By Gleb L. Kotkin, V. G. Serbo

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2 For the stress field of Fig. 1, change the known data to σxx = 2000 psf, σyy = 3000 psf, and σn(AA) = 2500 psf. Compute σxy and the shear stress on plane AA. 866 ≈ − 289 lbf/ft 2 Ans. (a) Solutions Manual • Fluid Mechanics, Fifth Edition 62 In like manner, solve for the shear stress on plane AA, using our result for σxy: å Ft,AA = τ AA L − (2000 cos30 ° + 289sin 30°)L sin 30° + (289 cos30° + 3000 sin 30°)L cos30° = 0 Solve for τ AA = 938 − 1515 ≈ − 577 lbf/ft 2 Ans. (b) This problem and Prob.

We need to estimate the derivative (du/dy) at the two values of y, then compute τ = µ(du/dy). Method 1: Use a Newton-Raphson three-point derivative estimate. At three equally-spaced points, du/dy|yo ≈ (−3uo + 4u1 − u2 )/(2∆y). 104 kg/m ⋅ s) ≈ 70 Pa Ans. 104 kg/m ⋅ s) ≈ 34 Pa Ans. (b) Method 2: Type the six data points into Excel and run a cubic “trendline” fit. 015 gives du/dy ≈ 314 s , τ ≈ 33 Pa Ans. (a) Ans. 8 A mechanical device, which uses the rotating cylinder of Fig. 6, is the Stormer viscometer [Ref.

872 kJ/kg·K. Raise and lower the pressure 1 kPa at the same entropy. 87565 kg/m3. 86849 kg/m3. 00358 kg/m 3 )} = 529 m/s Ans. (b) Again, as in Prob. 34, the ideal gas approximation is within 2% of a Steam-Table solution. 98 Estimate (a) its speed of sound, and (b) its bulk modulus at 1 atm. Solution: For a crude estimate, we could just take differences of the first two points: ì (500 − 1)(2116) lbf/ft 2 ü ft m a ≈ (∆p/∆ρ ) ≈ í ≈ 3760 ≈ 1150 Ans. 87E7 2 ≈ 895 MPa Ans. (b) ft For more accuracy, we could fit the data to the nonlinear equation of state for liquids, Eq.

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