## Transcribed Text

heat transfer to mercury-filled tube
A mercury-filled tube (see figure at left) is moved suddenly from ambient conditions, T = 20◦C, into
cold air at Tair = −10◦C. The heat transfer coefficient is reasonably constant at h = 9.7 W/m2K (the
mercury is a good conductor of heat, and may be treated as isothermal), so that the flow of heat from
the air to the mercury is given by
Q = hA(Tair − THg)
Normally this would be a simple system, but here you will consider the impact of the varying heat
capacity and density of the mercury. The heat capacity of mercury is well-represented by
CP,Hg(THg) = 139.6829 − 0.030701THg + 5.361 × 10−5THg
2
where CP,Hg has units of J/kg · K and THg has units of ◦C. The density of mercury is well-represented
by
ρHg(THg) = 13595.08 − 2.4689THg + 3.76 × 10−4THg
2 + 5.4 × 10−8THg
3
where ρHg has units of kg/m3 and THg has units of ◦C. The varying density changes the contact area,
so that
A(THg) = 1.9 cm2 + Aw0
ρHg(20◦C)
ρHg(THg)
where Aw0 = 19 cm2
is the wall contact area at the initial temperature THg(t = 0) = 20◦C (the 1.9 cm2
is the unchanging contact area at the bottom of the tube). The varying temperature of the mercury is
given by the heat balance
mHgCP,Hg
dTHg
dt = Q
where here mHg = 0.099 kg.
(a) First solve the problem analytically, assuming that CP,Hg, ρHg, and A are constant at their THg =
20◦C values (note: THg is not constant at 20◦C, just the three properties listed). Find THg(t =
300 s) for this simplified case.
(b) Now solve the problem with all parameters varying (with THg, and hence also with time) as given
above. Find THg(t = 300 s) using Runge-Kutta fourth-order (RK4) method over a single time
step. Show all intermediate values and clearly indicate their computation.
(c) Repeat, still with parameters varying, using two RK4 steps to determine THg(t = 300 s) more
accurately.
5.2 multi-chambered water tank
The water tank shown on the following page has multiple chambers connected by relatively small openings, along
with a steady inflow and an unsteady outflow. Initially the first chamber is filled with considerably more water than
the adjoining chambers (the second and third chambers), and then it gradually drains into them. This transient
behaviour eventually leads to a steady state where the levels decrease towards the outlet.
The overall tank is rectangular — the chambers are all 3 m in width (W = 3 m), and the tank is 0.8 m across
(perpendicular to the page, as drawn; L = 0.8 m)). Initially, the level in the first chamber is H1(t = 0) = 2.5 m, in
the second chamber, H2(t = 0) = 0.4 m, and in the third chamber, H3(t = 0) = 0.2 m (all relative to the outlet).
The inlet flow rate is fixed at Q0 = 0.009 m3/s. The discharge coefficients and areas of the three opernings are as
follows: Cd1 = 0.61 and A1 = 0.0084 m2
; Cd2 = 0.42 and A2 = 0.0034 m2
; Cd3 = 0.34 and A3 = 0.0030 m2
.
The rates of change in height in each chamber are given via the flow rates between chambers:
Q1 = Cd1A1
p
g(H2 − H1) W L · dH1/dt = Q0 − Q1
Q2 = Cd2A2
p
g(H2 − H3) W L · dH2/dt = Q1 − Q2
Q3 = Cd3A3
p
g(H3) W L · dH3/dt = Q2 − Q3
where g = 9.81 m/s
2
. As there are three variables here, it is useful to make the substitutions u(t) = H1(t),
v(t) = H2(t), and w(t) = H3(t). The above equations can then be simplified, to
du
dt = q0 − q1
√
u − v
dv
dt = q1
√
u − v − q2
√
v − w
dw
dt = q2
√
v − w − q3
√
w
with the initial conditions u(0) = 2.5, v(0) = 0.4, and w(0) = 0.2, where the coefficients are given by
q0 =
Q0
W L
q1 =
Cd1A1
√g
W L
q2 =
Cd2A2
√g
W L
q3 =
Cd3A3
√g
W L
(a) Compute H1(t), H2(t), and H3(t) using Euler method with a step size of h = 60 s, to t = 240 s. The exact
solution for t = 240 s is H3 = 0.4217455830 m. How does Euler method compare?
(b) Repeat, using RK2 method, again with a step size of h = 60 s. How does RK2 method compare to the exact
solution for t = 240 s?
(c) Repeat, using RK4 method, again with a step size of h = 60 s. How does RK4 method compare to the exact
solution for t = 240 s?
(d) Using RK4 method with a step size of h = 30 s gives H3 = 0.4217425160 m at t = 240 s. Without performing
the corresponding RK4 computations, compare the amount of error reduction to that (theoretically) expected
for RK4, gained by halving step size.
Tank system for above problem:
W
H1
– H3
H2
– H3
Cd1 & A1 Cd2 & A2 Cd3 & A3
W W
Q0
5.3 non-linear induction circuit
Many chaotic systems were studied for decades (or even centuries) before chaos was identified as a characteristic of
dynamic systems. The simplest such systems are known as jerk systems, because they contain third derivatives in
time, and in mechanics the time derivative of acceleration is associated with jerk forces. Typically, simple chaotic
systems oscillate with frequencies that are not clearly defined – in a phase plot this can look like a slightly offset or
blurry repeating loop.
Consider a mechanical system described by the third order ODE
...
x +¨x
3 + 2.97x
2x˙ + 0.49x = 0
where ...
x= d
3x/dt3
, ¨x = d
2x/dt2 and ˙x = dx/dt (the dimensionless parameter x is proportional to position, where
both position and time have been made dimensionless here).
From the initial conditions, x(0) = 0.08, ˙x(0) = 0.00, and ¨x(0) = 0, plot the first 1000 steps, in the coordinates
(x, x˙), taking steps of h = 0.10 in RK4 method. Plot the coordinates after each step as discrete points, not as a
line or curve. The resulting plot shows distinct patterns that are relatively unique to your particular problem, and
in most cases the result is relatively chaotic (rather than periodic). If your solution diverges to infinity, then it is
incorrect — check for syntax errors in your method.
5.4 heat transfer within a rectangular block
T = T1 T = T2
T = T1
T = T1
H
W
insulated T = T2
T = T1
T = T1
H
W
case 1:
case 2:
1
One surface of a copper block in a heat exchange assembly is heated to a temperature T2, above the ambient
temperature T1. The remainder of the assembly is either open to ambient (case 1) or is insulated on one side (case
2). The entire block is shown above, where the surfaces are either insulated or open to heat sources. The contact
surface is the insulated surface on the left. The third dimension can be ignored here, as heat transfer perpendicular
to the illustrated plane is negligible. Therefore the steady system is defined by the ODE
∂
2T
∂x2
+
∂
2T
∂y2
= 0
The block is W = 4.8 cm wide and H = 4.8 cm high. For this problem, divide the block into four sections across by
four sections up, as shown below:
x1
x2
x3
x4
x5
y1
y2
y3
y4
y5
There is no heat flux on the insulated side, so that the boundary condition there is
∂T
∂x
x=0
= 0
The right side of the block is maintained at fixed temperature T(W, y) = T2 = 205◦C, so that T(W, y) = 205. In
case 1, the other three sides of the block are all at the ambient temperature, T1 = 12◦C: T(x, 0) = 12, T(0, y) = 12,
and T(x, H) = 12. In case 2, the other three sides are insulated or ambient as illustrated: T(x, 0) = 12, ∂T /∂x|x=0 =
0, and T(x, H) = 12.
(a) There are unknown temperatures at nine nodes — write out (by hand) nine explicit equations, one for each
node, in terms of only the nine unknowns and explicit temperatures.
(b) Organize the equations into a matrix problem (you may use Matlab, Excel, or another program for this; most
of the elements in the matrix will be zero).
(c) Solve the matrix problem for the unknown temperatures — write them out explicitly in the form T(xi
, yj ),
using the coordinates specified above. As a check, the solutions at T(x3, y2) are 48.19◦C for case 1, and 53.66◦C
for case 2.
(d) Compare the two sets of results and note that insulating surfaces increases the influence of the heated surface
(by effectively providing fewer routes for the heat to escape): you should find higher temperatures at every one
of the nine nodes in case 2 (relative to case 1)

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