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Problem 1: A motor-driven fan (or blower) is mounted on a steel frame which is rigidly attached to the floor as shown in Figure 1. The rotor is unbalanced, and forced rotation at angular velocity, ω, induces a dynamic force on the main shaft. Assume you are given net weight/mass of the blower, M , and you can measure static deflection of the frame when the fan is mounted on it. You might also be given or can determine the ‘eccentric’ mass on the rotor, say m, as well its radial eccentricity, e, from the shaft center.
Figure 1: Unbalanced fan/blower
a. Draw a schematic that represents how you would model this problem; i.e., using ideal model elements (masses, spring elements, damp- ing, etc.). List the specific constitutive relations for all model elements. Include an input forcing due to the unbalanced rotor rotation. Discuss assumptions you would make, and list informa- tion you would need to have. Justify all physical effects you include in your model. Solution: Model as a rigid mass suspended by a spring with damping (mechanically in parallel), as il- lustrated below, part (a). Forces applied include Fk, due to spring-like forces from base, damping forces, Fd, and applied force, F (t), due to the unbalanced mass, m, rotating at N rev/min, or ω = 2πN/60 rad/sec.
b. Prove that you can model the force applied by the unbalanced rotor by a one-dimensional dynamic force in the vertical direction (call it z). Also show that this force can be quantified by, F (t) = Fosin(ωt), and determine how Fo and ω are related to the physical parameters of the problem (parameters for your model, speed of rotation, etc.). Solution: For the unbalanced mass, m, rotating at ω rad/sec, the total force magnitude, Fo = meω^2. Referring to part (b) in the diagram below, the component in the vertical direction is F (t) = Fosinθ, where θ = ωt, with ω = 2πN/60.
c. The most fundamental model of the vertical motion of the total fan mass can be repre- sented by a second order differential equation. Derive this mathematical model. Solution: From the free-body diagram, p˙ = mx¨ =
F = −Fk − Fd + F (t), and if Fk = kx and Fd = b x˙, mx¨ + b x˙ + kx = F (t).
R.G. Longoria, Fall 2012 ME 383Q, UT-Austin
Figure 2: Basic belt drive
Problem 2: Consider the system shown in Fig- ure 2. A stepper motor drives pulley A, which has moment of inertia JA, the timing belt has total mass m, and pulley B has moment of in- ertia, JB. There is static friction acting in the rotational elements (which we can assume move together) that has been measured at the input shaft (at A) as, Tf. Assume that the shafts and the belts are very stiff (negligible compliance).
a. To simplify the model, develop an expression for the effective rotational inertia, Jeff , seen at the motor shaft. Use energy and speed relations, and assume that the two pulleys have equal di- ameters. Solution: Begin by referring all iner- tia to the belt velocity, V. This is strictly done by considering the total kinetic co-energy of the pulleys and belt^1 ,
TωAωB Vb =
JAω^2 A +
JB ω B^2 +
mV 2.
Assume no slip, so the belt velocity is V = rωA, and ωB = ωA, and,
TV =
JAV 2 /r^2 +
JB V 2 /r^2 +
mV 2 =
[
m + (JA + JB )/r^2
]
V 2.
b. Develop a model that will enable you to derive an expression for the torque the motor must generate to accelerate the load of mass m to a velocity Va over a given time interval ta. Solution: Ignore any compliance in the belt, so the translational dynamics of the belt mass is expressed as,
p˙ =
[
m + (JA + JB )/r^2
] ˙
V =
F = (Tm − Tf )/r,
where Tm is the required motor torque, and Tf is the friction. Solving for acceleration of the belt, V˙ = (Tm − Tf )/ [mr + (JA + JB )/r] ,
which allows solving for velocity at time, ta, given zero initial conditions at t = 0 and with constant applied torques; i.e.,
Va =
(Tm − Tf )ta [mr + (JA + JB )/r]
So the (constant) motor torque required to accelerate the belt to Va over time ta is,
Tm =
ta
[mr + (JA + JB )/r] Va + Tf.
(^1) Note, as in BP notes, kinetic co-energy is T with relevant velocities as subscripts.
R.G. Longoria, Fall 2012 ME 383Q, UT-Austin
Problem 4: The diagram in Figure 4 represents a hydro-electric power generation system. In a later assignment, we will focus on simulation of this problem. In this problem, build a simplified model of the hydraulic system up to and including the valve. Assume that the surge tank pressure-volume relation, P = P (–V ) is known. Also assume that the reservoir level does not change significantly over the time of interest (i.e., when we might be studying potentially harmful transient effects), and that the turbine presents a fixed pressure at the valve outlet.
Figure 4: Power generation using a water turbine. From F.E. Cellier Continuous System Modeling, Springer-Verlag, New York, 1991.
Solution: Identify the flow circuit from Pr, an assumed constant reservoir pressure input, to the pressure at bottom of the surge tank, Ps, down to the inlet of the valve, Pv, and to the exit of the valve, which is assumed to have a pressure, Po, specified by the turbine inlet.
We identify three equations that can be used to track flow rate in the pressure tunnel, Qt, volume in the surge tank, –Vs, and flow rate in the pressure pipe, Qp. From the unsteady Bernoulli equation, the rate of change of fluid momentum, Γt, in the pressure tunnel is,
Γ˙t = It Q˙t = Pr − Ps − Pt,
R.G. Longoria, Fall 2012 ME 383Q, UT-Austin
where Pt is the pressure drop due to losses. A similar relation for the pressure pipe momen- tum, Γp, is, Γ˙p = Ip Q˙p = Ps + ρgzo − Pp − Pv − Po,
except we add the effect of elevation^2. Finally, for the volume in the surge tank,
Completing the mathematical model requires knowledge of the specific constitutive relations. In particular, for losses in the pipes, Px = RQ if laminar, or Px = R′Q|Q| if laminar, where R and R′^ are distinct parameters for the different flow conditions, and also R′^ is generally a function of Q and likely friction factors for the pipe. The valve constitutive law takes the form, Pv = KvQ|Q|, where Kv also may be depend on Q. The complete ‘state’ equations are composed in the form,
x ˙ =
Γ^ ˙t Γ^ ˙p
−Ptl − Ps + Pr +Ps + ρgzo − Ppl − Pv − Po Qt − Qp
or,
x ˙ =
Γ^ ˙t Γ^ ˙p
−Pt − Ps(–V (^) s) − Pr Ps(–V (^) s) + ρgzo − Pp − KvΓp |Γp|
I p^2 Γt/It − Γp/Ip
where, Px =
RxΓx/Ix, if laminar R′ xΓx |Γx|/I x^2 , if turbulent
with either x = t for the ‘tunnel’ or x = p for the ‘pipe’.
(^2) One way to get the ‘sign’ right on the pressure terms is to ask: “does it increase/decrease the momen- tum?”
R.G. Longoria, Fall 2012 ME 383Q, UT-Austin
