CH E 314 Midterm 1 (2019) PDF

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ChE 314 Heat Transfer Winter 2019 Midterm Wednesday, Feb. 13, 5:00-6:50 pm ETLC E1-007 Please remember to 1. Count the pages. There must be 11 in total. 2. Read the statement carefully, and make sure you understand what is asked 3. Provide a direct answer to the questions asked 4. Make clear schematics, diagrams, and graphs: a. arrows and dimensions must point precisely to the intended locations b. arrows and dimensions cannot finish too far or point inaccurately c. straight, parallel, and perpendicular lines must be drawn as such 5. Express the results of calculations clearly and in the units requested 6. Write clearly the steps taken in calculations or derivations. State explicitly the fundamental principles or definitions used. The questions are very direct. Typically, if you know the answer it will take you little time to know what to answer. If you feel unfamiliar with the question, moving on to other familiar questions and leaving the unfamiliar to the end might be a good strategy.

Total%points:% 1.

Consider a composite wall made of a layer of brick of 10 cm thickness covered by a layer of stucco of 0.6 cm thickness. The wall separates air at different temperatures: T∞,1=25℃ inside and T∞,2=-10℃ outside. The cross sectional area of the wall is 15 m2. The convection coefficient inside is h1 =10 W/m2K, and outside is h2 =100 W/m2K. Thermal conductivity of brick is kbrick =0.64 W/mK and that of stucco is kstucco =0.73 W/mK. Neglect the effect of radiation. 1.a

Calculate the heat flux through the composite wall in W/m2. (7 pt.)

Answer (number and units):

% 2

1.b

Calculate the overall heat transfer coefficient in W/m2K. For this question, assume the heat flux is 120 W/m2. (2 pt.)

Answer (number and units):

1.c

%

Calculate the temperature at the surface of the stucco layer in ℃. Assume the heat flux is 120 W/m2. (2 pt.)

Answer (number and units):

% 3

2.

Consider a large regular Tim Hortons coffee (𝑇" = 80℃) sitting in a room with cooler air at 𝑇&'( = 𝑇∞ = 20℃. The thickness of the paper cup is 0.4 mm. Assume the cup is a perfect cylinder of 80 mm diameter and 150 mm height filled all the way to the top and without a lid. The table is thermally insulating (adiabatic) and absorbs no heat. Neglect heat losses by evaporation. The natural convection coefficient to the outside is h=10 W/m2K, and the thermal conductivity of the paper cup is 𝑘+,+-( = 0.05 W/mK. The external area of the top is 50.27 cm2, and the external area of the side is 377.0 cm2. 2.a

Considering separate paths for heat flow for the top and the side, make a schematic of the equivalent circuit to study heat transfer between the hot coffee inside and the environment. Include convection, radiation, and conduction in the schematic where it exists. Indicate the coffee and surrounding temperatures in the equivalent circuit. (8 pt.)

4

2.b

Estimate R”cond for the paper cup side. Discuss relevant approximation made or that could be made for this case. (3 pt.)

Answer (number and units):

% 5

2.c

Simplify the equivalent circuit of 2.a by considering only external convection. Make a schematic of the simplified equivalent circuit indicating relevant resistances. Indicate the coffee and air temperatures in the equivalent circuit. Determine if there is a dominant thermal resistance and identify it. (7 pt.)

6

3.

Consider a 4-cm thick wall with one side maintained at Ts,1 = 60℃, the other side at Ts,2 = 20℃, and all its edges insulated. The wall is in steady state and there and no heat is generated within the wall. 3.a

Draw the isotherms and adiabats corresponding to this problem. Explain briefly the reasoning behind the lines you drew. (5 pt.)

Insulated

Ts,i ? Ts,1=60 ℃

Ts,2=20 ℃

12mm

0

x

Insulated 4cm

7

3.b

Draw the temperature profile through the thickness of the wall. Identify relevant temperatures and locations. (4 pt.)

8

4.

Determine the shape factor S for heat transfer through the wall of a spherical shell with inner radius r1 and outer radius r2. Hint: Use the tabulated thermal resistance of a sphere. (3 pt.)

9

5.

Determine the heat exchange rate in W between a steel sphere of diameter D=8 cm at temperature T1=40℃ buried 2 m in soil with a surface at temperature T2=20℃. Thermal conductivity of soil and the steel sphere are ksoil=0.5 W/mK and ksteel= 60 W/mK. (5 pt.)

Answer (number and units):

% 10

6.

The coils of induction heaters generate heat by electrical resistance that must be absorbed by a cooling fluid. Consider an induction heating coil as a long tube with a uniform heat generation q’’’ inside the material of the tube wall. The outer radius r2 is insulated, and the inner radius r1 contains cooling fluid at temperature T∞ and convection coefficient h. Consider that the thermal conductivity of the tube k is constant and isotropic and the tube is in steady state.

r1

q'''

h r2

6.a

State the appropriate heat diffusion equation by applying the relevant simplifications to the general case. (5 pt.)

6.b

For the simplified case, determine the independent variables, dependent variables, type of equation, boundary conditions, and independent parameters. (9 pt.)

11

Formula Sheet ChE 314 Heat Transfer, Winter 2019

Energy Balance Qin − Qout + Qgen = Qst qin − qout + qgen = qst qloss = m ˙ (i2 − i1 )

Qst |21 = m2 i2 − m1 i1 d(mi) qst = dt di c= dT

Heat Transfer Mechanisms qx′′ = −k

dT dx

′′ qconv = h (Ts − T∞ ) ′′ qcont = hcont (Ts,A − Ts,B )

∂T ∂T ∂T ′′ ) qcond = −k▽T = −k(i +j +k ∂z ∂x ∂y   4 4 q ′′rad = εσ Ts − Tsur

σ = 5.670 × 10−8 W/m2 K4

Heat Diffusion Equation Cartesian coordinates ′′′ q gen ∂T = α∇2 T + ρc ∂t Cylindrical coordinates     ′′′ qgen ∂T 1 ∂ ∂T 1 ∂2T ∂2T + =α r + 2 2 + ρc r ∂r ∂z 2 ∂t ∂r r ∂φ

α=

k ρc

Spherical coordinates      q ′′′ 1 ∂T ∂T 1 1 ∂ ∂2T ∂ ∂T gen + sin θ r2 + 2 2 =α 2 + ∂t ∂r r 2 sin θ ∂θ r ∂r ρc ∂θ r sin θ ∂φ2 Thermal Resistance ∆T q R′′ = AR

R=

ln (r2 /r1 ) Rcond,cylinder = 2πLk 1 Rconv = Ah   2 hrad = εσ T 2s + T sur (Ts + Tsur) Rfin =

1 Afinηfinh

Rseries = R1 + R2 1 1 R= Sk As q = k q ∗ss ∆T Lc

R′ = LR L Ak (1/r1 ) − (1/r2 ) = 4πk

Rcond,wall =

Rcond,sphere 1 Rrad = Ahrad 1 U= Aref Rtot 1 Rarray = Aη0 h   −1 −1 Rparallel = R−1 1 + R2 qcond = kS ∆T r As Lc = 4π

Figure 1: Conduction shape factors and dimensionless conduction heat rates for selected systems.

∗ Figure 2: Dimensionless conduction heat rates [q = qss kAs (T1 − T2 )/Lc Lc = (As /4π)1/2 ]

Figure 3: Efficiency of common fin shapes.

Figure 4: Efficiency of straight fins (rectangular, triangular, and parabolic profiles).

Figure 5: Efficiency of annular fins of rectangular profile....