IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Background 1. Energy conservation equation If there is no friction

IIT-Madras, Momentum Transfer: July 2005-Dec If there is frictional loss, then In many cases Background

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Q. Where are all frictional loss can occur ? in pipe, in valves, joints etc First focus on pipe friction In pipe, Can we relate the friction to other properties ? Background

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Example for general case: At the normal operating condition given following data 0 gauge pressure Example What should be the pressure at inlet ?

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Solution : taking pressure balance Example (continued) For pipe, Force balance Hence we can find total pressure drop

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 We have said nothing about fluid flow properties However, Normally we do not know the Usually they depend on flow properties and fluid properties Flow properties Empirical

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 In general we want to find  f is a measure of frictional loss higher f implies higher friction This is Fanning-Friction factor f f Friction Factor: Definition

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 So we write Friction factor This is for pipe with circular cross section

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Here f is function of other parameters For laminar flow, don’t worry about f, just use For turbulent flow, Is it possible to get expression for shear ? Friction factor: Turbulent Flow Using log profile

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Because original equation Equation for Friction Factor

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 This is equivalent of laminar flow equation relating f and Re (for turbulent flow in a smooth pipe) Equation for Friction Factor

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Friction Factor: Laminar Flow For laminar flow

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Use of f is for finding effective shear stress and corresponding “head loss” or “ pressure drop” In the original problem, instead of saying “normal operating condition” we say Pressure drop using Friction Factor Laminar or turbulent?

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 For turbulent flow We can solve for f, once you know f, we can get shear Pressure drop using Friction Factor Once you know shear, we can get pressure drop If flow is laminar, ( i.e. Re < 2300 ), we use

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 And original equation becomes, In above equation the value of f can be substitute from laminar and turbulent equation Laminar flow – straight forward Turbulent flow – iterative or we can use graph Pressure drop using Friction Factor 0 gauge pressure

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Determination of Q or D Given a pipe (system) with known D and a specified flow rate (Q ~ V), we can calculate the pressure needed i.e. is the pumping requirement We have a pump: Given that we have a pipe (of dia D), what is flow rate that we can get? OR We have a pump: Given that we need certain flow rate, of what size pipe should we use?

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Determination of Q or D We have a pump: Given that we have a pipe (of dia D), what is flow rate that we can get? To find Q i.e. To find average velocity (since we know D) Two methods: (i) Assume a friction factor value and iterate (ii) plot Re vs (Re 2 f) Method (i) Assume a value for friction factor Calculate V av from the formula relating  P and f Calculate Re Using the graph of f vs Re (or solving equation), re-estimate f; repeat

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Determination of Q or D Method (ii) From the plot of f vs Re, plot Re vs (Re 2 f) From the known parameters, calculate Re 2 f From the plot of Re vs (Re 2 f), determine Re Calculate V av

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 We take original example, assume we know p, and need to find V and Q Let us say Iteration 1: assume f = gives V = 1.73m/s, Re = 3.5x10 5, f = Iteration 2: take f = gives V = 1.15m/s, Re = 2.1x10 5, f = Iteration 3: take f = gives V = 1.04 m/s, Re = 2.07x10 5, f =

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 If flow is laminar, you can actually solve the equation

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 If you are given pressure drop and Q, we need to find D

IIT-Madras, Momentum Transfer: July 2005-Dec 2005 Iteration 1: Assume f = 0.01 Iteration 2: take f = and follow the same procedure Solving this approximately (how?), we get