Euler’s Equation of Motion
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Euler’s Equation Of Motion For Inviscid Flow
1) A sphere of radius R, whose centre is at rest, vibrates radially in an infinte incompressible fluid of density ρ, which is at rest at infinity. If the pressure at infinity is Π, show that the pressure at the surface of the sphere at time t is
Π+12ρ{d2R2dt2+(dRdt)2}.
[2019, 15M]
Here the motion of the fluid will take place in such a manner so that each element of the fluid moves towards the centre. Hence the free surface would be spherical. Thus the fluid velocity v′ will be radial and hence v′ will be function of r′ (the radial distance from the centre of the sphere which is taken as origin), and time t only. Let p be pressure at a distance r′.
Let P be the pressure on the surface of the sphere of radius R and V be the velocity there. Then the equation of continuity is:
r′2v′=R2V=F(t)…(1) From (1) ∂v′∂t=F′(t)r′2…(2) Again equation of motion is: ∂v′∂t+v′∂v′∂r′=−1ρ∂p∂r′ ⟹ −F′(t)r′2+∂∂r′(12v′2)=−1ρ∂p∂r′, (using (2)) … (3)
Integrating with respect to r′, (3) reduces to: −F′(t)r′+12v′2=−pρ+C, C being an arbitrary constant When r′=∞, then v′=0 and p=Π so that C=Π/ρ. Then, we get −F′(t)r′+12v′2=Π−pρ ⟹p=Π+12ρ[2F′(t)r′−v′2]…(4) But p=P and v′=V when r′=R. Hence (4) gives:
P=Π+12ρ[2R{F′(t)}r′=R−V2]…(5) Also V=dR/dt. Hence using (1), we have
{F′(t)}r′=R=ddt(R2V)=ddt(R2dRdt)=ddt(R2⋅dR2dt)=R2d2R2dt2+12dR2dtdRdt=R2d2R2dt2+R(dRdt)2Using the above values of V and {F′(t)}r′=R,(5) reduces to
P=Π+12ρ[2R{R2d2R2dt2+R(dRdt)2}−(dRdt)2]=Π+12ρ[d2R2dt2+(dRdt)2]Hence, proved.
2) A stream is rushing form a boiler through a conical pipe, the diameters of the ends of which are D and d. If V and v be the corresponding velocities of the stream and if the motion is assumed to steady and diverging from the vertex of the cone, then prove that vV=D2d2e(u2−v2)/2K, where K is the pressure divided by the density and is constant.
[2017, 15M]
Let AB and A′B′ be the ends of the conical pipe such that A′B′=d and AB=D. Let ρ1 and ρ2 be densities of the stream at A′B′ and AB. By principle of conservation of mass, the mass of the stream that enters the end AB and leaves at the end A′B′ must be the same. Hence the equation of continuity is π(d/2)2vρ1=π(D/2)2Vρ2
so that vV=D2d2×ρ2ρ1…(1)
By Bernoulli’s theorem (in absence of external forces like gravity), we have
∫dpρ+12q2=C…(2)
Given that p/ρ=k so that
dp=kdρ…(3)∴ (2) reduces to
k∫dρρ+12q2=C (using (3))
Integrating above, we get
klogρ+q2/2=C, C being an arbitrary constant
When q=v,ρ=ρ1 and when q=V,ρ=ρ2.
Hence (4) yields
klogρ1+v2/2=C and klogρ2+V2/2=C
Subtracting, k(logρ2−logρ1)+(V2−v2)/2=0
⟹log(ρ2/ρ1)=(v2−V2)2k or ρ2/ρ1=e(v2−V2)2k
Using (5), (1) reduces to v/V=(D2/d2)×e(v2−V2)/2k
Hence, proved.
3) An infinite mass of fluid is acted on by a force μr3/2 per unit mass directed to the origin. If initially the fluid is at rest and there is a cavity in the form of the sphere r=C in it, show the cavity will be filled up after an interval of time (25μ)12⋅C54.
[2009, 30M]
MethodI:− At any time t, let v′ be the velocity at distance r′ from the centre. Again, let I be the radius of the cavity and v its velocity. Then the equation of continuity yields r′2v′=r2v When the radius of the cavity is r, then Kinetic energy =∫∞r12(4πr′2ρdr′)⋅v′2[∵ Kinetic energy =12× mass ×( velocity )2]=2πρr4v2∫∞rdr′r′2, using (1)=2πρr3v2 The initial kinetic energy is zero. Let V be the work function (or force potential) due to external forces. Then, we have −∂V∂r′=μr′3/2 so that V=2μr1/2 ∴ the work done =∫crVdm,dm being the elementary mass =∫cr(2μr′1/2)⋅4πr′2dr′ρ=8πμρ∫cTr′3/2dr′=165πρμ(c5/2−r5/2) We now use energy equation, namely, Increase in kinetic energy = work done This ⇒2πρr3v2−0=(16/5)×πρμ(c5/2−r52) ∴ v=drdt=−(8μ5)1/2(c5/2−r5/2)1/2r3/2 wherein negative sign is taken because r decreases as t increases. Let T be the time of filling up the cavity. Then (2) gives ∫T0dt=−(58μ)1/2∫0Cr3/2dr√(c5/2−r5/2) or T=(58μ)1/2∫c0r3/2dr√(c5/2−r5/2) Put r5/2=c5/2sin2θ so that (5/2)×r3/2dr=2c5/2sinθcosθdθ∴T=(58μ)1/2∫π/2045c5/4sinθdθ=(25μ)1/2c5/4 Second Method:- Here the motion of the fluid will take place in such a manner so that each element of the fluid moves towards the centre. Hence the free surface would be spherical. Thus the fluid velocity v′ will be radial and hence v′ will be function of r′ (the radial distance from the centre of the sphere which is taken as origin) and time t. Also, let v be the velocity at a distance r. Then the equation of continuity is r′2v′=F(t)=r2v∂v′∂t=F′(t)r′2. From (1), The equation of motion is of ∂v′∂t+v′∂v′∂r′=−μr3/2−1ρ∂p∂r′F′(t)r′2+∂∂r′(12v′2)=−μr′3/2−1ρ∂p∂r′, using (2) Integrating (3) with respect to r′, we have −F′(t)r′+12v′2=2μr1/2−pρ+C,C being an arbitrary constant When r′=∞,v′=0,p=0. So from (4),C=0. Then (4) becomes −F′(t)r′+12v′2=2μr′1/2−pρ Now when r′=r,v′=v and p=0. So (5) reduces to −F′(t)r+12v′2=2μr1/2 Now, (1) ⇒F(t)=r2v⇒F′(t)=2rv(dr/dt)+r2(dv/dt) or F′(t)=2rvdrdt+r2dvdrdrdt=2rv2+r2vdvdr, as drdt=v Hence (6) gives
−1r[2rv2+r2vdvdr]+v22=2μr1/2 or rvdvdr+32v2=−2μr1/2 Multiplying both sides by 2r2, the above equation can be written as 2r3vdv+3r2v2dr=−4μr3/2dr or d(r3v2)=−4μr3/2dr Integrating. r3v2=−(8μ/5)r5/2+D,D being an arbitrary constant When r=c,v=0. So (7) gives D=(8μ/5)c5/2. Hence (7) reduces to or r3v2=(8μ/5)×(c5/2−r5/2)v=drdt=−(8μ5)1/2(c5/2−r5/2r3)1/2 taking negative sign for dr/dt since velocity increases as r decreases. Let T be the time of filling up the cavity, then T=−(58μ)1/2∫0cr3/2dr(c5/2−r5/2)1/2 Let r5/2=c5/2sin2θ so that (5/2)×r3/2dr=c5/2sinθcosθdθ or T=45(58μ)1/2∫π/20c5/2sinθcosθc5/4cosθdθ=4c5/45(58μ)1/2∫π/20sinθdθT=(2/5μ)1/2×c5/4
4) Liquid is contained between two parallel planes, the free surface is a circular cylinder of radius a whose axis is perpendicular to the planes. All the liquid within a concentric circular circular cylinder of radius b is suddenly annihilated; prove that if P be the pressure at the outer surface, the initial pressure at any point on the liquid distant r from the centre is plogr−logbloga−logb.
[2006, 30M]
5) State the conditions under which Euler’s equation of motion can be integrated. Show that −∂ϕ∂t+12q2+V∫dpρ=F(t), where the symbols have their usual meaning.
[2005, 30M]
6) An infinite mass of fluid is acted on by a force μr3/2 per unit mass directed to the origin. If initially the fluid is at rest and there is a cavity in the form of the sphere r=C in it, show the cavity will be filled up after an interval of time (25μ)12⋅C54.
[2003, 30M]
MethodI:− At any time t, let v′ be the velocity at distance r′ from the centre. Again, let I be the radius of the cavity and v its velocity. Then the equation of continuity yields r′2v′=r2v When the radius of the cavity is r, then Kinetic energy =∫∞r12(4πr′2ρdr′)⋅v′2[∵ Kinetic energy =12× mass ×( velocity )2]=2πρr4v2∫∞rdr′r′2, using (1)=2πρr3v2 The initial kinetic energy is zero. Let V be the work function (or force potential) due to external forces. Then, we have −∂V∂r′=μr′3/2 so that V=2μr1/2 ∴ the work done =∫crVdm,dm being the elementary mass =∫cr(2μr′1/2)⋅4πr′2dr′ρ=8πμρ∫cTr′3/2dr′=165πρμ(c5/2−r5/2) We now use energy equation, namely, Increase in kinetic energy = work done This ⇒2πρr3v2−0=(16/5)×πρμ(c5/2−r52) ∴ v=drdt=−(8μ5)1/2(c5/2−r5/2)1/2r3/2 wherein negative sign is taken because r decreases as t increases. Let T be the time of filling up the cavity. Then (2) gives ∫T0dt=−(58μ)1/2∫0Cr3/2dr√(c5/2−r5/2) or T=(58μ)1/2∫c0r3/2dr√(c5/2−r5/2) Put r5/2=c5/2sin2θ so that (5/2)×r3/2dr=2c5/2sinθcosθdθ∴T=(58μ)1/2∫π/2045c5/4sinθdθ=(25μ)1/2c5/4 Second Method:- Here the motion of the fluid will take place in such a manner so that each element of the fluid moves towards the centre. Hence the free surface would be spherical. Thus the fluid velocity v′ will be radial and hence v′ will be function of r′ (the radial distance from the centre of the sphere which is taken as origin) and time t. Also, let v be the velocity at a distance r. Then the equation of continuity is r′2v′=F(t)=r2v∂v′∂t=F′(t)r′2. From (1), The equation of motion is of ∂v′∂t+v′∂v′∂r′=−μr3/2−1ρ∂p∂r′F′(t)r′2+∂∂r′(12v′2)=−μr′3/2−1ρ∂p∂r′, using (2) Integrating (3) with respect to r′, we have −F′(t)r′+12v′2=2μr1/2−pρ+C,C being an arbitrary constant When r′=∞,v′=0,p=0. So from (4),C=0. Then (4) becomes −F′(t)r′+12v′2=2μr′1/2−pρ Now when r′=r,v′=v and p=0. So (5) reduces to −F′(t)r+12v′2=2μr1/2 Now, (1) ⇒F(t)=r2v⇒F′(t)=2rv(dr/dt)+r2(dv/dt) or F′(t)=2rvdrdt+r2dvdrdrdt=2rv2+r2vdvdr, as drdt=v Hence (6) gives
−1r[2rv2+r2vdvdr]+v22=2μr1/2 or rvdvdr+32v2=−2μr1/2 Multiplying both sides by 2r2, the above equation can be written as 2r3vdv+3r2v2dr=−4μr3/2dr or d(r3v2)=−4μr3/2dr Integrating. r3v2=−(8μ/5)r5/2+D,D being an arbitrary constant When r=c,v=0. So (7) gives D=(8μ/5)c5/2. Hence (7) reduces to or r3v2=(8μ/5)×(c5/2−r5/2)v=drdt=−(8μ5)1/2(c5/2−r5/2r3)1/2 taking negative sign for dr/dt since velocity increases as r decreases. Let T be the time of filling up the cavity, then T=−(58μ)1/2∫0cr3/2dr(c5/2−r5/2)1/2 Let r5/2=c5/2sin2θ so that (5/2)×r3/2dr=c5/2sinθcosθdθ or T=45(58μ)1/2∫π/20c5/2sinθcosθc5/4cosθdθ=4c5/45(58μ)1/2∫π/20sinθdθT=(2/5μ)1/2×c5/4