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Solutions Manual for Electromechanical Dynamics
For any use or distribution of this solutions manual, please cite as follows:
Woodson, Herbert H., James R. Melcher, and Markus Zahn. Solutions Manual for
Electromechanical Dynamics. vol. 3. (Massachusetts Institute of Technology: MIT
OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative
http://ocw.mit.edu/terms
ELECTRO:ECHANTCS OF COMPRESSIBLE, INVISCID
FLUIDS
PROBLE4 13.1
In static equilibrium, we have
= 0
-Vp - pgi,
Since p = pRT, (a) may be rewritten as
RT
Solving,
- + pg = 0
dx
we obtain
p = p
-L x
RT
1
a
e
PROBLEM 13.2
Since the pressure is a constant, Eq. (13.2.25) reduces to
Ov dv = dz
JB
(a)
y
where we use the coordinate system defined in Fig. 13P.4.
Now, from Eq. (13.2.21)
we obtain
Jy =
(Ey + vB)
(b)
If the loading factor K, defined by Eq. (13.2.32) is constant, then
Thus, J
y
Then
KvB =
(c)
avB(1-K)
=
pv dv
dv
or
+ E
SP dv
jdz
avB (1-K)
_-
=
(d)
2
(e)
B
GB2(1-K)
_
=
-
o(I-K)
Ai
A(z)
(f)
From conservation of mass, Eq. (13.2.24), we have
PiviAi =
pA(z)v
(g)
Thus
PiviAi
v 2
dv
-a(l-K)Bi Ai
dz =
(h)
Integrating, we obtain
- a(l-K)Bi
9n v =
z + C
(i)
-a
v=
whereZ
d
vie
d
Oi
vi
Pi Vi
a(l-K)B
v = v. at z = 0.
1
(j)
and we evaluate the arbitrary constant b
y realizing that
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.3
Part a
We assume T, Bo, w, a, cp and c
Since the electrodes are short-
E = 0, and so
circuited,
J
are constant.
=
y
vB
(a)
o,
We use the coordinate system defined in Fig. 13P.4.
Applying conservation of energy,
Eq. (13.2.26), we have
v2) = 0, where we have set h = constant.
v
Thus, v is a constant, v = vi.
Conservation of momentum, Eq. (13.2.25), implies
2
- ViBo
io
dz
Thus, p =
(b)
(c)
- viBz + p
(d)
The mechanical equation of state, Eq. (13.1.10) then implies
v.B 2 z +i
V B 2 z
RT
RT
-
=
RT
i
(e)
From conservation of mass, we then obtain
(-
Piviwdi =
+
vi wd(z)
(f)
Thus
d(z)
=
pld
(
(g)
viBoz
Part b
Then
v.B
p(z) =
Pi
2
z
1RT
(h)
RT
i
PROBLEM 13.4
Note:
There are errors in Eqs. (13.2.16) and (13.2.31).
1
d(M 2 )
S dx
{(Y-1)(1+y1M2 )E 3 + y[2 +(y-1)M 2 ]v 1B 2 } J3
-M
r
(13.2.16)
and
i2
M2
2 ) d(M
d
)
dx
(1M) )
(1-
[2 +
S
(
[(y-I)(I1+yM2)E
+
3
'2
+
2 }
3
(y-1)M
v1 2]
P
1 2 ypv
(y-1)M2 ]dA
A
dx
(13.2.31)
Part a
We assume that
0, y , Bo, K and.M are constant along the channel.
the corrected form of Eq. (13.2.31), we must have
Then, from
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
2
[(y-l) (l+M2)(-K) + y(2+(y-1)M )]
1-
0 =
-K)
[2 +(y-1)M ]
(a)
Now, using the relations
and
v2
M2 yRT
p
pRT
v
M2
Yp
pv
we write
(b)
B 2 0C(l-K)M 2
Thus, we.obtain
[(y-)(l+yM2 )(-K) + y(2+ (y-1)M2 )]
2 + (y - 1)M 2
1 dA
A2 dz
pVA
(c)
From conservation of mass,
pvA =
(d)
piviAi
A
Using (d), we integrate (c) and solve for
Ai
to obtain
A(z)
Ai
1
1 - ýz
(e)
where
)]B2M2 (1-K)
2
yM12 )(-K) + y(2 +(y-l)M
[(y-1)(1+
Sivi[ 2
+ (y-l)M2 ]
We now substitute into Eq. (13.2.27) to obtain
vB 2 (1-K)o
1
1 dv
S
(-M)
-v
dz
o
[(y-l)(-K) + y]
(-M2)
Yp
1 dA
A dz
(f)
A dz
Thus may be rewritten as
d
=
v dz
(-M
2
)
[(y-l)(-K)
+ y]
iiAi
(1-M2)
p viAi
A
(g)
Ai
Solving, we obtain
£n v =
or
v(z)
- 82
(1 -
=
Vi
where
8 z) + en vi
8 z)2/
(h)
1
(i)
1
i
82 =
n(l -
(-i
2
)
2
(1-K)M 2 [(y-l)(-K) +y]Bo
i vi
01
Now the temperature is related through Eq. (13.2.12), as
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
M2 yRT =
Thus
v2
(j)
2
T(z)i
(k)
From (d), we have
P(z)
Pi
vi Ai
v A
=
(
Thus, from Eq. (13.1.10)
T
Vi Ai
p(z) =
Pi
v
A
Ti
(m)
Since the voltage across the electrodes is constant,
w(
or
w(z)
Thus,
- Kv(z)B
KviBoi
Kv(z)B
Kv(z)B0
=
w(z)
wi
(n)
Vi
wi
v(s)
(o)
vi
v(z)
Then
d(z)
A(z)
di
i
Ai
(q)
()
w(z)
Part b
We now assume that a, y,
Bo, K and v are constant along the channel.
Then, from
Eq. (13.2.27) we have
0
=
1-)
Cl-M 2 )
But, from Eq.
--
C(y-l) (-K) + y]viB2
1l
i0
1 ­
(l-K)v. B2 z
=
ov.
i
3z
(s)
2
B
1 0
Pi
Substituting the results
A(z)
1-
Pi
(l-K)
=
1 A
A dz
(13.2.25) we know that
Pi
where
(-K)
yp
_
i
A(z)
of
b),
(-Ka)
pi into
and solvin
for
we obtain
(u)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
and so, from Eq. (13.1.10)
=
T(z)
T.
p(z) Pi
Pi p(z)
(v)
As in (p)
w(z)
wi
vi
=
v(z)
=
1
(w)
Thus
d(z)
di
A(z)
A
=
(x)
Part c
We wish to find the length Z such that
C T(i) +
]2
1 [v
C
p
z
C T(o) +
.9 =
(y)
[v(o)]
For the constant M generator of part (a), we obtain from (i) and (k)
2
1
C
C
[
Ti
p
()
]2
1
Ti
C
p T(z)
i
1 2
Vi
1
221
-28
1
/Bl
9
+1 2
(Z
2i
Reducing, we obtain
-
)
(1 -
26 /8
2
.9
(aa)
Substituting the given numerical values, we have
,1 =
We then solve
2/8
and
.396
(aa) for
2
1
=
- 7.3 x 10
-2
Z, to obtain
% 1.3 meters
k
For the constant v generator of part (b), we obtain from (s),
CT i
pI
+
)
C
v
i
2
(1 -
B )
(u) and (v)
i
+ S= 2
T
p
2
(t),
(bb)
.9
i
or
C T.
C
(1- 8 4/ )
3 1 2
v2
1
2
C T + L v
1
2
p i
=
.9
Substituting the given numerical values, we have
83
(cc)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.4 (continued)
= B
and
.45
=
B /B
.857
Solving for Z, we obtain
Z
1.3 meters.
PROBLEM 13.5
We are given the following relations:
E(z)
wi
di
B.
E.
w(z)
d(z)
I
/2
(z)
Ai'
B(z)
0, y, and K are constant.
and that v,
Part a
From Eq.
(13.2.33),
J =
(1-K)ovB
(a)
For constant velocity, conservation of momentum yields
dp
dz
=
(1-K)ovB 2
-
(b)
Conservation of energy yields
pvC
pdz
=
-
K(1-K)G(yB)
2
(c)
Using the equation of state,
=
p
pRT
(d)
we obtain
or
T
Thus,
(-K) (1-K)ovB 2
+
dz
C
p
2
vB (1-K)
=
T
(l-K)
-K
OvBB
R
p dT
dz
T dP
-- +
T dz +
dz
(1-K)OvB 2
R
(f)
(g)
+
(g)
B.i2 (A.)
1
1
A(z)
2
and
.A. =
and
(e)
(e)
+
R
Also
2
p(z)A(z)
11
d =
T d-
Therefore
and
pc
dT d =
p dz
ovB'fdi(1-K)(-2
+ --- )
P(!)
Bp
-K(1-K)av
(i)
Pi
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.5
(continued)
and so
K(1-K)OvB2
dT
i
dz
()
pic p
Therefore
T
2
OvB i
- K(1-K) z + T
PiCp
=
Let
(k)
2
Let
-K(1-K)avBi
=
Pi Cp
(i)
Then
T =
az
C-
Ti
+ 1)
(m)
1
+ ovB (1-K)(
i
S
K
cp
i
R
dz
(n)
pi (az + Ti)
We let
S vB(-K)(
K
1
Cp
R
P.
a
Pi
e
KR
Integrating (n), we then obtain
in p =
8 in(az + Ti)
+ constant
or
p=
=z
i
-
+
1 )
8
(o)
Therefore
A
A(z)
=
()
Ti
Part b
From (m),
T()
at
T(a)
+
T
i
.8
Ti
Ti
or
-l-
.
-. 2
Ti
Now
­ K(1-K)viBi
E
Ti
Pic Ti
But
R Ti
c T
pi
2
=
(1)
Pi
=
y
pi(1­
i
y
=
2.5 x 10 6
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.5 (Continued)
Thus
Thus
- .5(.5)50(700)16
Ti
-2
8.0 x 10-2
-
.7(2.5 X 10 6)
Solving for R,
we obtain
•2x 102
z=
1.25 meters
=
8
Part c
pi
p=
( (Xz
1)
•.+
Numerically
8=
- 1
KR
1
1
(l1)K
6.
%
Thus
p(z)
.7(1 -
=
.08z)
6
Then it follows:
7
Ti(l -
=
T(z)
Pi(1 -
pRT =
p(z) =
7
= 5 x 10S(l - .08z)
.08z)
.08z)
From the given information, we cannot solve for Ti, only for
v
Pi
RT. =
I
Pi
Now
=
2
-i
s
7 x 10
yM
vi
yRT(z)
v2
yp(z)
.5
1 -
.08z
Part d
The total electric power drawn from this generator is
VI =
p
= -
E
Thus
pe
-E(z)w(z)J(z)£d(z)
E(z)(1-K)GvB(z)£d(z)w(z)
=
- Eiwi(1-K)ovB diz
=
-KvB
i
= K(vB )2
w diO (-K)
=
.5(700)216(.5)50(.5)1.25
=
61.3 x 106 watts =
61.3 megawatts
86
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
T (z)
T
(z)
Ti
ST
(1 -
.08z)
1.25
p(z)
p(z) =
p(z)
7ý(1
- .08z)
1.25
p(z) =
5 x 10
5
(1 -
1.25
m
6
(z)
M2 (z) =
1.25
(1-
.5
.08z)
.08z)
7
ELECTROILECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.6
Part a
We are given that
4 Vo
E--i
x3
x
X
Lj/3
and
4
e
Eo VO
9
L'3 x
The force equation in the steady state is
dv
ix
pv
m xx d•
=
Since pe/p
pe E
=
q/m =
constant, we can write
Vd O
2 v2A
dx2Vx
Solving for v
/3
X!3
V4
=
L3
m)
we obtain
x
Part b
The total
force per unit volume acting on the accelerator system is
F = peE
Thus, the total force which the fixed support must exert is
f
=
total
-
FdVi
x
EV 2
-16 E0 0
=-f
L8
27
-3
x
x
0
f total
8
=
8
9
oo
2
Ai
x
PROBLEM 13.7
Part a
We refer to the analysis performed in section 13.2.3a.
The equation of motion
for the velocity is, Eq. (13.2.76),
2
v
2 a 2(a)
a ax 2
•-22V
3-ý
=
The boundary conditions are
v(-L) =
V
v(O)
0
=
cos Wt
We write the solution in the form
(a)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.7 (continued)
Re[A ej(wt-kx )+ B ej(wt+kx l
v(x t) =
) ] (b)
a
where
a
Using the boundary condition at x 1 = 0, we can alternately write the solution as
Re[A sin kx e j
v =
t
I
Applying the other boundary condition at x = - L, we finally obtain
V
v(xt) = - sin kL sin kx cos wt.
(d)
1
sin kL
1
The perturbation pressure is related to the velocity throughEq. (13.2.74)
Pov'3
at
-x,
a(e)'
Solving, we obtain
0Vo
oL
s sin kx sin wt =
sinkL
1
x
(f)
1
p V
0 k
k 0 sin
kL cos kx 1 sin wt
or
p
f)
-
=
(g)
where po
0 is the equilibrium density, related to the speed of sound
a, through
Eq. (13.2.83).
Thus, the total pressure is
p Vw
+ k sin kL cos kx
P= + p
and the perturbation pressure at x, =
p'(-L, t) =
sin at
(h)
- L is
00Voa
sin k cos kL sin wt sin kL
(i)
Part b
There will be resonances in the pressure if
sin kL =
0
n = 1,2,3....
n7
kL =
or
(j)
(k)
Thus
w
(a)
a
L
PROBLEM 13.8
Part a
We carry through an analysis similar to that performed in section 13.2.3b.
We assume that
E = iE
2
(xl,t)
J= i2J2 (x,
t)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.8
B =
i
[P 00
H +
3
i
H' (x 1 ,t)]
03
Conservation of momentum yields
Dv
D 1 =
Dtax
.x
+J2
o(Ho + H3)
20
1
(a)
0
Conservation of energy gives us
D
p Dt
2
1
(u +
v ) =
1
2
ax
pv
1
(b)
+ J E
22
We use Ampere's and Faraday's laws to obtain
aH'
=
- - JJ
[email protected]
ax
and
(c)
2
P oaH
DE
a2
x
(d)
at
3x,
while
Ohm's law yields
J
Since
=
2
-+
([E
- v B
2
13
i
(e)
oo
E
=
vB
2
(f)
1 3
We linearize, as in Eq. (13.2.91), so E
2
'
v p H
100
O
av,
alH'3
Linearization of the conservation of mass yields
at
=
o
)v
(h)
Thus, from (g)
,H'
3
ooy_
p
o at
at
()
Then
Ho Po
p'
H'3
Linearizing Eq.
(13.2.71), we obtain
=
p
Dt
o
p
D'
Dt
(k)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.8 (continued)
Defining the acoustic speed
as
\ 0o/
where p
is the equilibrium pressure,
(oH
2
= P0o
p
2
we have
p'
=
as22
(p)
Linearization of convervation of momentum (a) yields
av
p o at 1 =
-
9H'
H o(m)
x3 o
oHo
T
3xl
(m
1
or, from (j) and (Z),
Iv,
ao
0
a2
(n)
-s
Po at=
Differentiating (n) with respect to time, and using conservation of mass (h), we
finally obtain
°H
S2v
2V=
2v
( a2
(0 )
Defining
0
1
2
a2
a2
p1 H
+-0
s
we have
(p)
p o
a2 v
3 v
I
1
2(V)
a2V
Part b
We assume solutions of the form
(
Re [A ej (Ct-kx )+ A e j wt+
V =
1
where
k =
1
(r)
a
V(-L,t) =
V
= - L is
cos wt =
s
V Re e jwt
(s)
s
= 0
M dViot) =
p
dt
'A'
+ poH H
oo
1=0
From (h),
)] ­
The boundary condition at x
and at x
kx
2
(j)
and
1-g__ =
a2
s
1
A (t)
1=0
(t),
P
v
at
(o
u)
ax
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.8 (continued)
H'
a
Ho
Po
Thus
dv,(0,t)
dt
M
2
aoH
=A
From (u), we solve for p'j
+
2
az-T-
I p' =A
to obtain:
x =0
Substituting into (s) and (t), we have
a 2/Poak
Mjw(A + A)
1 2
=
A(-
~)•
\
) (A
­
1
/
a'\W
A2)
Ale+jk£+ A2e-jki
Solving for A
A
1
1
and A , we obtain
2
(Mjw + Aap o)V
s
=
2(-Mw sin kZ + Aap cos k£)
(Aap
- Mjw)V
s
2(-Mw sin kZ + Aa %cos kt)
A2
Thus, the velocity of the piston is
v (O,t) =
[A
Re
j
wt
+ A ]e
S1
v (0,t)
2
Aap V
w s
+ Aaos
Wt
-Mw
Elsin kk +Aap cos kZCOWt
=
1
(aa)
PROBLEM 13.9
Part a
The differential equation for the velocity as derived in problem 13.8 is
32v
1
Ftwhere
a
2
2
a
-
=
a
2
+
s
a
a
with
=
S
a2v
1o
aX 1 2
(a)
00
Po
2
(YPo2
where p
-•oP
Po
=
0
Part b
We assume a solution of the form
p
1
p H
2
2
2
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.9 (continued)
where k =
[De J(t-kx1)]
Re
V(x ,t) =
1
­
a
We do not consider the negatively traveling wave, as we want to use this system as
a delay line without distortion.
V(-L,t) =
and at x
The boundary condition at x
=
- L is
Re V ejWt
s
= 0 is
dV
BV (O,t)+
p'(O,t)A -
=
dt
;,
(b)
A
joHoH'
From problem 13.8, (h), (j) and (9)
H'
av
pa
s
p
',
-
=t
at
o ax
and
1
-
H
=
a o
S 0
Thus, (b) becomes
ta 2
t
-BDej
+
(-)
where
=0
p'A
S
PoD(- jk)
pI
=w
x0O
j
(c)
t
2
(d)
se
w
(d)
Thus, for no reflections
2 Ap0a2
- B +
(a)
a
0
a
(e)
S
or B =
Aapo
(f)
PROBLEM 13.10
The equilibrium boundary conditions are
T[-
(L
1
(L
T[-
+ L
+ A),t]
2
+ A),t]As =
=
T
0
- PoAc
Boundary conditions for incremental motions are
T[-
(L
2) - T[-
(L
1)
1
+ L
2
+ A),t] =
+ A),t]A
v (-Ll ,t) =
3)
and 4 )
v
(O,t)
=
- p(-L
Ve[-(L
0
T
s
(t)
d
,t)Ac = M
+ A),t]
-
v
(-L ,t)
since the mass is rigid
since the wall at x=0 is fixed.
PROBLEM 13.11
Part a
We can immediately write down the equation for perturbation velocity, using
equations (13.2.76) and (13.2.77) and the results of chapters 6 and 10.
93
ELECTROMECHANTCS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.11 (continued)
We replace 3/at by 3/at + v-V
(-+
Letting
V
t
)v'
o
=
x
to obtain
a
2
sax
2V
= Re V j(t-kx)
v'
we have
(W- kV S)2 =
a2k
2
as
Solving for w, we obtain
=
W
k(V
± as )
Part b
Solving for k, we have
k =
V±a
V+a
o
s
For Vo > as, both waves propagate in the positive x- direction.
PROBLEM 13.12
Part a
We assume that
E =
i
J =
iz
z
z
i
y
B=
E (x,t)
z
J (x,t)
z
o1[H
+ H'(x,t)]
y
We also assume that all quantities can be written in the form of Eq. (13.2.91) .
Vx
p' -
Jz
Ho
(conservation of momentum
(a)
linearized)
The relevant electromagnetic equations are
aH'
and
3E
3x
(b)
J z
ax Yx =
aH'
0o----
z
at
(c)
and the constitutive law is
Jz =
z
(E + VxpoH )
z
xoo
We recognize that Eqs. (13.2.94), (13.2.96) and (13.2.97) are still valid, so
1
I3' P
at
ax
(d)
eav
(e)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM' 13.12 (continued)
and
a 2 P'
p'
(f)
S
Part b
We assume all perturbation quantities are of the form
v
Using (b),
x
=
^
Re[v e
j(wt-kx)
]
(a) may be rewritten as
(g)
+ jkp + p0HojkH
pojwv =
and (c) may now be written as
(h)
jojwH
- jkE =
Then, from (b) and (d)
o(E
- jkH =
+ V oHo
(i)
) Solving (g) and (h) for H in terms of v, we have
vOVoH
H =
(j)
-
jk+po 9
From (e) and (f), we solve for p in terms of v to be
p=
(k)
kP asv
Substituting (j) and (k) back into (g), we find
2 jk(po Ho)2
v
L2
-
Lk
0)
00jk+ 'w
Thus, the dispersion relation is
k
j(MoHo)20
k,
(02 - k2a2) -_
(+ 2
=
(m)
0
+ j11w)po
We see that in the limit as a + m, this dispersion relation reduces to the lossless
dispersion relation
_-2 k2(a2 +
=
(n)
0
Part c
If a is very small, we can approximate (m) as
W2-
k2a2
_ j(
w
s fo for which we can rewrite (o) as
2
O
kr
)H = (0o)
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.12 (continued)
kl
s
k2
-
O
2
2
(
=
2
, we obtain
Solv ing for k
2
0p
()02_ j
2
0
k2=
2
22a
OHo
Po
2
)
w°
0 02
2)
O+
o
a
s
2
2a s
a
s
Since a is very small, we expand the radical in (q) to obtain
2
[12
O A oo
_
2 a .
O2]
S
L
2
Thus, our approximate solutions for k
EW
are
(IIoHo)2
k2
a
(
o0
0
O
s
H )
a (11
2
The wavenumbers for the first pair of waves are approximately:
while for the second pair, we obtain
+
o
0
k~+~lo~op
The wavenumbers from (u) represent a forward and backward traveling wave, both with
amplitudes exponentially decreasing.
Such waves are called 'diffusion waves'.
The
wavenumbers from (v) represent pure propagating waves in the forward and reverse
directions.
ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS
PROBLEM 13.12 (continued)
Part d
If 0 is very large, then (m) reduces to
H2
W2 - k 2 a 2 -
o
0 ;
k
p0
a2 = a 2
Ow
s
H2
O
+
(w)
Po
This can be put in the form
a
2
a
where
H 2 k0
f(w,k), =-
As
(x)
f(wk)
k2=
(Y)
2
pOwa
a becomes very large, the second term in (x) becomes negligible, and so
k
2
-
2
However, it is this second'term which represents the damping in space; that is,
k
+
,a
f(w,k)
j2a
(z)
Thus, the approximate decay rate, ki, is
ki
ki
2a w
2
or
i
H2 k
o
f(w,k)a
2p
a0
a
2poaw•
k
H2
w
a
2pa
0
(aa)
```
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