Fluid Mechanics 2 Notes (2024) PDF

APPLICATIONS OF LINEAR MOMENTUM BALANCE IN INTEGRAL FORM Acquire the concepts of simple flow analysis for simple flow problems and its application to determine forces associated in a flow system and for balance. FLOW IN A REDUCING PIPE BENDS PROBLEM 1 Problem: Determine the force exerted on a p...

APPLICATIONS OF LINEAR MOMENTUM BALANCE IN INTEGRAL FORM Acquire the concepts of simple flow analysis for simple flow problems and its application to determine forces associated in a flow system and for balance. FLOW IN A REDUCING PIPE BENDS PROBLEM 1 Problem: Determine the force exerted on a pipe associated to a steady fluid flow in a reducing pipe bend as Inlet $" shown in the figure. 1 !# !! Control volume Outlet 2 !" FLOW IN A REDUCING Y 1 PIPE BENDS PROBLEM 1 > u Pipe wall Problem: Inlet Control volume Determine the force exerted on a pipe associated to a steady fluid Pipe wall flow in a reducing pipe bend as Inlet $" shown in the figure. Outlet 1 !# !! Outlet 2 !"  FLOW IN A REDUCING PIPE BENDS PROBLEM 1 Linear momentum equation General form of linear momentum balance: 1 Net force acting on c.v. , " # = % &! ! ( )* + % !&). ,- # $ 2 Net outflow of momentum from Assumptions c.v. Rate of accumulation 1. Steady state m , p , of momentum within c.v. 2. Incompressible flow/fluid P 3. Uniform/constant velocities at inlet V and outlet FLOW IN A REDUCING PIPE BENDS PROBLEM 1 Flow in %-direction and &-direction 0̇ % = 0̇ & = 0̇ &% = && = & 2&,( < Flow only in 1 %-direction 1 2&,) Y !" V !! 2 Outlet Inlet Velocities with both Velocity with only x-component and y-component x-component Analyze the properties and flow Cross section area, *& Cross section area, *% components: 0̇ = &*&!& 0̇ = &*%!% FLOW IN A REDUCING The two-dimensional problem can be solved by separating PIPE BENDS PROBLEM 1 them into their respective components: @ - 3 ? Write the expressions for momentum at - 1 inlet and outlet. !" !&,( !! = !!,* 14 !&,) 1 3 − direction 6 − direction 2 Inlet ̇ %,( (−) 0! 0 Outlet 0! ̇ &,) cos 1 0(! ̇ &,) sin 1) (−) FLOW IN A REDUCING PIPE BENDS PROBLEM 1 Determine forces involved and their expression. C&,( 1 1 B! C&,) B" D 2 Force 3 − direction 6 − direction Inlet #+ #+%,( = B%,( *% 0 Forces Outlet #+ #+&,( = −B&,( *& cos 1 #+&,) = B&,) *& sin 1 Pressure difference, #+ Pressure and shear Weight #, 0 −D= Gravitational force, #, stress by the pipe wall. mg Pipe A A( A) Unknown force, A  FLOW IN A REDUCING PIPE BENDS PROBLEM 1 Write the sum of forces/net force and the net momentum Only at outflow based on their respective components: the outlet % − direction 1 & − direction ↓ Net force: Net force: " #( = C%,( *% − C&,( *& cos 1 + E( " #) = C&,) *& sin 1 − D + E) 2 Net momentum outflow: ↳ Net momentum outflow: % &!( ! ( )* = ∑ 0̇ - !-./ + ∑(0̇ - !- )012 % &!) ! ( )* = 0(−! ̇ & sin 1) # # = 0! ̇ &,( cos 1 − 0! ̇ %,( = &*&!& −!& sin 1 = &*&!&,( !&,( cos 1 − &*%!%,( !%,( = −&*&2&& sin 1 = &*&2&& cos 1 − &*%2%& FLOW IN A REDUCING % − direction PIPE BENDS PROBLEM 1 B%*% − B&*& cos 1 + E( = &*&!&& cos 1 − &*%!%& 1 E( = &*&!&& cos 1 − &*%!%& −B% *% + B&*& cos 1 ()*+,-./ 1.2+) : Equal magnitude but acting on the opposite direction I( = −E( 2 I( = &*%!%& − &*&!&& cos 1 +B% *% − B&*& cos 1 Write the linear momentum balance for the respective & − direction components–substitute into the general form. B&*& sin 1 − G + E) = −&*&!&& sin 1 Take note that A is the force acted on the fluid (our c.v.) E) = −&*&!&& sin 1 − B&*& sin 1 + G by pipe while the question asked for the pressure acted on the pipe by fluid. Let H to be the force acted on pipe ()*+,-./ 1.2+) : by water. I) = −E) H = I( J + I) K I) = &*&!&& sin 1 + B&*& sin 1 − G FORCE BALANCE ON FLUID FLOW PROBLEM 2 A jet of fluid exits a nozzle and strikes a vertical plane surface as shown in the figure. Determine the force, F required to hold the plate stationary if *782 = 0.005 0& the jet is composed of: i. Water, &3 = 1000 MN/04 #.782 = 12 0/T ii. Air, &5.6 = 1.206 MN/04 " @ !? FORCE BALANCE ON FLUID FLOW PROBLEM 2 4./,2.5 6.578) Define the control volume for the fluid system: U. 2. To determine the force to keep the plane at stationary, to perform force balance on the plane, the force by jet on the *782 = 0.005 0& plane should be determined. Hence, the control volume must cut/include the area where the jet is in contact with plane. # Assumptions.782 = 12 0/T 1. Steady state " @ 2. Incompressible 3. Uniform/constant velocities !? at inlet and outlet FORCE BALANCE ON FLUID FLOW PROBLEM 2 9.2+) :*5*/+) on % − ;-2)+,-./ n U. 2. To keep the plate I stationary, the forces on the plane must be equally balanced: *782 = 0.005 0& Only % − direction is involved. ^WXU_ b@ ℎWXYZW[-\] ^WXU_ X_`aYX_) -ℎ_ d_- ^]We = -W b_ \cc]Y_) W[ -ℎ_ c]\[_ # #782,( = −#.782 = 12 0/T Acting to the left – opposite to the direction of #7 " @ !? # + #782,( = 0 #!"#,% FORCE BALANCE ON FLUID FLOW PROBLEM 2 Wa- (6 − )YX_U-YW[) Linear momentum balance in ? − )YX_U-YW[: ?@$ , 0 Y[ (3 − )YX_U-YW[) " @ , " # = % &! ! ( )* + % !&). ,- # $ !? Wa- (6 − )YX_U-YW[) #782,( = " 0! ̇ 012 − 0! ̇./ C.,)D = 0 − 0! ̇./ The control volume/jet is exposed to atmosphere = −0! ̇ 782 hence there no force due to pressure gage different. The force by atmospheric pressure is cancelled on the left and right side. FORCE BALANCE ON FLUID FLOW PROBLEM 2 9.2+) ,. E) ) F F5*/) D,*,-./*2& #782,( = −0! ̇ 782 For water jet: = −(&*782 !782 )!782 # = −(&3 *782 !782 )!782 Force from jet/control volume to plane (reaction force): " # = −720 g @ #782,( = (&*782 !782 )!782 !? For air jet: From force balance: # = −(&5.6 *782 !782 )!782 #9:;,* # # + #782,( = 0 # = −0.868g # = −#782,( # = −(&*782 !782 )!782