Derive the differential equation for the transient temperature field of an infinitely long
cylindrical and isotropic body taking into account local changes of the conductivity λ . The
density ρ and the specific heat capacity c remain constant. There are heat generation
sources Φ "' in the body. Their intensity varies with their position in the body .
The so-called Poens device is used to determine the conductivity of solids of up to 2.5
W/m ⋅K . The measurement is done on the basis of the equations for the steady-state one-
dimensional conduction in plates.
The device for single plates is setup as follows:
The plane probe is placed between two copper plates. One is cooled down to a homogenious
and constant temperature,the other is electrically heated with constant power. In order to
achieve one-dimensional heat flow and to let the entire power of the heated plate pass through
the probe, the system of probe and plate is surrounded by edge bands and auxiliary heating to
minimize heat loss. To control the auxiliary heating, thermocouples are placed in the insulated
space between the auxiliary heating and the heating plate. As well as on the probe and the
edge strips. The auxiliary heating is adjusted properly when the thermo couples that lie in the
same plane give equal values. The driving temperature gradient between the upper and lower
side of the probe is also measured by thermo couples.
heat source ( electric heat source or hot water flow)
. .. . .. . .. . .
. . . . Hot plate, Qel
. . . .
the probe material
Probe plate ( circular, approx. 10 cm in
diameter or square with
30 x 30 cm 2 or 50 x 50 cm 2 )
Measurement data: geometrical data δ Probe ,A Probe
temperatures Tup , Tlow
a) What is the measurement error caused by an air gap of δ A = 0.05 mm between both the
probe and the cooling plate and the probe and the heating plate, if the thermocouples show
the temperature of the heating or cooling plate ?
1. for concrete: λ = 1.2 W/m ⋅ K
2. for cork: λ = 0.051 W/m ⋅ K
The thickness of the probe is δ P = 20 mm , the conductivity of the air in the air gaps is
λ L = 0.026 W/m ⋅ K .
Hints: Determine the ratio of the conductivity of the probe without the air gap ( λ P ) and
the conductivity when the air gap is taken into account λ P,Corr = λ P,actual . )
b) Name other possible sources of measurment errors and suggest ways to avoid them.
Both sides of a plane wall are kept at a constant temperature T1 and T2, where T1 > T2 .
Describe qualitatively the temperature profile for steady state conditions, if,
a) the conductivity remains constant and
b) the conductivity is a function of the temperature according to the equation
λ = λ 0 (1 + γ (T − T0 )) with γ > 0 ( λ 0 = conductivity at the reference temperature T0).
Explain the sketched profiles.
The measurement of the surface temperature of a plane wall, consisting of a layer of material
A with thickness δ A = 12.5 cm and material B, thickness δ B = 20 cm , yielded a temperature
of TA = 260°C at the outer side of layer A and TB = 32°C for layer B After applying an
insulation layer of thickness δ ins = 2.5 cm (λins = 0.075 W/m ⋅ K) at the outer surface of layer
B, the following temperatures were measured:
Surface temperature of layer A TA * = 305°C , temperature of the contact surface between
layer B and insulation layer (formerly the outer surface of layer B) TB * = 219°C, surface
temperature of the insulation layer Tins = 27°C.
Calculate the heat flux per unit area q" through the wall for the steady-state with and without
The wall thickness of a furnace is to be minimized for the steady state case, considering the
following steady-state boundary conditions: The material for the wall can be chosen from the
table below. The outer surface of the oven should consist of a 8 mm thick steel plate acting as
a mechanical shield. The surface temperature at the outer surface of the furnace wall (steel
plate) must not exceed TSt = 60°C , and that of the inner surface is Ti = 1000°C. The allowed
maximum heat flux loss is q ′′ = 1.5 kW/m 2 .
The following table presents two cases (A) and (B) for the possible material combinations.
Conductivity λ [ W / mK ]
Material Allowed Case (A) Case (B)
temperature 60°C 800°C 1200°C 60°C 800°C 1200°C
Firebrick 1500°C 0.9 1.3 1.5 0.9 1.3 1.5
Refractory brick 1200°C 0.4 0.6 0.7
Insulating stone 800°C 0.2 0.25
Steel 45 45
Determine the composition of the wall and the thickness of each layer for both cases, aming
for the lowest possible total thickness.
Calculate the total heat lost through the outer surfaces of the sketched apartment at an outer
temperature of Ta = −15°C and a room temperature of Ti = 20°C. Consider one-dimensional
heat flow. The temperatures of the neighbouring rooms and their dimensions (in cm) are
shown in the diagram. The structure of the walls and the ceilings as well as the conductivities
of the building materials are listed in the table.
Heat transfer coefficients (both convection and radiation superimposed):
Outer side of the room: α a = 20 W / m 2 K
⎧8 W/m 2 K
⎪ wall surfaces, windows, ceiling if heat flow is directed upwards
αi = ⎨
⎪6 W/m K
⎩ ceiling if heat flow is directed downwards
Heat transfer coefficient of the wooden double-paned window:
kF = 2.9 W / m2K
cm W 20˚C 20˚C
plaster 1,5 0,87 6˚C
thermal insulation 2,0 0,09 20˚C
brickwork 34 1,05
plaster 1,5 0,87 500
plaster 1,5 0,87
thermal insulation 2,0 0,09
concrete ceiling 20 1,40
screed 4 0,70 20˚C 20˚C
wooden floor 1 0,2
all dimensions im cm x 150
A hollow cylinder with inner radius ri , outer radius ra and length L is heated in such a way
that its inner and outer surfaces reach a constant and evenly distributed temperature Ti and To,
respectively. The cylinder material has a conductivity that changes as a function of
temperature according to the following equation:
λ = λ o (1+γ ( T-To ) )
λ o = conductivity at reference temperature To .
a) Derive an equation for the heat flow rate through the wall (mantle) of the cylinder.
Compare this equation to the one for the case of constant conductivity. At which mean
temperature Tm would the conductivity λ have to be introduced in the equation for the
heat flow rate for constant conductivity, if this λ was to be used to calculate the heat flow
rate for the case of a linear temperature-dependent conductivity?
b) State the equation describing the temperature distribution in the hollow cylinder.
Brine (salt water) at an average temperature of TS = −20°C flows through a room in an
insolated pipe made of steel with inner diameter di = 50 mm and wall thickness δ = 5 mm .
The room temperature of the air and that of the surfaces enclosing the room is TR = 20°C .
Heat transfer coefficients:
inner side of the pipe α i = 2300 W / m 2 K
outer side of the insulated pipe αa = 6W / m2 K
steel λ R = 54 W / m K
insulation λins = 0.042 W / m K
a) Determine the thickness of the insulation δins so that at the surface of the insulation, even
at the maximum dew temperature of the surrounding air of Tτ = 15°C , the water does not
b) Calculate the heat flow q ' that the brine absorbs per unit pipe length and unit time under
Water with an average temperature of TW = 80°C , flows through a copper warm-water pipe
placed in a room of TR = 20°C . The copper pipe ( λ = 372 W/m ⋅ K ) has an inner diameter of
di = 6 mm and wall thickness δ=1 mm.
Heat transfer coefficients:
inner surface of pipe αi = 2300W / m 2 K
outer surface of pipe αa = 6 W / m2K
a) Calculate the heat transfered q ε ' per unit pipe length m for
- non-insulated pipe, and
- an insulated pipe with a cork layer of thickness t = 4 mm (λ = 0.042 W/m ⋅ K) .
b) Sketch qualitatively the profile of the heat emission qε ' as a function of the insulation
thickness for different conductivities of the insulation material and explain the underlying
c) What is the value of the conductivity λins of the insulation material, if a general reduction
of heat loss is to be achieved?
Assumption: Changes of the outer heat transfer coefficients as a function of the outer diameter
are not to be considered.
A cork-insulated brine pipe is fastened to the ceiling by steel bands welded on the pipe. The
outer wall temperature of the brine pipe is Ts = −23,5°C , the room temperature TR = 20°C.
outer diameter of the pipe d o = 50 mm
thickness of the insulation δins = 40 mm
cross section of the steel bands A c = 25 mm×6 mm
length of the iron strips L = 290 mm
Heat transfer coefficients at the surface of the steel bands: α = 6 W / m2 K
conductivity of the steel bands material λ = 58 W / m 2 K
a) Calculate the heat Q absorbed by the salt water from one iron strip?
b) Up to what height h0 does frost form on the steel band ( h0 is the distance from the surface of
the brine pipe ), if the steam content of the air in the room is above the saturation vapour
pressure for the maximum steel band temperature?
- Homogeneous temperature distribution in the cross section of the steel band
- Heat transfer from the iron strips to the ceiling and from the strips to the pipe
insulation is negligible.
A fin of rectangular cross-section A Q = δ i L and length L is to be designed to maximize the
amount of heat transfered through it for a given amount of material m (mass)
a) Determine the equation relating the fin height h and the thickness δ , so that the above
condition is fulfilled. Temperature differences in the cross section of the fin and heat
transfer at the top of the fin can be neglected.
b) Discuss the advantages and disadvantages of a constant fin cross section.
c) What are the heights of the optimized rectangular fin of thickness δ =1 mm, if the fin is
- steel λSt = 58 W / m ⋅ K
- aluminum λ Al = 229 W / m ⋅ K
- copper λ Cu = 372 W / m ⋅ K
and the heat transfer coefficient at the surface of the fin is α = 120 W / m 2 K ?
d) The following temperatures were measured on a fin exposed to air flow, with air
temperature TA = 20°C
fin base: TB =155°C ;
fin head: TH = 110°C
Investigate whether the fin design is correct with respect to the material needed. If this is
not the case, state whether the height of the fin should be shortened or extended.
To avoid reaching the dew point temperature at the inner side of the rear window of a car, thin
electrical heating wires are used. These can be utilized at low outer temperatures to add so
much energy to the rear window, that the dew temperature will not be reached at any point on
Determine the necessary heating power per unit conductor length q 'H needed to avoid
reaching the dew point temperature under the following conditions:
Rear window data:
thickness δ = 5 mm
distance between the heating wires t = 30 mm
conductivity λ = 1.16 W/m.K
Air temperatures and heat transfer coefficients:
outer side of the window: To = 5°C
αo = 30 W / m 2 ⋅ K
inner side of the window: Ti = 20°C
αi = 3 W / m 2 i K
Allowed minimum temperature of the window = dew point temperature
Tτ = 13°C
t Heating wire
Hints and instructions:
1. Steady-state conditions
2. Boundary influences are neglected. Due to symmetry , it is enough to consider a section of
the window with two heating wires– ref. to diagram.
3. The problem is to be considered one-dimensional. Therefore only temperature changes in
the x-direction are to be considered.
4. Homogeneous heat flow rate over the thickness of the window δ .
5. The origin of the coordinate system is to be placed between two heating wires.
a) State the differential equation of a steady-state temperature distribution in a thin circular
fin with varying thickness δ (r), assuming constant conductivity λ of the fin material and
constant heat transfer coefficient α at the surface of the fin. The surrounding temperature
TA is constant and known. The arc element ds of the fin flank may be substituted by a
radial element dr for simplicity. The temperature differences in the cross section of the fin
are assumed negligible.
δ (r) λ
b) Under the given conditions, determine the profile δ (r) of a thin-edged circular fin, with
even thermal loads, i.e. constant heat flux q" in the radial direction. This form of a fin
requires minimum material for a pre-defined heat transfer power.
The inner diameter of the fin is 2 ⋅ ro , the outer 2 ⋅ R.
The tip of the fin has the same temperature as the surroundings.
c) What is the heat that the fin, designed as given in b), can transfer if the temperature at the
base of the fin is TB ?
The following figure shows a section of a double-walled cylindrical container. The walls of
the container are made of stainless steel and are interconected by ligaments of the same
material as shown in the fig. below. The distance between two ligaments is so large, that we
can assume there is no influence between them. The space between the two walls is filled with
wool. The container is filled with a fluid of constant temperature. The surrounding
temperature is also constant .
d δW W
inner diameter of the container d = 3mm
thickness of the walls of the container δ W = 3mm
thickness of the ligaments δ t = 1.5 mm
thickness of the insulation δins = 80 mm
conductivity of the container
and ligament material λ C = 16 W/ mK
conductivity of the insulation λins = 0.04 W/ mK
Heat transfer coefficients:
outer side of the container αo = 8 W / m2 K
inner side of the container αi = 400 W / m 2 K
liquid TL = − 40°C
air TA = 20°C
a) Derive an equation that can be used to calculate the heat transferredvia the ligament to the
liquid. The heat exchange between the ligament and the insulation can be neglected. The
radial thermal resistance of the container walls may also be neglected.
What is the heat flow rate to the liquid under the given conditions?
b) How great is the temperature decrease TC,o − TC,i between the outer and inner wall of the
container at the ligaments?
c) What is the minimum distance between the ligaments, so that the assumption of a ‘no
influence between the ligaments remains valid? Criteria for the evaluation should be the
decrease of the temperature difference between the container wall and the surroundings.
The field of influence of one ligament is to be limited to the point where the temperature
difference has been reduced to 1% of the maximum.
d) Estimate the total heat flow rate to the liquid and hence the necessary cooling power per
meter container length needed to preserve the temperature of the fluid, if the distance
between two ligaments is 1.5 m. Assume that the heat flow rate through the walls of the
container is not influenced by the ligaments.
The fuel of a reactor fuel rod for high-temperature reactors is stored between a graphite sphere
of 25 mm outer diameter and another graphite hollow sphere with 60 mm outer diameter and
31 mm inner diameter. The fuel rod has a heating power of Q R = 2.5 kW.
Conductivity of graphite: λ G = 126 W/mK (electrode graphite), conductivity of the fuel layer
λ F = 12 W/mK.
What is the temperature difference between the center of the fuel rod and its surface?
Both ends of a copper rod with length 2 L and a diameter d are kept at the same temperature
To . The left half of the rod is totally insulated against radial heat loss. An electric current
generates Joule’s heat of density Φ "' inside. Air of temperature TA flows around the non-
insulated right side, where a heat transfer coefficient α is given. The conductivity of the rod
is λ .
a) Derive the equation for the temperature profile in the rod.
b) Determine the value of Φ "' so that the temperature in the middle of the rod equals the
temperature To at its ends.
c) Calculate Φ "' using the following data:
L = 1 m; d = 5.2 mm; To = 120°C; TA = 100°C;
α = 6 W/ m K λ = 372 W/mK for the conditions of part b).
d) Determine the extremes of the temperature distribution for the given values giving their
position and value and also sketch the temperature profile.
Hint: Place the origin of the coordinate system at the center of the rod.
An explosive, which is stored in form of a plate, is surrounded by a shielding and insulation
mantle and placed on an adiabatic base. The conductivity of the explosive is λ = 0.85 W/mK,
the overall heat transfer coefficient between the surface of the explosive and the surroundings
TS is k = 0.2 W/ m 2 K .
The inevitable reaction rate of the explosive causes steady heat production which depends on
the temperature as given by the following relationship:
Φ "' = ΦS ⋅ (1 + γ (T − TS ))
T = temperature of the explosive ΦS = 0.3 W/ m3
TS = surrounding temperature γ = 0.2 1/K
Determine the thickness of the explosive layer tcrit that will certainly lead to explosion?
Instructions: Drive the equation for the steady state temperature profile in the explosive for
an arbitrary layer thickness s. Explosion occurs only if the temperature is allowed to increase
TS surface of explosive
t λ, Φ"'
A copper sphere of diameter d = 1 cm and initial temperature To =15°C is suddenly placed in
an airflow with temperature TA . The heat transfer coefficient at the surface of the sphere is
α = 50 W/m 2 K. The specific heat capacity of copper is c = 0.419 kJ / kg K and the density
ρ = 8300 kg/ m3.
a) After what time interval does the temperature of the sphere rises to TS = 17.5°C, if the air
temperature remains constant at TA = 20°C?
b) There is a linear temperature rise with time at a rate of κ = 360 K / h with a starting value
of TAo = 20°C
What is the temperature difference between the air and the copper sphere TA − TS after
Sketch the temperature of the air and the copper sphere as a function of time.
c) The temperature of the air oscillates as a sine wave with the amplitude ΘA max = 5 K and
a frequency of f = 1/6 min −1 (angular frequency ω = 2πf ) around a mean value of
TAm = 20°C.
Derive an equation for the temperature of the copper sphere as a function of time.
Determine the amplitude of the temperature oscillation of the sphere ΘSmax .
Find the time lag between the temperature oscillation of the sphere and the oscillation of
Sketch qualitatively the temperature of air and the sphere as a function of time.
Water of temperature T0 = 20˚C flows in a cylindrical container of total volume
Vtotal = 27 m3 at a rate of V0 = 9 dm 3 per minute. At time t = 0, the container is empty. The
water in the container is mixed by an agitator so that the temperature distribution in the
container is homogeneous. A coiled tube with surface area A contains water vapor to heat the
contents of the container. The relevant surface area for the heat transfer is proportional to the
current amount of water in the container. The coiled tube has n = 10 coils and a diameter of
D = 1.2 m. The outer diameter of the tube is d = 25 mm, the saturation temperature of the
vapor is TS = 105˚C and the heat transfer coefficient of the coil (on the water side) is
α = 600 W / m 2 K.
a) Derive the equation for the water temperature as a function of time.
b) Find the water temperature TWe , when the container is full?
Saturated steam at TS
N∼0 . Water
. A(t) Q=0
The following can be assumed:
1. The vapor temperature in the coiled tube remains constant.
2. The substance properties can be considered to be independent of the temperature.
3. The energy added by the agitator is negligible.
4. The heat transfer coefficient is constant.
5. The thermal resistance of the coiled tube is negligible.
6. The walls of the container are adiabatically insulated.
The volume VC of a compensation container (for the compensation of temperature
fluctuations) is to be determined. A flow of V = 1 m3 / h enters the container at a temperature
Ti oscillating in a sine-wave manner. The period of oscillation is t S = 0.25 h, the mean
temperature Tm and the amplitude Θi max . The volume of the container is to be designed in a
way that the maximum temperature fluctuation Θo max at the outlet is only 0.1 % of the
maximal temperature fluctuation Θi max at the inlet.
VC To (t) = T(t)
The following may be assumed:
1. The fluid in the container is mixed by an agitator so that there are no temperature
differences in the container.
2. The added energy by the mixer is P. A heat flow Q W enters the container from the
3. Steady-state-like conditions are to be considered, i.e. after a time interval, when the
oscillation has reached its final state( t → ∞ ).
Hint: A particular solution of the linear differential equation with constant coefficients can be
obtained by using a complimentary function in for of the perturbation term:
Θp = k1 sin(ωt + ϕ) + k 2
The following sketch shows the structure of a simple electric night-storage heater . The heater
consists of a core storage unit, which is surrounded by an insulating layer (thickness δins ,
conductivity λins ). An electric heating element (heating power Qel ) is placed in the storage
element (mass of the storage unit including the heating element is mC , specific heat capacity
α (for the whole
surface of the oven)
heating element . .
. δ ins
A = surface of the oven
(without the base surface)
a) The oven is operated in cycles. Determine the core temperature TC ( t ) as a function of
time (final state) during the heating and cooling cycle, if the warm-up time of the oven is
t L , and during this period the storage unit is charged with constant electrical power Qel .
The total cycle time is t g = t L + t E = 24 h (t E = cool-down time). The room
temperature is TR , the material properties cK , λins and the heat transfer coefficient α at
the surface are known and assumed constant over all cycle periods. mK, A and δins are
The following assumptions are made for the calculation:
1. The thermal resistance of the storage unit including the heating element is negligible
compared to the other thermal resistances, so that no temperature differences occur in
the storage unit.
2. The capacity of the insulating layer is approximately zero.
3. Plane heat flow in the insulating layer is to be considered, and the surface of the
storage unit A (axcluding the base surface) is to be taken as the relevant surface area
for the heat flux.
Instructions: Set the start of the charging process to t = 0!
b) Sketch the qualitative core temperature profile for 2 cycles.
c) What charging time t L is necessary, if the required daily (ie per one complete cycle)
energy is QH = 13.5 kWh and the power of the heating element is Qel = 1.5 kW?
What is the temperature difference between core and room at the beginning and ending
phase of a charge cycle?
m K = 90 kg α = 10 W/m 2 ⋅ K
c K = 1.2 kJ/kg ⋅ K λ ins = 0.08 W/m ⋅ K
δins = 0.04 m
A = 0.8 m 2
d) Determine the percentage of the heat transfered during one charging and one discharging
cycle in relation to the daily energy transfered QH .
In order to cool a fluid of mass m F = 0.5 kg in an adiabatic container, n = 10 identical thin
plastic spheres are filled with ice at TI = 0°C and dropped in the fluid. The spheres have a
diameter of d = 20 mm and a wall thickness of δ = 0.5 mm. The initial temperature of the
fluid is equal to the room temperature TA = 20°C. The constant, mean heat transfer coefficient
between the surface of the sphere and the fluid is α = 400 W/m 2 K.
α n=10 spheres
δ d ρ. c = 0
fluid ≡ water
density ρF = ρW = 1000 kg / m3
spec. heat capacity cF = cW = 4180 J / kg
density ρE = 917 kg / m3
spec. melting enthalpy r = 333 . 103 J / kg
plastic (sphere material)
conductivity λ = 0.2 W / m K
a) Determine the temperature of the fluid and the temperature of content of the spheres.
- at time t m , after all the ice has melted ( TFm , TKm ) , and
- after a very long time (TF∞ , TK∞ ) ?
b) Find the equation for the temperature of the content of the spheres TSp (t) and the fluid
TF (t) , starting at the point of dropping the spheres in the fluid t0, until the time when all
the ice has melted.
c) After what time t m has the ice completely melted?
d) Formulate the differential equations for the temperature of the fluid and that of the content
of the spheres for the time interval t m ≤ t ≤ ∞ and state the required initial conditions.
e) Determine the equation governing the temperature profile of the fluid in the spheres for
the time interval t m ≤ t ≤ ∞ .
f) Sketch the temperature as a function of time in the time interval 0 ≤ t ≤ ∞ .
The following assumptions may be used:
1. Temperature differences inside the fluid and in the plastic balls are rapidly recreasing.
2. The heat transfer resistance at the inner side of the plastic balls is negligible.
3. The heat capacity of the spherical plastic balls and that of the liquid container may be
4. Since δ < < d , the curvature of the spheres can be neglected.
5. During the melting process the content of the spheres is a mixture of water and melting
Two bodies of semi-infinite extension with different material properties λ, ρ, c and different,
yet standardized temperatures T10, and T20 are suddenly put in contact (contact resistance = 0).
a) How does the temperature change as a function of the location and time in both bodies?
b) Which temperature TM is reached in the contact plane?
After the forming procedure a rolled sheet of steel of width W and thickness 2 ⋅ δ , material
properties λ, ρ, c , is driven through an oil bath of constant temperature TB with a constant
velocity v. Before entering the bath, the steel sheet has an evenly distributed temperature T0.
The heat transfer coefficient α between the surface of the sheet and the oil, is constant over
the entire submerged length L. The front sides of the sheet are to be considered adiabatic.
2. δ v
Calculate the velocity of the sheet v, if the temperature difference between the center of the
steel sheet and the oil bath TS − TB is reduced after exiting the bath to half the value at the
entrance. The following two cases shall be considered:
a) - Heat conduction in the x-direction is negligible.
- The temperature over the cross section of the sheet 2δB is constant.
b) - Heat conduction in the x-direction is negligible.
- Temperature distribution in y-direction is to be considered (using the following diagram)
The curvature of the steel sheet may be neglected in both cases.
Dimensions: W = 1 m, 2 δ = 20 mm, L = 5 m
Heat transfer : α = 400 W / m2 K
Material properties: λ = 40 W / m K, c = 0.5 kJ / kg K, ρ = 8000 kg / m3
0,1 0,5 1 5 10 50 100 500 1000
A homogenous rotationally symmetric volume with mass m1 = 60 g is suddenly submerged
in a liquid bath of constant temperature TB . The submersion time is t1 = 4 sec .
a) A body of similar geometry made from the same material with equal initial temperature
and mass m 2 = 80 g is submerged in the same bath. After what time t 2 does it reach the
same temperature in the center of the body, if the heat transfer between liquid and body is
very good ( α = ∞ ) ?
b) Consider the case of finite heat transfer ( α ≠ ∞ ). What is the condition for the heat
transfer to yield similar temperature fields?
A concrete wall of thickness δ = 0.5 m has a sine-shape temperature distribution at time t = 0
with surface temperatures T(x = 0) = T(x = δ ) = TSo = 20°C and in the center of the wall
T(x = δ/2) = TMo = 100°C.
Find the temperature distribution 1 hour later. Determine the temperature TM at the center of
the concrete wall at that time, if the surface temperatures of the wall are kept at TO = 20°C.
Material properties of concrete: conductivity λ = 1.4 W / m K , specific heat capacity
c = 0.88 kJ / kg K , density ρ = 2300 kg / m3.
Two plastic plates, each with thickness δ = 3 cm, are to be glued together by a thermoplastic
glue. The glue operates at a minimum temperature TM = 118°C. The plates of initial
temperature T0 = 20°C are fixed in a press with an upper clamp temperature of TS1 = 230°C
and a lower clamp temperature of TS2 = 200°C . The temperatures of the clamps are constant.
After what time is the required temperature of 118°C reached at the gluing point?
The problem is to be solved using the Binder-Schmidt method. A step width of Δx = 1 cm
and a magnifying factor of 2 : 1 is to be selected.
Material properties of the plastic: λ = 0.58 W / m K ρ = 1300 kg / m3
c = 1.12 kJ / kg K
A room is kept at a constant temperature of Ti = 20°C and separated from the atmosphere by a
concrete wall of thickness δ = 20 cm. After a certain point in time, the wall is exposed to the
sun for 2 hours, during which the heat absorbed per unit area is q" = 600 W / m 2 .
What are the surface temperatures TWi and TWo of the wall 1 hour after the exposure to the
sun. Before the exposure the wall had a uniform temperature of To = 20°C and the
temperature of the atmosphere during this period remains constant at Ta = 20°C. The heat
transfer coefficients are constant α o = 20 W / m 2 K , and αi = 8 W / m 2 K at the outer and
inner side of the wall, respectively.
temperature conductivity of concrete: a = 0.0025 m2 / h
heat conductivity of concrete: λ = 1.2 W/m K.
Hint: Use the Binden-Schmidt difference method and choose a step width of Δx = 5 cm.