Mechanical - Heat and Mass Transfer - Radiation

**1. A black body at 3000 K emits radiation. Calculate the following:**

**i) Monochromatic emissive power at 7 ****m****m wave length.**

**ii) Wave length at which emission is maximum.**

**iii) Maximum emissive power.**

**iv) Total emissive power,**

**v) Calculate the total emissive of the furnace if it is assumed as a real surface having emissivity equal to 0.85. Given: Surface temperature T = 3000K**

**2.The energy received by a 2 **Â´**2 m solar collector whose normal is inclined at 45**Â°**to the sun. The energy loss through the atmosphere is 50% and the diffuse radiation is 20% of direct radiation.**

**Given: **Surface temperature T = 6000 K

Distance between earth and sun R = 12 Â´1010 m

Diameter on the sun D1 = 1.5 Â´109 m

Diameter of the earth D2 = 13.2 Â´106 m

3. **A furnace wall emits radiation at 2000 K.Treating it as black body radiation, calculate**

**1. ****Monochromatic radiant flux density at 1**m**m wave length.**

**2. ****Wave length at which emission is maximum and the corresponding emissive power.**

**Total emissive power**

**Given: **Temperature T = 2000 K;** **l= 1** **mm = 1** **Â´10-6

**Solution:**

**1. Monochromatic emissive power (Eb**l**):**

4. **Calculate the heat exchange by radiation between the surfaces of two long cylinders****having radii 120mm and 60mm respectively. The axis of the cylinder is parallel to each other. The inner cylinder is maintained at a temperature of 130**Â°**C and emissivity of 0.6. Outer cylinder is maintained at a temperature of 30**Â°**C and emissivity of 0.5. (AU 2012)**

**Given: **r1** **= 60 mm** **= 0.060 m

r2= 120 mm

= 0.12

T1 = 130Â°C + 273

= 403

e1 = 0.6

T2= 30Â°C + 273 = 303 K

e2 = 0.5

**To find: **Heat exchange (Q)

**Solution: **Heat exchange between two large concentric cylinders is given by

**5. Emissivity's of two large parallel plates maintained at 800**Â°**C and 300**Â°**C are 0.5 respectively. Find net radiant hat exchange per square meter for these plates. Find the percentage reduction in heat transfer when a polished aluminums radiation shield of emissivity 0.06 is placed between them. Also find the temperature of the shield. (AU 2010)**

**Given:** T1 = 800Â°C + 273

= 1073 K

T2= 300Â°C + 273

= 573 K

e1= 0.3 e2 = 0.5

Shield emissivity e3 = 0.06

**To find:**

1. Net radiant heat exchange per square meter. (Q/A)

2. Percentage reduction in heat loss due to radiation shield.

3. Temperature of the shield (T3).

**Solution: **Heat exchange between two large parallel plates without radiation shield is given by

6. **Two circular discs of diameter 20 cm each are placed 2 m apart. Calculate the radiant heat exchangefor these discs if there are maintained at 800**Â°**C and 300**Â°**C respectively and the corresponding emissivity are 0.3 and 0.5.**

Given: D1 = 20 cm = 0.2 mD2 = 0.2 m

T1 = 800Â°C + 273

= 1073 K

T2 = 300Â°C + 273

= 573 K

e1= 0.3e2 = 0.5

To find: Heat exchange (Q)

7. **A long cylindrical heater 30 in diameter is maintained at 700**Â°**C. It has surface emissivity** **of 0.8. The heater is located in a large room whose wall is 35**Â°**C. Find the radiant heat transfer. Find the percentage of reduction in heat transfer if the heater is completely covered by radiation shield (**e**= 0.05) and diameter 40 mm.**

**Given: **Diameter of cylinder D1=30mm=0.030 mm

Temperature T1=700Â°C + 273 = 973 K

Emissivity e1 = 0.8

Room temperature T2 = 35Â°C + 273 = 308 K

Radiation Shield:

Emissivity e3 = 0.05

Diameter D3 = 40 mm = 0.040 m

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