Question

3. For this horizontal T through which oil (SG =0.85) is flowing, what external force in the x-y plane (through the bolts or other supporting devices) is needed to hold the T in place? The following data are given: Q1 = 0.2m/s, Q2 = 0.1m/s, P1 = 99kPa, P2 = 83kPa, P3 = 468kPa, D1 = 10cm, D2 = 7cm, D3 = 10cm

Answer 1

This is the case of Application of momentum Equation - Net et Farro Force on Body = Rate of change of lineast Momentum Calculation of Fareo in se Direction - Direction in otts Force on Fluid in a the H Divoction F Not External Fareo on Fluid - P AR-P2 Az & Fe fate of change of linear momentom= & OzV₂ - SO, V, Vi = A = t (0)2 = 0.007854m2 A2 = I (0.07J2 = 0.003848 m2 Az = q (0:1)2 = 0.007854m2 Vi= 0 = 25.465 mis | V2 = 02 = 25.987 m/s Oi = 0₂ +03) Oz= 0.2-0.1 = 0.1 m315 V = 0 3 = 12-732m15 Pian - Pz Az + Fx = 9gX0.007854 - +Fx 468 Xo.007854 - 2298.13 303 V3-80. VE 850 0.18 12.732 -0.8%25.465) = -3246-03-0 forom os -2894-13 + Fu=-3246.83 1 Fe = a -348 74 Answer Forco on Fluid in re n oloo etion Fe = 3 Fare on supporting devices in the se disection = 8. AN EN 3447N

calculation of Force on y Direction. Let's force on Fluid in Downwoorld Direction Fy ( Not External Force = fate of change of Linoor momentum P2 A₂ -o + Fy = 10242 - 0 Fy = 85070.1X25.899 – 8341000 X.003848 Fy = 1881.86 M (Downward) Answer hence Force on Support Fy= 1881.86 Cupwards) Fy Resultent 99999922 Lo Fu Not Resultent Farce on suppor Fart Fy2 FR = Fort Fyz FR = 1913.89M Ans tond = Fg Fr 10 = 79.50° Angle from the de Diovetion Anneer
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