Express these readings in gage or vacuum pressure, whichever is appropriate. standard day, pressure gage A reads 8 kpa and gage B reads 105 kpa. To ( n 1) B.5 Denver, Colorado, has an average altitude of 500 ft.
If the starting point is not at z = 0, then replace z by δz: n 1 Bδ z To << 1, or : δ z << Ans. Then the series may be rewritten as follows: n 1 Bz p = pa ρa gz( ) T oģ Chapter Pressure Distribution in a Fluid 7 For the linear law to be accurate, the nd term in parentheses must be much less than unity. (.0), and note that p a nb/t o = (p a /RT o )gz = ρ a gz. where n = g RB Multiply by p a, as in Eq. (.0) can be expanded into a series: Bz (1 ) T o n = 1 n Bz T o + n( n 1) Bz ( )! T o. (.0), into a power series and show that the linear approximation p p a - ρ a g z is adequate when To g δ z < <, where n = ( n 1) B RB Solution: The power-law term in Eq. P.4 For gases over large changes in height, the linear approximation, Eq.
The applied pressure is estimated to be p = γhpress = (9790 N/m )(0. The capillary rise in the tube, from Example 1.9 of the text, is h cap Ycosθ (0.07 Nm / )cos(0 ) = 0.00 m γ R (9790 Nm / )( m) Then the rise due to applied pressure is less by that amount: hpress = 0.5 m 0.0 m = 0. Solution: For water, let Y = 0.07 N/m, contact angle θ = 0, and γ = 9790 N/m. After correcting for surface tension, estimate the applied pressure in Pa. When a pressure is applied, water at 0 C rises into the tube to a height of 5 cm. A vertical clean glass piezometer tube has an inside diameter of 1 mm. (b) AA This problem and Prob.1 can also be solved using Mohr s circle. (a) xy In like manner, solve for the shear stress on plane AA, using our result for σxy: F = τ L (000cos0 + 89sin0 )Lsin0 t,aa AA + (89cos sin0 )L cos0 = 0 Solve for τ = lbf/ft Ans. Solution: Sum forces normal to and tangential to AA in the element freebody above, with σn(aa) known and σxy unknown: Fn,AA = 500L ( σ xy cos sin 0 )L sin 0 ( σ sin cos0 )L cos0 = 0 xyĢ 7 Solutions Manual Fluid Mechanics, Fifth Edition Solve for σ = ( )/ lbf/ft Ans. Compute σxy and the shear stress on plane AA. P.1, change the known data to σxx = 000 psf, σyy = 000 psf, and σn(aa) = 500 psf. P.1 Solve for t,aa σ AA AA 68 lbf/ft Ans. Now sum forces normal and tangential to side AA. Solution: Make cut AA so that it just hits the bottom right corner of the element. P.1, let σ xx = 000 psf σ = 000 psf xy yy σ = 500 psf Find the shear and normal stresses on plane AA cutting through at 0. 1 Chapter Pressure Distribution in a Fluid.1 For the two-dimensional stress field in Fig.
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