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Page 1 Control-Volume Analysis of Mass,Momentum and Energy Control-Volume Analysis of Mass, Momentum and Energy is an important topic of Fluid mechanics which deals with topics such as control mass, control volume, momentum equation, continuity equation and Impact of Jets on planes and vanes. Control Mass • A fixed mass of a fluid element in the flow-field is identified and conservation equations for properties such as momentum, energy or concentration are written. • The identified mass moves around in the flow-field. • Its property corresponds to the same contents of the identified fluid element may change from one location to another. Control Volume • This approach is popular and widely applied in the analysis. • An arbitrarily fixed volume located at a certain place in the flow-field is identified and the conservation equations are written. • The property under consideration or analysis may change with time. The Momentum Equation • It relates the sum of the forces to the acceleration or rate of change of momentum Page 2 Control-Volume Analysis of Mass,Momentum and Energy Control-Volume Analysis of Mass, Momentum and Energy is an important topic of Fluid mechanics which deals with topics such as control mass, control volume, momentum equation, continuity equation and Impact of Jets on planes and vanes. Control Mass • A fixed mass of a fluid element in the flow-field is identified and conservation equations for properties such as momentum, energy or concentration are written. • The identified mass moves around in the flow-field. • Its property corresponds to the same contents of the identified fluid element may change from one location to another. Control Volume • This approach is popular and widely applied in the analysis. • An arbitrarily fixed volume located at a certain place in the flow-field is identified and the conservation equations are written. • The property under consideration or analysis may change with time. The Momentum Equation • It relates the sum of the forces to the acceleration or rate of change of momentum • From conservation of mass, mass into face 1 = mass out of face 2 • The rate at which momentum enters face 1 is P\A\U\U\ =thu\ • The rate at which momentum leaves face 2 is P2A2 ^ 2 ~ ^ 2 • Thus the rate at which momentum changes across the stream tube is P2A2ii2u2 ~ P \A \U \U\ = W M 2 • The Momentum equation is: F = m (u 2 - u j F = Qp(u2 - ux) • This force acts on the fluid in the direction of the flow of the fluid • If the Motion is not one dimensional l , We consider the forces by resolving in the directions of the co-ordinate axes. The force in the x-direction Page 3 Control-Volume Analysis of Mass,Momentum and Energy Control-Volume Analysis of Mass, Momentum and Energy is an important topic of Fluid mechanics which deals with topics such as control mass, control volume, momentum equation, continuity equation and Impact of Jets on planes and vanes. Control Mass • A fixed mass of a fluid element in the flow-field is identified and conservation equations for properties such as momentum, energy or concentration are written. • The identified mass moves around in the flow-field. • Its property corresponds to the same contents of the identified fluid element may change from one location to another. Control Volume • This approach is popular and widely applied in the analysis. • An arbitrarily fixed volume located at a certain place in the flow-field is identified and the conservation equations are written. • The property under consideration or analysis may change with time. The Momentum Equation • It relates the sum of the forces to the acceleration or rate of change of momentum • From conservation of mass, mass into face 1 = mass out of face 2 • The rate at which momentum enters face 1 is P\A\U\U\ =thu\ • The rate at which momentum leaves face 2 is P2A2 ^ 2 ~ ^ 2 • Thus the rate at which momentum changes across the stream tube is P2A2ii2u2 ~ P \A \U \U\ = W M 2 • The Momentum equation is: F = m (u 2 - u j F = Qp(u2 - ux) • This force acts on the fluid in the direction of the flow of the fluid • If the Motion is not one dimensional l , We consider the forces by resolving in the directions of the co-ordinate axes. The force in the x-direction Fx = w(« 2 cos & 2 - « 1 COS^I ) = m{t42 x - u lx ) or Fx = p Q {u j cos Q j - u \ cos 6^) ~ P Q (U 2 x ~ u\x) • And the force in the y-direction: F y = /«{m2 sin # > - wj sin 6^) - m { u 2 y -U ]y) or F y = pQ ( u 2 sin O i - u \ sin 0 \) = p d ^ l y - • The resultant force can be found by combining these components : ^ ^Resultant And the angle of this force: (£= Ian-1 f — KFX) Application of the Momentum Equation Force due the flow around a pipe bend • A converging pipe bend lying in the horizontal plane turning through an angle of 0 As the fluid changes direction a force will act on the bend. Page 4 Control-Volume Analysis of Mass,Momentum and Energy Control-Volume Analysis of Mass, Momentum and Energy is an important topic of Fluid mechanics which deals with topics such as control mass, control volume, momentum equation, continuity equation and Impact of Jets on planes and vanes. Control Mass • A fixed mass of a fluid element in the flow-field is identified and conservation equations for properties such as momentum, energy or concentration are written. • The identified mass moves around in the flow-field. • Its property corresponds to the same contents of the identified fluid element may change from one location to another. Control Volume • This approach is popular and widely applied in the analysis. • An arbitrarily fixed volume located at a certain place in the flow-field is identified and the conservation equations are written. • The property under consideration or analysis may change with time. The Momentum Equation • It relates the sum of the forces to the acceleration or rate of change of momentum • From conservation of mass, mass into face 1 = mass out of face 2 • The rate at which momentum enters face 1 is P\A\U\U\ =thu\ • The rate at which momentum leaves face 2 is P2A2 ^ 2 ~ ^ 2 • Thus the rate at which momentum changes across the stream tube is P2A2ii2u2 ~ P \A \U \U\ = W M 2 • The Momentum equation is: F = m (u 2 - u j F = Qp(u2 - ux) • This force acts on the fluid in the direction of the flow of the fluid • If the Motion is not one dimensional l , We consider the forces by resolving in the directions of the co-ordinate axes. The force in the x-direction Fx = w(« 2 cos & 2 - « 1 COS^I ) = m{t42 x - u lx ) or Fx = p Q {u j cos Q j - u \ cos 6^) ~ P Q (U 2 x ~ u\x) • And the force in the y-direction: F y = /«{m2 sin # > - wj sin 6^) - m { u 2 y -U ]y) or F y = pQ ( u 2 sin O i - u \ sin 0 \) = p d ^ l y - • The resultant force can be found by combining these components : ^ ^Resultant And the angle of this force: (£= Ian-1 f — KFX) Application of the Momentum Equation Force due the flow around a pipe bend • A converging pipe bend lying in the horizontal plane turning through an angle of 0 As the fluid changes direction a force will act on the bend. This force can be very large in the case of water supply pipes. The bend must be held in place to prevent breakage at the joints Taking Control Volume, • In the x-direction: FTx =PQ (U2x - u1 x ) U2X = u2 c o s ^ F Tx = P Q { U2 COS0 — U\ ) • In the y-direction: F Ty = P Q \ U2 y - U\y) U\y = «| sinO = 0 »2y = u2 sin61 F Ty = P Q U2 S'n ^ Using Bernoulli Equations here 2 2 P\ « 1 P2 u 2 # — + — + 2 1 = — + — + Z7 +A/- Pg 2g pg 2g • where hf is the friction loss (this can often be ignored, hf=0) • As the pipe is in the horizontal plane, z-|=Z2 And with continuity, Q= u- iA-i U 2A2 Pi ^ Pi + • p Q1 \____1 _ a2 > 2 \A 2 Ay) • Knowing the pressures at each end the pressure force can be calculated, Fp = pressure force at 1 - pressure force at 2 Fp = p ] Aj cosO- p2 A2 cos0 = Pi Ai - p2 A2 cosO Fpy = p\ A\ sin 0 - p 2 A2 sin 6 = — p 2 A2 sin 0 Page 5 Control-Volume Analysis of Mass,Momentum and Energy Control-Volume Analysis of Mass, Momentum and Energy is an important topic of Fluid mechanics which deals with topics such as control mass, control volume, momentum equation, continuity equation and Impact of Jets on planes and vanes. Control Mass • A fixed mass of a fluid element in the flow-field is identified and conservation equations for properties such as momentum, energy or concentration are written. • The identified mass moves around in the flow-field. • Its property corresponds to the same contents of the identified fluid element may change from one location to another. Control Volume • This approach is popular and widely applied in the analysis. • An arbitrarily fixed volume located at a certain place in the flow-field is identified and the conservation equations are written. • The property under consideration or analysis may change with time. The Momentum Equation • It relates the sum of the forces to the acceleration or rate of change of momentum • From conservation of mass, mass into face 1 = mass out of face 2 • The rate at which momentum enters face 1 is P\A\U\U\ =thu\ • The rate at which momentum leaves face 2 is P2A2 ^ 2 ~ ^ 2 • Thus the rate at which momentum changes across the stream tube is P2A2ii2u2 ~ P \A \U \U\ = W M 2 • The Momentum equation is: F = m (u 2 - u j F = Qp(u2 - ux) • This force acts on the fluid in the direction of the flow of the fluid • If the Motion is not one dimensional l , We consider the forces by resolving in the directions of the co-ordinate axes. The force in the x-direction Fx = w(« 2 cos & 2 - « 1 COS^I ) = m{t42 x - u lx ) or Fx = p Q {u j cos Q j - u \ cos 6^) ~ P Q (U 2 x ~ u\x) • And the force in the y-direction: F y = /«{m2 sin # > - wj sin 6^) - m { u 2 y -U ]y) or F y = pQ ( u 2 sin O i - u \ sin 0 \) = p d ^ l y - • The resultant force can be found by combining these components : ^ ^Resultant And the angle of this force: (£= Ian-1 f — KFX) Application of the Momentum Equation Force due the flow around a pipe bend • A converging pipe bend lying in the horizontal plane turning through an angle of 0 As the fluid changes direction a force will act on the bend. This force can be very large in the case of water supply pipes. The bend must be held in place to prevent breakage at the joints Taking Control Volume, • In the x-direction: FTx =PQ (U2x - u1 x ) U2X = u2 c o s ^ F Tx = P Q { U2 COS0 — U\ ) • In the y-direction: F Ty = P Q \ U2 y - U\y) U\y = «| sinO = 0 »2y = u2 sin61 F Ty = P Q U2 S'n ^ Using Bernoulli Equations here 2 2 P\ « 1 P2 u 2 # — + — + 2 1 = — + — + Z7 +A/- Pg 2g pg 2g • where hf is the friction loss (this can often be ignored, hf=0) • As the pipe is in the horizontal plane, z-|=Z2 And with continuity, Q= u- iA-i U 2A2 Pi ^ Pi + • p Q1 \____1 _ a2 > 2 \A 2 Ay) • Knowing the pressures at each end the pressure force can be calculated, Fp = pressure force at 1 - pressure force at 2 Fp = p ] Aj cosO- p2 A2 cos0 = Pi Ai - p2 A2 cosO Fpy = p\ A\ sin 0 - p 2 A2 sin 6 = — p 2 A2 sin 0 • There are no body forces in the x or y directions, so FR x = FR y = 0 • The body force due to gravity is acting in the z-direction so need not be considered Fp = pressure force at 1 - pressure force at 2 Fp = p^A] cos0 -/>2^2 cos# = p\A\ - P2A2 cos# Fpy = p\ A \ sin 0 - P2A2 sin 9 = — p2 ^2 sin 0 — PQu2 sin 0 + P2 A2 sin 6 And the resultant force on the fluid is given by • And the direction of application is Impact of a Jet on a Plane A jet hitting a flat plate (a plane) at an angle of 90° The reaction force of the plate, i.e. the force the plate will have to apply to stay in the same positionRead More
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