exercicios p2
56
Problem 4.56
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Transcript of exercicios p2
Problem 4.56
Problem 4.64
Problem 4.65
k x0 L sin q( )Ö-( )Ö r V2Ö AÖ sin q( )Ö=Hence
Fspring k xÖ= k x0 L sin q( )Ö-( )Ö=But
Fspring V sin q( )Ö r VÖ AÖ( )Ö=
Momentum
Governing equation:
x0 1 mÖ=k 1NmÖ=L 2 mÖ=A 0.005 m2Ö=r 999
kg
m3Ö=
The given data are
Solution
Find: Deflection angle as a function of speed; jet speed for 10o deflection
Given: Data on flow and system geometry
2 is deflected by a hinged plate of length 2 m supported by a spring with spring constant k = 1 N/m and uncompressed length x0 = 1 m. Find and plot the deflection angle θ as a function of jet speed V. What jet speed has a deflection of 10°?
Problem 4.68
Solving for θ q asink x0Ö
k LÖ r AÖ V2Ö+
åææç
õöö÷
=
For the speed at which θ = 10o, solve
Vk x0 L sin q( )Ö-( )Ö
r AÖ sin q( )Ö=
V1
NmÖ 1 2 sin 10( )Ö-( )Ö mÖ
999kg
m3Ö 0.005Ö m2Ö sin 10( )Ö
kg mÖ
N s2ÖÖ=
V 0.867ms
=
The deflection is plotted in the corresponding Excel workbook, where the above velocity is obtained using Goal Seek
Problem 4.68 (In Excel)
A free jet of water with constant cross-section area 0.005 m2 is deflected by a hingedplate of length 2 m supported by a spring with spring constant k = 1 N/m and uncompressedlength x 0 = 1 m. Find and plot the deflection angle θ as a function of jetspeed V . What jet speed has a deflection of 10°?
Given: Geometry of systemFind: Speed for angle to be 10o; plot angle versus speed
Solution
ρ = 999 kg/m3
x o = 1 m To find when θ = 10o, use Goal SeekL = 2 m
k = 1 N/m V (m/s) θ (o)A = 0.005 m2 0.867 10
V (m/s) θ (o)0.0 30.00.1 29.20.2 27.00.3 24.10.4 20.90.5 17.90.6 15.30.7 13.00.8 11.10.9 9.521.0 8.221.1 7.141.2 6.251.3 5.501.4 4.871.5 4.33
Deflection Angle vs Jet Speed
0
5
10
15
20
25
30
35
0 2 4 6 8 10 12 14 16 18 20
V (m/s)
θ (d
eg)
The equation for q is q asink x0Ö
k LÖ r AÖ V2Ö+
åææç
õ
÷=
Problem 4.69
Problem 4.70
Problem 4.71
Problem 4.72
Problem 4.73
Problem 4.74
Problem 4.75
Problem 4.76
Problem 4.77
Problem 4.78
Problem 4.79
Problem 4.80
F p r U2 20
1rr u r( )2Ö
òóô
dÖ-åææç
õöö÷
Ö=
F- U r- pÖ 12Ö UÖ( )Ö0
1ru r( ) rÖ 2Ö pÖ rÖ u r( )Ö
òóô
d+=
Applying this to the horizontal motion
Momentum
Governing equation:
Solution
Find: An expression for the drag
Given: Data on wake behind object
where r is the non-dimensional radial coordinate, measured perpendicular to the flow. Find an expression for the drag on the object.
r 1>u r( ) U=
r 1¢u r( ) U 1 cosp rÖ2
åæç
õö÷
2-
åæç
õö÷
Ö=
The horizontal velocity in the wake behind an object in an air stream of velocity U is given by
Problem 4.81
F p r U2Ö 1 2
0
1
rr 1 cosp rÖ2
åæç
õö÷
2-
åæç
õö÷
2
Ö
òóóóô
dÖ-
èéééê
øùùùú
Ö=
F p r U2Ö 1 2
0
1
rr 2 rÖ cosp rÖ2
åæç
õö÷
2Ö- r cos
p rÖ2
åæç
õö÷
4Ö+
òóóô
dÖ-
åæææç
õööö÷
Ö=
Integrating and using the limits
F p r U2Ö 138
2
p2+å
æç
õö÷
-èéê
øùú
Ö=
F5 pÖ8
2p
-åæç
õö÷rÖ U2Ö=
Problem 4.82
Problem 4.83
Problem 4.86
Problem *4.91
Problem 4.107
Problem 4.108
Vr QÖ2 kÖ
-r QÖ2 kÖ
åæç
õö÷
2 r QÖ VjÖ
k++=Solving for V
k V2Ö r QÖ VÖ+ r QÖ VjÖ- 0=
k V2Ö r QÖ VjÖ r QÖ VÖ-=Hence
Fdrag V r- QÖ( )Ö Vj r QÖ( )Ö+=
Applying the horizontal component of momentum
Momentum
Governing equation:
CV in boat coordinatesSolution
Find: Formula for boat speed; jet speed to double boat speed
Given: Data on jet boat
A jet boat takes in water at a constant volumetric rate Q through side vents and ejects it at a high jet speed Vj at the rear. A variable-area exit orifice controls the jet speed. The drag on theboat is given by Fdrag = kV2, where Vspeed V. If a jet speed Vj = 25 m/s produces a boat speed of 10 m/s, what jet speed will be required to double the boat speed?
Problem 4.109
Let ar QÖ2 kÖ
=
V a- a2 2 aÖ VjÖ++=
We can use given data at V = 10 m/s to find α V 10ms
Ö= Vj 25ms
Ö=
10ms
Ö a- a2 2 25Öms
Ö aÖ++=
a2 50 aÖ+ 10 a+( )2= 100 20 aÖ+ a2+=
a103
ms
Ö=
Hence V103
-1009
203
VjÖ++=
For V = 20 m/s 20103
-1009
203
VjÖ++=
1009
203
VjÖ+703
=
Vj 80ms
Ö=
Problem 4.110
Problem 4.112
Problem 4.113
Problem 4.114
Problem *4.168
Tshaft A
r
Vxyz½½
³ rÖ Vxyz½½
Öòóóô
d=
For no rotation (ω = 0) this equation reduces to a single scalar equation
Governing equation: Rotating CV
d 0.005 mÖ=ri 0.05 mÖ=ro 0.25 mÖ=D 0.015 mÖ=
mflow 15kgs
Ö=r 999kg
m3Ö=The given data is
Solution
Given: Data on rotating spray system
Water flows in a uniform flow out of the 5 mm slots of the rotating spray system as shown. Theflow rate is 15 kg/s. Find the torque required to hold the system stationary, and the steady-statespeed of rotation after it is released.
Problem *4.171
or Tshaft 2 dÖri
rorr VÖ rÖ VÖ
òóô
dÖ= 2 rÖ V2Ö dÖri
rorr
òóô
dÖ= r V2Ö dÖ ro2 ri
2-åç
õ÷Ö=
where V is the exit velocity with respect to the CV
V
mflowr
2 dÖ ro ri-( )Ö=
Hence Tshaft r
mflowr
2 dÖ ro ri-( )Ö
èééê
øùùú
2
Ö dÖ ro2 ri
2-åç
õ÷Ö=
Tshaftmflow
2
4 rÖ dÖ
ro ri+( )ro ri-( )Ö=
Tshaft14
15kgs
Öåæç
õö÷
2³
m3
999 kgÖ³
10.005 mÖ
³0.25 0.05+( )0.25 0.05-( )
³=
Tshaft 16.9N mÖ=
For the steady rotation speed the equation becomes
Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d- A
r
Vxyz½½
³ rÖ Vxyz½½
Öòóóô
d=
The volume integral term Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d- must be evaluated for the CV.
The velocity in the CV varies with r. This variation can be found from mass conservation
For an infinitesmal CV of length dr and cross-section A at radial position r, if the flow in is Q, the flow out is Q + dQ, and the loss through the slot is Vδdr. Hence mass conservation leads to
Q dQ+( ) V dÖ drÖ+ Q- 0=
dQ V- dÖ drÖ=
Q r( ) V- dÖ rÖ const+=
At the inlet (r = ri) Q Qi=mflow
2 rÖ=
Hence Q Qi V dÖ ri r-( )Ö+=mflow
2 rÖ
mflow2 rÖ dÖ ro ri-( )Ö
dÖ ri r-( )Ö+=
Qmflow
2 rÖ1
ri r-
ro ri-+
åæç
õö÷
Ö=mflow
2 rÖ
ro r-
ro ri-
åæç
õö÷
Ö=
and along each rotor the water speed is v r( )QA
=mflow2 rÖ AÖ
ro r-
ro ri-
åæç
õö÷
Ö=
Hence the term - Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d becomes
Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d- 4 rÖ AÖ wÖri
rorr v r( )Ö
òóô
dÖ= 4 rÖ wÖ
ri
ro
rrmflow
2 rÖÖ
ro r-
ro ri-
åæç
õö÷
Öòóóóô
dÖ=
or
Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d- 2 mflowÖ wÖ
ri
ro
rrro r-
ro ri-
åæç
õö÷
Öòóóóô
dÖ= mflow wÖro
3 ri2 2 riÖ 3 roÖ-( )Ö+
3 ro ri-( )ÖÖ=
Recall that A
r
Vxyz½½
³ rÖ Vxyz½½
Öòóóô
d r V2Ö dÖ ro2 ri
2-åç
õ÷Ö=
Hence equation Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d- A
r
Vxyz½½
³ rÖ Vxyz½½
Öòóóô
d= becomes
mflow wÖro
3 ri2 2 riÖ 3 roÖ-( )Ö+
3 ro ri-( )ÖÖ r V2Ö dÖ ro
2 ri2-å
çõ÷Ö=
Solving for ω w3 ro ri-( )Ö rÖ V2Ö dÖ ro
2 ri2-å
çõ÷Ö
mflow ro3 ri
2 2 riÖ 3 roÖ-( )Ö+èê
øúÖ
= w 461 rpm=
Tshaft 2 dÖri
rorr VÖ rÖ VÖ
òóô
dÖ=or
Tshaft A
r
Vxyz½½
³ rÖ Vxyz½½
Öòóóô
d=
For no rotation (ω = 0) this equation reduces to a single scalar equation
Governing equation: Rotating CV
d 0.005 mÖ=ri 0.05 mÖ=ro 0.25 mÖ=D 0.015 mÖ=
mflow 15kgs
Ö=r 999kg
m3Ö=The given data is
Solution
Given: Data on rotating spray system
If the same flow rate in the rotating spray system of Problem 4.171 is not uniform but instead varies linearly from a maximum at the outer radius to zero at a point 50 mm from the axis, find the torque required to hold it stationary, and the steady-state speed of rotation.
Problem *4.172
where V is the exit velocity with respect to the CV. We need to find V(r). To do this we use maconservation, and the fact that the distribution is linear
V r( ) Vmaxr ri-( )
ro ri-( )Ö=
and 212Ö VmaxÖ ro ri-( )Ö dÖ
mflowr
=
so V r( )mflowr dÖ
r ri-( )ro ri-( )2
Ö=
Hence Tshaft 2 rÖ dÖri
rorr V2Ö
òóô
dÖ= 2mflow
2
r dÖÖ
ri
ro
rrr ri-( )
ro ri-( )2
èééê
øùùú
2
Ö
òóóóô
dÖ=
Tshaftmflow
2 ri 3 roÖ+( )Ö
6 rÖ dÖ ro ri-( )Ö=
Tshaft16
15kgs
Öåæç
õö÷
2³
m3
999 kgÖ³
10.005 mÖ
³0.05 3 0.25Ö+( )0.25 0.05-( )
³=
Tshaft 30N mÖ=
For the steady rotation speed the equation becomes
Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d- A
r
Vxyz½½
³ rÖ Vxyz½½
Öòóóô
d=
The volume integral term Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d- must be evaluated for the CV.
The velocity in the CV varies with r. This variation can be found from mass conservation
For an infinitesmal CV of length dr and cross-section A at radial position r, if the flow in is Q, the flow out is Q + dQ, and the loss through the slot is Vδdr. Hence mass conservation leads to
Q dQ+( ) V dÖ drÖ+ Q- 0=
dQ V- dÖ drÖ=
Q r( ) Qi d-
ri
r
rmflowr dÖ
r ri-( )ro ri-( )2
Öòóóóô
dÖ= Qi
ri
r
rmflowr
r ri-( )ro ri-( )2
Öòóóóô
d-=
At the inlet (r = ri) Q Qi=mflow
2 rÖ=
Q r( )mflow
2 rÖ1
r ri-( )2
ro ri-( )2-
èééê
øùùú
Ö=Hence
and along each rotor the water speed is v r( )QA
=mflow2 rÖ AÖ
1r ri-( )2
ro ri-( )2-
èééê
øùùú
Ö=
Hence the term - Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d becomes
4 rÖ AÖ wÖri
rorr v r( )Ö
òóô
dåææç
õöö÷
Ö 4 rÖ wÖ
ri
ro
rmflow
2 rÖrÖ 1
r ri-( )2
ro ri-( )2-
èééê
øùùú
Ö
òóóóô
dÖ=
or
2 mflowÖ wÖ
ri
ro
rr 1ro r-( )2
ro ri-( )2-Ö
èééê
øùùú
Ö
òóóóô
dÖ mflow wÖ16
ro2Ö
13
riÖ roÖ+12
ri2Ö-åæ
çõö÷
Ö=
Recall that A
r
Vxyz½½
³ rÖ Vxyz½½
Öòóóô
dmflow
2 ri 3 roÖ+( )Ö
6 ro ri-( )Ö rÖ dÖ=
Hence equation Vr
2 wÖ Vxyz
½½³å
çõ÷³ rÖ
òóóô
d- A
r
Vxyz½½
³ rÖ Vxyz½½
Öòóóô
d= becomes
mflow wÖ16
ro2Ö
13
riÖ roÖ+12
ri2Ö-åæ
çõö÷
Ömflow
2 ri 3 roÖ+( )Ö
6 ro ri-( )Ö rÖ dÖ=
Solving for ω wmflow ri 3 roÖ+( )Ö
ro2 2 riÖ roÖ+ 3 ri
2Ö-åç
õ÷ ro ri-( )Ö rÖ dÖ
=
w 1434 rpm=
Problem *4.175
![Apostila - Comandos El_tricos - Exercicios[1]](https://static.fdocument.org/doc/165x107/5571fcdb49795991699810b6/apostila-comandos-eltricos-exercicios1.jpg)
