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First Semester Examination 2021/2022 Academic Session

February/March 2022

EMH 441 – Heat Transfer

Duration : 3 hours

Please ensure that this examination paper contains SEVEN (7) pages and FOUR (4) question before you begin the examination.

Instructions : Answer ALL FOUR (4) questions.

Answer all questions in English OR Bahasa Malaysia OR a combination of both.

Each question must begin from a new page.

...2/-

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1. a) A steel pipe (k = 45.0 W/mK) having a 0.05 m outer diameter is covered with a 0.05 m thick layer of magnesia (k = 0.07 W/mK) which in turn covered with a 0.02 m layer of fiberglass insulation (k = 0.048 W/mK).

The pipe wall outside temperature is 95 oC and the outer surface temperature of the fiberglass is 30 oC. Calculate:

(i) the interfacial temperature between the magnesia and fiberglass.

(ii) the steady state heat transfer.

(40 marks) b) A large steel frying pan of thickness 0.5 cm, initially at 20 oC, is placed

on a stove. The bottom of the pan is subjected to a uniform heat flux 300 W/m2 and the top is exposed to cool ambient at 20 oC. The heat transfer coefficient between the pan and the ambient air is 30 W/m2K. Calculate the temperature of the pan at 5 minutes and 10 minutes after the heating starts (For steel, thermal conductivity k = 70 W/mK, density ρ = 7840 kg/m3 and heat capacity Cp = 450 J/kgK).

(60 marks) 2. Water enters a tube with diameter of 0.01 m. The velocity (m/s) and temperature

(K) can be expressed as

𝑒𝑒(π‘Ÿπ‘Ÿ) = 0.2[(1βˆ’(π‘Ÿπ‘Ÿ/𝑅𝑅)2] 𝑇𝑇(π‘Ÿπ‘Ÿ) = 280 + 100(π‘Ÿπ‘Ÿ/𝑅𝑅)3

where r and R are radial position and the radius of the tube, respectively.

a) Derive the average velocity and the mixed mean water temperature from the given velocity and temperature profiles.

(60 marks) b) Calculate the hydrodynamic and the thermal entrance length. The water

properties can be taken to be density of 1000 kg/m3, dynamic viscosity of 1 Γ— 10βˆ’3 kg/ms and thermal diffusivity to be 1.6 Γ— 10βˆ’7m2/s.

(40 marks) 3. Water is boiled at 100 0C at atmospheric pressure of 1atm using a 25 cm long

and 5 cm diameter electric heating element, as depicted in Figure 3. The cylindrical heating element is made of platinum. As indicated by the wattmeter, the electrical power consumed during the boiling process is 2 kW.

...3/-

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Figure 3

a) Assume that the efficiency of the electrical system is 100%, no loss of energy in the system and the electrical power is fully used to boil water in the pan, calculate the temperature on the surface of the heating element. Assume the boiling occurs in the nucleate boiling regime.

(60 marks) b) Calculate the peak heat transfer rate.

(40 marks) 4. a) As shown in Figure 4(a), two large spheres (D1 and D2) have a diameter

of 5 m and 2 m, respectively. Calculate:

(i) The view factor from surface D1 to surface D1.

(20 marks) (ii) Radiation heat transfer for D1 to D2, assume all sphere are black body condition. Take Stefan Boltzmann coefficient = 5.67 x 10-8 W/m2K4.

(20 marks)

...4/-

Heating Element

2π‘˜π‘˜π‘˜π‘˜

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Figure 4 (a)

b) As shown in the Figure 4(b), the effective surface area = 20 m2. Calculate overall heat transfer coefficient. Take cP for air and oil are 1000 J/kg.K and 2100 J/kg.K. The overall heat transfer coefficient = 350 W/m2K.

Calculate outlet temperatures of air and oil.

Figure 4 (b)

(60 marks)

-oooOooo-

...5/- D1

Diameter = 5m

Temperature = 1000K

D2

Diameter = 2m Temperature = 700K

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APPENDIX 1

Formulation

1. General properties

2. External forced convection heat transfer formulations:

Flow in a tube

Average velocity

π‘‰π‘‰π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž= 2

𝑅𝑅2οΏ½ 𝑒𝑒(π‘Ÿπ‘Ÿ)π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘…π‘… Mixed mean temperature 0

π‘‡π‘‡π‘šπ‘š = 2

π‘‰π‘‰π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘…π‘…2οΏ½ 𝑇𝑇(π‘Ÿπ‘Ÿ)𝑒𝑒(π‘Ÿπ‘Ÿ)π‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘Ÿπ‘…π‘…

0

Hydrodynamic entrance length πΏπΏβ„Ž,π‘™π‘™π‘Žπ‘Žπ‘šπ‘šπ‘™π‘™π‘™π‘™π‘Žπ‘Žπ‘™π‘™= 0.05𝑅𝑅𝑅𝑅𝑅𝑅 Thermal entrance length 𝐿𝐿𝑑𝑑,π‘™π‘™π‘Žπ‘Žπ‘šπ‘šπ‘™π‘™π‘™π‘™π‘Žπ‘Žπ‘™π‘™= 0.05π‘…π‘…π‘…π‘…π‘…π‘…π‘Ÿπ‘Ÿπ‘…π‘…

3. Natural convection:

Grashof number:

πΊπΊπ‘Ÿπ‘Ÿ

𝐿𝐿

=

π‘Žπ‘Žπ‘”π‘”(π‘‡π‘‡π‘ π‘ πœˆπœˆβˆ’π‘‡π‘‡2∞)𝐿𝐿3𝑐𝑐

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4. Boiling and condensation:

π‘žπ‘žΜ‡π‘™π‘™π‘›π‘›π‘›π‘›π‘™π‘™π‘›π‘›π‘Žπ‘Žπ‘‘π‘‘π‘›π‘› =πœ‡πœ‡π‘™π‘™β„Žπ‘“π‘“π‘Žπ‘ŽοΏ½π‘”π‘”(πœŒπœŒπ‘™π‘™βˆ’ 𝜌𝜌𝜈𝜈)

𝜎𝜎 �

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�𝐢𝐢𝑝𝑝𝑙𝑙(π‘‡π‘‡π‘ π‘ βˆ’ π‘‡π‘‡π‘ π‘ π‘Žπ‘Žπ‘‘π‘‘) πΆπΆπ‘ π‘ π‘“π‘“β„Žπ‘“π‘“π‘Žπ‘Žπ‘…π‘…π‘Ÿπ‘Ÿπ‘™π‘™π‘™π‘™ οΏ½

3

π‘žπ‘žΜ‡π‘šπ‘šπ‘Žπ‘Žπ‘šπ‘š=πΆπΆπ‘›π‘›π‘™π‘™β„Žπ‘“π‘“π‘Žπ‘Ž[πœŽπœŽπ‘”π‘”πœŒπœŒπœˆπœˆ2(πœŒπœŒπ‘™π‘™βˆ’ 𝜌𝜌𝜈𝜈)]1/4 π‘žπ‘žΜ‡π‘šπ‘šπ‘™π‘™π‘™π‘™ = 0.09πœŒπœŒπœˆπœˆβ„Žπ‘“π‘“π‘Žπ‘ŽοΏ½πœŽπœŽπ‘”π‘”(πœŒπœŒπ‘™π‘™βˆ’ 𝜌𝜌𝜈𝜈)

(πœŒπœŒπ‘™π‘™ +𝜌𝜌𝜈𝜈)2 οΏ½

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