A. Introduction & Definitions 1. Density: mass per unit volume  = M/V; units = kgm -3 (XIII.A.1) e.g.,  H 2 O) = 1000 kgm -3 ;  Fe) ~ 8000 kgm Pressure: Perpendicular Force per unit area P = F perp /A; units = Nm -2 = Jm -3 = Pa(XIII.A.2) 3.“Gauge Pressure” XIII. Fluid Mechanics

1.Pressure & gravity in uniform fluids Assuming equilibrium, then the force due to pressure must be equal to the force due to gravity: “Hydrostatic Equilibrium” dh (p+dp)A -(p)A dW =  gdV =  gAdh -pA + (p + dp)A –  gAdh = 0; dp/dh =  g (XIII.B.1) XIII. Fluid Mechanics B. Hydrostatic Equilibrium h

1.Pressure & gravity in uniform fluids Assuming equilibrium, then the force due to pressure must be equal to the force due to gravity: “Hydrostatic Equilibrium” dh (p+dp)A -(p)A dW =  gdV =  gAdh XIII. Fluid Mechanics B. Hydrostatic Equilibrium h For uniform density, ∫ dp =  g ∫ dh p = p 0 +  gh(XIII.B.2)

2.Pascal’s Law Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls containing the fluid. Consider two connected cylinders with different areas A1 and A2: P = F1/A1 = F2/A2 => F2 = (A2/A1)F1.(B.3) Pascal’s law demonstrates how force can be multiplied, e.g., hydraulics. XIII. Fluid Mechanics B. Hydrostatic Equilibrium

1. Archimedes Principle: When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. a)Consider raising a marble statue from the sea floor with V = 5 m 3 and  ~ 3000 kgm -3. What is the tension in the rope? Assume sea water. XIII. Fluid Mechanics C. Buoyancy TFBFB W  F = 0 = T + F B – W; T = W - F B.

1. Archimedes Principle: When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. m =  m V = 15,000 kg. |W| = mg ~ 150,000 N. |W SW |=  sw Vg = 50,000 N = |F B |. T = W - F B = 100,000 N; |T/W| ~ 67% XIII. Fluid Mechanics C. Buoyancy TFBFB W

b)In terms of densities: T = W - F B =  m Vg –  sw Vg; T = m statue g(1-  sw /  m ).(XIII.C.1) XIII. Fluid Mechanics C. Buoyancy TFBFB W

1. BE: Relates pressure, flow speed and height for an ideal fluid. 2. Derivation: Apply work-energy theorem to a section of fluid in a flow tube. dW = Fds. ds 1 = v 1 dt ds 2 = v 2 dt V= Ads. dW= (p 1 - p 2 )dV. (D.1) y1y1 y2y2 F1F1 F2F2 a) Work XIII. Fluid Mechanics D. Bernoulli’s Eqn.

1. BE: Relates pressure, flow speed and height for an ideal fluid. 2. Derivation: Apply work-energy theorem to a section of fluid in a flow tube. dK = d(1/2mv 2 ) ds 1 = v 1 dt ds 2 = v 2 dt = 1/2  dV(v v 1 2 ). (D.2) y1y1 y2y2 F1F1 F2F2 b) Kinetic Energy XIII. Fluid Mechanics D. Bernoulli’s Eqn.

1. BE: Relates pressure, flow speed and height for an ideal fluid. 2. Derivation: Apply work-energy theorem to a section of fluid in a flow tube. dU = d(mgh) ds 1 = v 1 dt ds 2 = v 2 dt =  dVg(y 2 - y 1 ) (D.3) y1y1 y2y2 F1F1 F2F2 c) Gravitational Potential Energy XIII. Fluid Mechanics D. Bernoulli’s Eqn.

1. BE: Relates pressure, flow speed and height for an ideal fluid. 2. Derivation: Apply work-energy theorem to a section of fluid in a flow tube. WET: dW = dK + dU =>  p = 1/2  dV(v v 1 2 ) +  dVg(y 2 - y 1 ), or(D.4) p +  gy + 1/2  v 2 = constant. (D.5) XIII. Fluid Mechanics D. Bernoulli’s Eqn.

Suppose that air streams past the wings such that the speed is 100 m/s over the top surface and 90 m/s past the bottom surface. If the plane has a mass of 1500 kg, and the wings are 2 m wide, what is the wingspan required for lift? Note: Density of air ~ 1 kgm -3. p +  gy + 1/2  v 2 = constant. Note:  gy = constant. p t + 1/2  v t 2 = p b + 1/2  v b 2 ;  p = 1/2  v b 2 - v t 2 ) ~ Pa. (pressure below is higher than pressure above) Example Problem #12

3.Example: Lift. Suppose that air streams past the wings such that the speed is 100 m/s over the top surface and 90 m/s past the bottom surface. If the plane has a mass of 1500 kg, and the wings are 2 m wide, what is the wingspan required for lift? Minimum lift is L = mg ~ 15,000 N. L = pA = p(Width)(Wingspan) = (950 Pa)(2 m)(WS); WS = 15,000 N/(1900 Pa-m) ~ 8 m. Example Problem #12