Fluid Statics Lecture 03: Pressure PDF
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Uploaded by FlatteringChromium403
Rensselaer Polytechnic Institute
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This document presents lecture notes on fluid statics, specifically focusing on pressure, its units, conversions, and variations. It covers concepts like absolute and gage pressure, pressure at a point in a fluid, pressure variation with depth, and pressure in connected fluids, illustrating theory with diagrams and equations.
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# Fluid Statics ## Lecture 03: Pressure **Pressure:** normal force exerted by the fluid per unit area. **Units:** * **SI:** * N/m<sup>2</sup> = Pa = kg m/s<sup>2</sup> = kg/m<sup>2</sup> * Common: MPa, kPa, 10<sup>3</sup> Pa * **USCU:** * 1 kgf/cm<sup>2</sup> * lbf/ft<sup>2</sup...
# Fluid Statics ## Lecture 03: Pressure **Pressure:** normal force exerted by the fluid per unit area. **Units:** * **SI:** * N/m<sup>2</sup> = Pa = kg m/s<sup>2</sup> = kg/m<sup>2</sup> * Common: MPa, kPa, 10<sup>3</sup> Pa * **USCU:** * 1 kgf/cm<sup>2</sup> * lbf/ft<sup>2</sup> = lb ft/s<sup>2</sup> = lb/ft<sup>2</sup> * **1 slug:** mass in USCU * **1 lbf:** 1 lbm x32.2 ft/s<sup>2</sup> * **1 lbf = (32.2 lbm/1 slug) x 1 ft/s<sup>2</sup>** **Pr units:** * psi: lbf/in<sup>2</sup> **Conversions:** * 1 MPa = 10<sup>6</sup> Pa * 1 psi = 1 lbf/sq in * 1 ban = 10<sup>5 </sup> = 0.1 MPa = 100 kPa * 1 atm = 101.325 kPa * 1 kgf/cm<sup>2</sup> = 9.807 N/cm<sup>2</sup> = 9.807 x 10<sup>4</sup> Pa = 0.9807 ban = 0.9679 atm ## Absolute Pressure vs Gage Pressure **Absolute Pressure:** with reference to vacuum **Gage pressure:** with reference to atmospheric pressure * P<sub>abs</sub> = P<sub>g</sub> + P<sub>atm</sub> * 0-123 MPa (abs) * 0.325 MPa (gage) * P<sub>vr</sub> = P<sub>r</sub> - P<sub>atm</sub> * P<sub>atm</sub> = absolute pressure **Pressure at a point** (Pascal Law) * Pressure at a point in the fluid is the same in all directions * No specific direction but only magnitude. * (scalar quantity for stationary fluid) - Gets its direction when acting on a surface ### Proof:  * Consider a differential element in stationary fluid. ∑F<sub>x</sub> = 0 * P<sub>1</sub> △z△y - P<sub>3</sub> △y l sinθ = 0 ∑F<sub>z</sub> = 0 * P<sub>2</sub> △x△y - P<sub>3</sub> △y l cos θ - 1/2 p<sub>g</sub> △x △y △z = 0 * Weight of element Note * △x = l cos θ * △z = l sin θ * 1) = P<sub>1</sub> △y l sin θ - P<sub>3</sub> △y l sin θ = 0 * P<sub>1</sub> - P<sub>3</sub> = 0 * P<sub>1</sub> = P<sub>3</sub> (la) * 2) → P<sub>2</sub> △y l cos θ - P<sub>3</sub> △y l cos θ - 1/2 p<sub>g</sub> l cos θ △y △z = 0 * P<sub>2</sub> - P<sub>3</sub> - 1/2 p<sub>g</sub> △z = 0 * As △z → 0 * P<sub>2</sub> - P<sub>3</sub> = 0 * P<sub>2</sub> = P<sub>3</sub> (2a) * Thus P<sub>1</sub> = P<sub>2</sub> = P<sub>3</sub> ## Pressure At A Point In Stationary Liquid * Pressure at a point in a stationary liquid is the same in all directions. ## Variation of Pressure with Depth * (In gravity) * Consider a rectangular differential element * ∑F<sub>z</sub> = 0 * → P<sub>1</sub> △x △y - P<sub>2</sub> △x △y - p<sub>g</sub> △x △y △z = 0 * △p = P<sub>2</sub> - P<sub>1</sub> = -p<sub>g</sub> △z * Assume △y is the thickness * P, S could be varying functions * As lim △z → 0 * dp/dz = - p<sub>g</sub> ## Pressure In A Connected Fluid * For varying P, g * △p = P<sub>2</sub> - P<sub>1</sub> = - ∫p<sub>g</sub>dz * Pressure in a 'connected fluid' is the same at same height from free surface.  * P<sub>abs</sub> = P<sub>atm</sub> + pgh ## Mechanical Advantage  * F<sub>1</sub> = P A<sub>1</sub> * F<sub>2</sub> = P<sub>2</sub> A<sub>2</sub> * P = F<sub>1</sub> / A<sub>1</sub> = P<sub>2</sub> = F<sub>2</sub> / A<sub>2</sub> * F<sub>2</sub> = A<sub>2</sub>/A<sub>1</sub> F<sub>1</sub> * A<sub>2</sub> > A<sub>1</sub> * Mechanical Advantage ## Barometer * Measures ambient (Atmospheric) pressure  * P<sub>atm</sub> = pgh * h... 760 mm of Hg ## Manometers: Pressure Measurement * Measurement (Gage Pressure)  * P<sub>gas</sub> = P<sub>1</sub> = P<sub>a</sub> = P<sub>atm</sub> + p<sub>g</sub>h * P<sub>acos</sub> negligible * Continuity (same height) * P<sub>mercury</sub> ## Layered Liquids  * P<sub>atm</sub> * a * P<sub>1</sub> h<sub>a</sub> = P<sub>a</sub>g h<sub>a</sub> * b * P<sub>2</sub> = P<sub>1</sub> + P<sub>g</sub> h<sub>b</sub> ## Example Problem  * h<sub>1</sub> = 0.1 m * h<sub>2</sub> = 0.2 m * h<sub>3</sub> = 0.35 m * P<sub>water</sub> = 1000 kg/m<sup>3</sup> * P<sub>oil</sub> = 850 kg/m<sup>3</sup> * P<sub>Hg</sub> = 13,600 kg/m<sup>3</sup> * P<sub>atm</sub>: 85.6 kPa * P<sub>1</sub> + P<sub>w</sub>g h<sub>1</sub> = P<sub>b</sub> * P<sub>b</sub> + P<sub>o</sub>g h<sub>2</sub> = P<sub>c</sub> * P<sub>c</sub> - P<sub>m</sub>g h<sub>3</sub> = P<sub>2</sub> * Find P<sub>1</sub> (absolute pressure) * Sol'n: * P<sub>1</sub> + P<sub>w</sub>g h<sub>1</sub> + P<sub>oil</sub> g h<sub>2</sub> - P<sub>Hg</sub> g h<sub>3</sub> = P<sub>2</sub> ## Apply * △P = P<sub>2</sub> - P<sub>1</sub> = - P<sub>g</sub> h * P<sub>1</sub> = P<sub>2</sub> + P<sub>g</sub> h * P<sub>1</sub> = P<sub>atm</sub> - (P<sub>w</sub>g h<sub>1</sub> + P<sub>oil</sub>g h<sub>2</sub> + P<sub>m</sub> g h<sub>3</sub>) * P<sub>1</sub> = P<sub>atm</sub> - g (P<sub>w</sub>h<sub>1</sub> + P<sub>oil</sub> h<sub>2</sub> - P<sub>m</sub> h<sub>3</sub>) * P<sub>1</sub> = 130 kPa (abs)
