Problems_II_5 - Problem 5.1 Apply[Clear,Names"Global`" Off[General:spell Off[General:spell1"SAE 40 at 160°F" T = 160 °F ρ =

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Unformatted text preview: (* Problem 5.1 *) Apply[Clear,Names["Global`*"]]; Off[General::spell]; Off[General::spell1]; "SAE 40 at 160°F" T = 160; (* °F *) ρ = 0.89 - 0.00035 (T - 60); Print["density ρ = ", ρ , " g/cm^3"]; SAE 40 at 160°F density ρ = 0.855 g ê cm^3 problem_II_5_1.nb 1 (* Problem 5.3 *) Apply[Clear,Names["Global`*"]]; Off[General::spell]; Off[General::spell1]; T = 93; (* °C *) t = 50; (* sec *) ρ = 0.89 - 0.00063 (T - 15.6); Print["density ρ = ", ρ , " g/cm^3"]; µ = (0.22 t - 180/t) ρ ; Print[" µ = ", µ , " mPa s"]; "from Figure II.5.3 => SAE 20" density ρ = 0.841238 g ê cm^3 µ = 6.22516 mPa s from Figure II.5.3 => SAE 20 problem_II_5_3.nb 1 (* Problem 5.5 *) Apply[Clear,Names["Global`*"]]; Off[General::spell]; Off[General::spell1]; "SAE 10"; T = 130; (* °F *) Diam = 4; (* in *); R = Diam/2; (* in *); c = 0.0015; (* in *) n = 2000; (* rpm *) L = 6; (* in *) m = 2.3*10^-6; (* lb sec / in^2 - from Fig.5.4*) Tf = (4 Pi^2 m n/60 L R^3)/c; Print["Friction torque = ", Tf, " lb in = ", Tf/12, " lb ft"]; H = (Tf/12 n)/5252; Print["Power loss = ", H, " hp"]; Friction torque = 96.8537 lb in = 8.07114 lb ft Power loss = 3.07355 hp problem_II_5_5.nb 1 (* Problem 5.7 *) Apply[Clear,Names["Global`*"]]; Off[General::spell]; Off[General::spell1]; "SAE 20"; T = 70; (* °C *) Diam = 200*10^-3; (* m *); R = Diam/2; (* m *); L = 100*10^-3; (* m *) c = 0.1*10^-3; (* m *) n = 1000/60; (* rps *) W = 20*10^3; (* N *) m = 12.5*10^-3; (* Pa sec - from Fig.5.3*) P = W / (Diam*L); S = ( m n / P) (R/c)^2; Print["Sommerfeld number = ", S]; Print["L/D = ", L/Diam]; "from Fig 5.9 h0/c = 0.32" h0 = 0.32*c; Print["minimum film thickness h0 = ", h0]; "from Fig 5.11 (R/c)f = 5.5" f = 5.5*c/R; Print["bearing coefficient of friction f = ", f]; "from Fig 5.14 P/pmax = 0.32" pmax = P/0.32; Print["maximum pressure within oil film pmax = ", pmax, " psi"]; "from Fig 5.10 the position angle of minimum film thickness f = 42°" "from Fig 5.15 terminating position of the oil film q p0 = 56°" "from Fig 5.15 angular position angle of maximum pressure q pmax = 17°" "from Fig 5.12 Q/RcnL = 5.2" Q = R c n L 5.2; Print["total flow Q = ", Q, " m^3/s"]; "from Fig 5.13 Qs/Q = 0.8. Of the volume of oil Q pumped by the rotating journal, an amount Qs flows out the ends. The side leakage that must be made up by the oil represents 80% of the flow. The remaining 20% of the flow is recirculated." Sommerfeld number = 0.208333 L ê D = 1 ÅÅÅÅ 2 from Fig 5.9 h0 ê c = 0.32 minimum film thickness h0 = 0.000032 from Fig 5.11 H R ê c L f = 5.5 bearing coefficient of friction f = 0.0055 from Fig 5.14 P ê pmax = 0.32 maximum pressure within oil film pmax = 3.125 μ 10 6 psi problem_II_5_7.nb 1 from Fig 5.10 the position angle of minimum film thickness f = 42° from Fig 5.15 terminating position of the oil film q p0 = 56° from Fig 5.15 angular position angle of maximum pressure q pmax = 17° from Fig 5.12 Q ê RcnL = 5.2 total flow Q = 0.0000866667 m^3 ê s from Fig 5.13 Qs ê Q = 0.8. Of the volume of oil Q pumped by the rotating journal, an0....
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This note was uploaded on 08/29/2011 for the course MECH 3230 taught by Professor Staff during the Spring '08 term at Auburn University.

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Problems_II_5 - Problem 5.1 Apply[Clear,Names"Global`" Off[General:spell Off[General:spell1"SAE 40 at 160°F" T = 160 °F ρ =

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