Question

Question 3 [20 marks] Water (density p1000 kg/m2; dynamic viscosity 0.001 Pa-s) flows steadily through a horizontal, straight pipe with circular cross section of diameter D=0.2 m. The volumetric flow rate is 0.01 m°/s. Argue that this is turbulent flow. [4 marksl а. Pressure drop in the pipe is due to friction. The pressure drop per unit length can be written as Др 4f L with U the average velocity in the pipe and fthe friction factor. Given the pipe wall roughness fcan be considered a constant: f 0.015 b. What is the average energy dissipation rate (symbol E) in the pipe? [4 marks Argue based on a dimensional analysis - that the expression for the Kolmogorov length C. /4 scale (the micro scale of turbulence) reads Cg=(v*/E with /p the kinematic viscosity. Determine based on the average dissipation rate in the pipe. [6 marks] K The shear stress at the inner pipe surface follows from a force balance in the streamwise direction over the liquid in the pipe: D^TDT. The wall shear velocity L relates to the wall shear T/p. We want to simulate the flow in the pipe with help of standard wall stress: functions. For this the distance of the first grid point next to the inner pipe wall needs to satisfy the condition 30<y* <100 d. Given this condition, what is the allowed range of distances of the first grid point from the inner pipe wall? [6 marks]

Answer 1

Given: The density of fluid p 1000 Kg/m3 Dynamic viscosity u0.001 Pa.s wwww wwwwww Circular cross section with diameter D - 0.2 m D2 0.03142 m2 mplies Area of Cross section A = = ww wwv 4 The Volumetric flow rate Øy- 0.01 m3/sec But: Volumetric flow rate is equal to A* V Where V is flow velocity meter per second. Hence flow velocity is given by: 0.01 m3/sec m 0.3183 = = 0.03142 m2 A sec a) Argue that the flow is turbulent? A flow is laminar or turbulent is decided with respect to magnitude of Reynold's mber. wwww If Re < 2000 flow is laminar 2000 Re< 4000: flow is in transition Re 4000 flow is turbulent. The above values are for flow in pipe. But, Re is defined as the relative magnitudes of the inertia and the viscous forces; hence it is a dimensionless quantity. w www 1000 0.3183 0.2 63660 Hence it is given as: RepVD 0.001 As Re> 4000, Flow in the pipe is Turbulent.

1 pu AP b) Pressure drop in the pipe = 4f where 2 D U is average velocity in the pipe. f is friction factor 0.015 By Bernoulli equation: P2U P. gZ2 h + Where subscripts 1 and 2 represent two flow evaluated points in the flow and h represents the energy dissipation of the flow (m2/sec) Therefore h (Pressure drop per unit density +kinetic energy drop potential due to height) Assume the points of consideration are at same elevation. Then above changes to h Pressure drop per unit density kinetic energy drop. Since the term of velocity in the pressure drop equation considered to be an average velocity in the flow over the sections. Assume changes in velocity to be zero. h pressure drop per unit density in the flow from point to point But pressure drop per unit length is given; hence energy dissipation per unit length is given by: U2 AP h 4f 0.0152 J/Kg per unit 4 0.0151 0.31832 2 D 0.2 length Energy dissipation rate is given by e h* average velocity E 0.0152 0.3183 =0.00484 m2/sec3 c) Kolmogorov length (k) = y3/ where y is kinematic viscosity = w/p By dimensional analysis: y is given by dimensions of (dynamic viscosity / density) dimensions of dynamic viscosity - (shear stress length)/velocity (Newton.second)/squre.meter Representing: Length by L Mass by M Time by S We get dimensional quantities for: y as (kg/sec.m)/(kg/m^3) = m2/sec = L2T-1 e as (m2/sec3)= 12T-3 las (m) L Assume l = yaeb yaeb Assume L= (L2T-1)a.(L2t-3)° Implies: 2a+2b 1 and -a -3b 0 Implies a -3b and 2(-3b)+2b 1 Implies b(1/4) and a (3/4) Implies y

3) Given: 4P TDTw where Tw is wall shear force 1 D2 W 4 L wwwwwww Tw where u, is shear wall velocity u wwwwwaoonn Shear stress and velocity are related as: dU T where u is dynamic viscosity and dy wwwwwwvvw du/dy Velocity gradient radially. wwwwwwww wwwww AP Therefore D 4 Tw = L 1 ρυ2 4f 1 D 4 Tw f 2 hence it is not dependent on diameter wwwwwwww pU2 T f 2 dU dy pdy f U2 dU 2u Integrating on both sides we get: 1 P (yx f y100) (: 3U3 2u 2 0.001 yx-100 1000 0.015 3*0.31833 100 1.3782 98.6218 mm it is in between 30 to 100 yx yr is the first grid point.