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Answer each question following the format

could provide the class note when needed

ENGM2022: Assignment 2: Winter Term 2021
NOTE: To receive any points, when applying the 4-step and 7-step
methods you are expected to label each step and complete it as we do in
the notes. Show your work. The sketch featues always need to be justified
and this is spelled out in each problem as clearly as possible. This assignment is passed in 1 minute before midnight, SATURDAY, JANUARY 30.
Late assignments are not accepted.
1. (16 pts)(This problem is from our textbook and follows the same algebra we
used in the Week2Class3 FridayJan15 notes on mixing tank problem)
Suppose an RC-series circuit has a variable resistor. If the resistance at time t
is given by R = k1 + k2 t, where k1 and k2 are given positive constants then our
mathematical model for the charge is
(k1 + k2 t)
1
dQ
+ Q = E0
dt
C
(1)
where C is the capacitance, E(t) is a ’perfect’ battery in the sense that E0 never
changes, and Q(0) = Q0 .
(a) (1 pt) What are the units of k1 and k2 ?
(b) (1 pt) Is this differential equation linear or nonlinear in the dependent
variable? State what you think and then prove it.
(c) (7 pt) Complete the 7-step method and Show that
k1
Q(t) = CE0 + (Q0 − CE0 )
k1 + k2 t
!1/(Ck2 )
(2)
In your sketch roughly justify what the solution looks like for Q0 > CE0
and Q0 < CE0 by considering the sign of terms in the provided solution.
(d) (3 pt) Solve the problem but this time assume that k2 = 0 and the circuit
resistance is constant. Identify the process time constant (PTC) and show
the evolution of the solution in terms of the PTC on your sketch for full
points.
(e) (4 pt) Find the limit as k2 → 0 in the provided solution and show that it
is the same as your k2 = 0 solution by completing the following steps
i. show that y = (k1 /(k1 + k2 t))1/(Ck2 ) may be rewritten as y = (1 +
k2 t/k1 )−1/(Ck2 )
ii. write down ln(y)
iii. note that ln(1 + X) ≈ X for X 1 with X = k2 t/k1 1 to
approximate ln(y) in the k2 → 0 limit
iv. exponentiate your ln(y) results to get y
v. replace the k2 → 0 limit for y back into the provided solution which
should now be the same as your k2 = 0 result
vi. Use L’Hopital’s rule to find the limit k2 → 0 of y = (k1 /(k1 +
k2 t))1/(Ck2 ) . Show that you get the same result as before.
2. (10 points) Refer to Week2Class3 FridayJan15 notes on mixing tank problem.
(a) (2 pt) Take limit b → a and show that we recover the b = a solution which
we wrote down in a previous class where Qin = Qout = Q and tank volume
stays at its initial value V = V0 . Use the same technique as you used in
the previous problem in (e) without L’Hopital’s rule.
(b) (8 pt) Sketch the full solution where a 6= b for a < b which corresponds to
Qin < Qout so that the tank empties out from its initial volume V0 > 0.
i. (1 pt) Show that the tank empties in a time t = V0 /(b − a).
ii. (1 pt) Factor -1 from a − b and rewrite the solution in terms of b − a,
e.g. the exponent −a/(a − b) is a/(b − a) and now we can see that
it is positive. Do the same thing and replace (a − b)t with −(b − a)t.
Show that
h
M (t) = V0 Cin [1 − (b − a)t/V0 ] 1 − (1 − (b − a)t/V0 )a/(b−a)
i
(3)
iii. (2 pt) Early Times Approximate the term (1 − (b − a)t/V0 )a/(b−a) at
small time using the Taylor Expansion (1 + X)a ≈ 1 + aX for aX 1
and setting X = −(b − a)t/V0 . Insert the approximation into the
provided solution and once the dust settles on the algebra and show
that at small times M ≈ atCin and remember that a = Qin is the
input flow rate. This little result means that the M is increasing at
small times as we expect and we know how rapidly that is happening
too!
iv. (2 pt) Nearly Empty As the tank empties of fluid all the mass goes with
it and M = 0 when the tank is empty. When the tank is nearly empty
1 − (b − a)t/V0 is small and given that the exponent a/(b − a) > 0 show
that M ≈ V (t)Cin where V (t) = V0 − (b − a)t is the tank volume.
v. (2 pt) Now you can sketch M (t). Be sure to explicitly label your sketch
with the results from above for full points. The only thing we have
not found out is an approximation to the solution near its peak. But
that is okay!
3. (15 points) Consider a model of a typical electric oven cooking an apple pie.
The supplies the ’oven ambient temperature’ T∞ (t) and the apple pie exchanges
heat with the oven.
Electric ovens typically use Pulse Width Modulation (PWM) to control the oven
ambient temperature. Over a fixed-duration ’duty cycle’, a constant current is
turned on in the oven element. If the duty cycle has 1 second duration, then
the oven reaches a ’steady’ temperature when the current is on for a constant
fraction of the 1 second duty cycle. For example, the oven might be on for the
first half of the duty cycle and off for the second half of the duty cycle. If you
think about this a little bit you will realize that the oven ambient temperature
must be cycling around an average temperature. It turns out the this average
temperature is the value that you ’dialed in’ on the stove to cook your apple
pie. As a cook you have no control over the fluctuations above and below this
average value - you get what you pay for! This cycling of the actual temperature
has maximum deviations from the ’dialed in’ average that are a function of the
duty cycle length - long duty cycles have larger fluctuations than smaller duty
cycles.
Let’s assume we turn on the stove and dial in an average oven ambient temperature T1 to cook our apple pie. We wait until the oven beeps at us after
reaching this average value and then we put in the pie to cook. Let’s assume
that the manufacturer has set the duty cycle to have a duty period d seconds
during which time the oven ambient temperature is T∞ = A + B cos(αt).
Assume that the exchange of heat between the apple pie and the oven ambient temperature T∞ (t) follows Newton’s Law of Cooling so that dT /dt =
−k(T − T∞ (t)) where T is the apple pie temperature, T∞ (t) is the oven ambient
temperature and the process time constant (PTC) is 1/k is determined by the
apple pie constituents (the stuff we eat and the pan). Finally, assume that the
pie initial temperature is at room temperature T0 before being placed into the
hot oven.
(a) (1 pt) Given T∞ = A + B cos(αt) set A, B, and α so that the average of
T∞ (t) over a cycle is T1 , the period is d, and the size of the fluctuations is
T2 .
(b) (1 pt) After the pie is removed from the oven it is still quite warm after
1.5 hrs but still not nearly at the room ambient temperature. What is a
reasonable estimate for the PTC and thus the rate constant k?
(c) (7 pt) Complete the first 6 steps of the 7-step method. Use our complex
variable approach to integrate the RHS in Step 5. Work with the variables
k and α in your model. Be sure to write out the full solution at the end
of Step 6 that satisfies the condition and show that
T (t) = T1 + T2 √
k
cos(αt − φ) + De−kt
2
2
k +α
(4)
where tan φ = α/k and D = T0 − T1 − k 2 /(k 2 + α2 )T2
(d) (1 pt) What is the transient solution?
(e) (1 pt) What is the steady state solution?
(f) (1 pt) Write the system diagram with ’Input Amplitude’, ’Transfer Function’, ’Output Amplitude’.
(g) (1 pt) Sketch transfer function in terms of input frequency α.
(h) (1 pt) Is the apple pie acting as a low-pass, band-pass, or high-pass filter
of the fluctuating oven ambient temperature?
(i) (1 pt) What is the time scale over which the initial condition T0 is ’forgotten’ ? Briefly explain - one sentence and refer to the steady state versus
the transient solutions.
4. (7 points) In a test for gestational diabetes a pregnant woman is given a disgusting sugary drink to see how well her body can process glucose. Let’s assume
that the drink causes a sharp and dramatic change in the ambient glucose load
that can be modelled as G∞ (t) = G1 exp(−αt) where the exponential drop is
due the effects of dilution before reaching the bloodstream. Once the glucose
enters her bloodstream her body releases insulin and drives down the glucose
levels - insulin causes cells to uptake glucose in a control response. Let’s model
this control response in terms of our simple first order process and assume that
blood glucose is dG/dt + kG = k1 G∞ (t) where G = G0 is her blood glucose
before ingesting the sugary drink. Note: our parameters α and k and k1 are
plus!
(a) (1 pt) What value of k would we associate with an instant and overwhelming insulin response?
(b) (1 pt) What value of k would we associate with no insulin response?
(c) (5 pt) Use the 7-step method to solve this problem and assume that α 6= k.
Show that
G(t) = G1
i
k1 h −αt
e
− e−kt + G0 e−kt
k−α
(5)
Sketch for the special case where the initial blood glucose is G0 = 0 (not
physically possible - oh well).
In your G0 = 0 sketch identify the small time approximation by noting
eX ≈ 1 + X, X 1 so that both αt and kt are small and show that
G(t) ≈ G1 kt at small times. For large times show that G → 0 from above
for both k > α and k < α. Now you can complete the G0 = 0 sketch and
be sure to explicitly label your early/large time results on the sketch for
full points!
5. (20 points) Find the solution of ÿ + 2cẏ + 2y = 0, y(0) = y0 , ẏ(0) = v0 .
(a) (1 pt) Choose c > 0 so that the model is critically damped.
(b) (3 pt) Solve the critically-damped model, but do not sketch, for y(0) = y0 ,
ẏ(0) = v0 and show that
y(t) = (y0 + (v0 + cy0 )t)) e−ct
(6)
(c) (3 pt) Now sketch critically-damped y(t) assuming that y0 = 1 and v0 =
−2. In your sketch you’ll need to determine if y can go below y = 0. To
receive any sketch points use pure logic without calculus to show whether y
goes below zero. Complete the sketch with that single piece of information
- and the facts we outlined in class about the number of zero crossings
available to damped solutions of second order differential equations.
(d) (4 pt) Assume that c is large enough to make the model over-damped.
Find the solution y(t) and do not sketch. Show that
y=
where λ1 = −c +
q
(y0 λ2 − v0 )eλ1 t − (v0 − y0 λ1 )eλ2 t
λ2 − λ1
c2 − 1/2 and λ2 = −c −
(7)
q
c2 − 1/2.
(e) (2 pt) Show that the over-damped solution satisfies the conditions y(0) =
y0 , ẏ(0) = v0 .
(f) (1 pt) Refer to the over-damped solution and find a formula to choose y0
and v0 so that the solution goes to zero as rapidly as possible - this will
require the coefficient of the slowest part of y to be zero.
(g) (6 pt) Assume that c is small enough to make the model under-damped.
Show that
y(t) = e
q
−ct
q
y02 + ((v0 + cy0 )/ω)2 cos(ωt − φ)
(8)
where ω = 1/2 − c2 and tan φ = ((v0 + cy0 )/ω)/y0 . In your sketch please
draw the solution envelope and identify the psuedo-period.
...

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