QUESTION 4Given ; ID = 0.02 m = 0.2 kg/se = 0.000045 m = 1.0 Pa.s = 1.0 N.s / = 3.0 Pa.s = 3.0 N.s / = 60 kg / = 1000 kg / Weight fraction = 0.149 = ?

The cross sectional area of the pipe ;A = = = 3.1416

The mass flowrate ; u = = = 10.6103 m/s = = 0.6366 m/s

The pipe roughness ; = = 0.00225

The Reynolds number ; = = = 1273236 (Turbulent flow, Re 4000) = = = 4244 (Turbulent flow, Re 4000)

From Figure 3.7,the pipe friction factor ; = 0.002875 = 0.005

The pressure drop (per unit length) ; = 4 (4)(0.002875)((60)(10.6103) = 366.0554 (N/(4)(0.005)((1000)(0.6366) = 636.6 (N/

The hold-up in terms of parameter X ;X = = = 1.3187

From Figure 5.4, turbulent-turbulent flow is used ; = 5.00 = 4.00Since X1, is used; =

The pressure gradient in the pipe (per unit length) ; = ( = (636.6) = 10185.6 (N/ = 10.1856 (kN/

QUESTION 2Given ; = 0.05 m = 0.1 m = 1000 kg/ Q = 2 / sSituation: Sudden expansion/ sudden enlargement = ? = ?

The cross-sectional area of the pipe ; A = = = 1.9635 = = 7.8540

The velocity of the liquid ; Q = uAu = = = = 1.0186 m/s= = = 0.2546 m/s

The head loss due to friction ; = = = 0.02975 m water

The pressure drop due to friction ; = = (1000) = 291.848 N/