … states that the time rate of change of momentum is equal to the net force applied in a given direction, so ( ) ma momentum d F = = ∑ dt Sum of external = Rate of momentum change of system (mass times 6 forces ( acceleration) emember the General Control Volume Equ. Remember the General Control Volume Equ....
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Lecture 5 - ater Water esources Resources ngineering...

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