Question

Q2 (a) A 12 mm diameter mild steel sphere (k = 42.5 W/m K) is exposed to cooling airflow at 27 "C resulting in the convective coefficient, h = 114 W/m' K. The relevant properties of mild steel are given as follows: Density p= 7850 kg/m . Specific heat c = 475 J/kg K and thermal diffusivity a = 0.043 m/hr Determine: (i) Time required to cool the sphere (lumped parameter system) from 540 °C to 95°C. [7 marks] (ii) Instantaneous heat transfer rate 2 minutes after the start of cooling (5 marks] (iii) Total energy transferred from the sphere during the first 2 minutes [5 marks] (b) Sun may be approximated as black body. It emits radiation having maximum intensity at Imax = 0.5 micron. Calculate: (1) The surface temperature of sun using Wien's displacement law [2 marks] (ii) The radiation heat flux at sun's surface using Stefan-Boltzmann Law. The Stefan-Boltzmann constant, o = 5.67 x 10-4 W/m² K4. [3 marks] (c) One side of a 15 cm thickness layer of insulation (k = 0.75 W/m 'C) has a temperature of 200°C while the other loses heat to an ambient at 27 °C at the rate of 600 W/m². Take A = 1.0 m². Calculate: (i) The air-film heat transfer coefficient (h) [2 marks] (ii) The skin temperature (say, T) of the insulation [2 marks] (d) What are the concept of 'Black body' and 'Gray body' Radiation??

Answer 1

Solution 2 ca) Diameter of Sphere D = 12 mm = 0.012 m & surface aree of sphere A = 4782 = TO? A = ITX (0.012)² m2 A = 4.528104 m² → Ambient heat air temperature Too = 27°C transfer coefficient in Air - hally w/m2K specific heat Cp = 475 Ilegik So time required to cool from T= 95°C To=540°C to or clumped parameter system following relation should be used T- To ie them to co Totus & m= mass gobject (sphere) mass of sphere = m = density & volume = 8V m=7850 x 4 T (2)3 kg

m= 7850 ** 3.14 x 6.012)” ks (m= 0.00709) kg Now we put all values 95 -27 in equation 0 - 114x4.52 x 164 475 X 0.00209 suo -27 7 0.13255 = e-0.0153t taking log natural both sides 32 In (0.13255) = -0.0153 + t: 72.0807 I seconds + 0.0153 Ite 132.072 seconds to Answer is time required to fool from 540°cto 95°C (11) Now - temperature after two minutes, (T2) To = 540° 1 t=2 min = 120 seconds Tz - Too - -0.0153 € то сте -0.0153x120 - 1.836 Tz - 27 = e or 3 -0.0153x12. se" suora2 = 9 Tam 27 = 0.1594 513

Tz- 513*0.1594 +27 °C [T2 = 108.79 °C • Tenger" lemperature of sphere after a minutes, • Instantaneous heat transfer rate Slamin = ha (To-To). after 2 minutes. Glom = 114 x4.52x104 (108-99- 827) W 2min ill Olamin = 4.21498 | w - Answer (1) b heat transfer rate after 2 min amount of Heat transferred during first two minutes Q = mu (To-T₂) Q = 0.00709x475 (540-108.79) J TG = 1452.20 J y Heat transferred in first two minutes.