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Introduction to Heat Transfer
- 1.26 Repeat Problem 1.25 but assume that the surface of the storage vessel has an absorbance (equal to the emittance) of 0.1. Then determine the rate of evaporation of the liquid oxygen in kilograms per second and pounds per hour, assuming that convection can be neglected. The heat of vaporization of oxygen at –183°C is .arrow_forwardConsider a silicon wafer positioned in a furnace that is zone-heated on the top section and cooled on the lower section. The wafer is placed such that the top and bottom surfaces of the wafer exchange radiation with the hot and cold zones respectively of the furnace. The zone temperatures are Tsur,h = 950 K and Tsur.c = 330 K. The emissivity and thickness of the wafer are e = 0.65 and d = 0.78 mm, respectively. With the ambient gas at T = 700 K, convection heat transfer coefficients at the upper and lower surfaces of the wafer are 8 and 4 W/m2-K. Find the steady-state temperature of the wafer, in K. i K Save for Laterarrow_forward! Required information Irradiation on a semi-transparent medium is at a rate of 640 W/m². If 160 W/m² of the irradiation is reflected from the medium and 130 W/m² is transmitted through the medium, Determine the absorptivity of the medium. The absorptivity of the medium is 0.75 Xarrow_forward
- Consider a silicon wafer positioned in a furnace that is zone-heated on the top section and cooled on the lower section. The wafer is placed such that the top and bottom surfaces of the wafer exchange radiation with the hot and cold zones respectively of the furnace. The zone temperatures are Tsur.h = 900 K and Tsur.c = 330 K. The emissivity and thickness of the wafer are ɛ = 0.65 and d = 0.78 mm, respectively. With the ambient gas at T, = 700 K, convection heat transfer coefficients at the upper and lower surfaces of the wafer are 8 and 4 W/m2-K. Find the steady-state temperature of the wafer, in K. Tw i Karrow_forward6. An electric hot plate is placed in a room which is maintained at a temperature of 297 K. The plate is maintained at a temperature of 403 K and has an emissivity of 0.8. If the plate surface resembles a circular disc of diameter 250 mm, electrical power consumed by the hot plate will be; a. 22 W b. 65 W c. 75 W d. 86 Warrow_forwardAt a distance of 10 meters we have a radiation intensity of 0.1 mSv/h after passing through 48 mm of a shielding material (TVL of 16 mm). If an Ir-192 source is used (SEC=0.15mSv/hr@1m per GBq), what is its activity? 5,333 GBq 666.67 GBq 800 Gbq 100 GBq 66.67 GBq 66,666 GBqarrow_forward
- Define the absorption of radiation incident on an opaque surface of absorptivity α.arrow_forward3.5m 12.0m 6m -3.5marrow_forwardA small sphere (emissivity =0.503 radius=r1) is located at the center of a spherical abestos shell ( thickness =1.74 cm, outer radius= r2; thermal conductivity of abestos is 0.090 J/ (sm c degrees) The thickness of the shell is small compared to the inner and outer radii of the shell. The temperature of the small sphere is 695 degrees Celsius while the temperature of the inner surface of the shell is 352 degrees Celsius, both temperatures remaining constant. Assuming that r2/r1 =8.75 and ignoring any air inside the shell, find the temperature in degrees Celsius of the outer surface of the shell.arrow_forward
- A small sphere (emissivity = 0.745, radius = r1) is located at the center of a spherical asbestos shell (thickness = 1.72 cm, outer radius = r2; thermal conductivity of asbestos is 0.090 J/(s m Co)). The thickness of the shell is small compared to the inner and outer radii of the shell. The temperature of the small sphere is 727 °C, while the temperature of the inner surface of the shell is 406 °C, both temperatures remaining constant. Assuming that r2/r1 = 6.54 and ignoring any air inside the shell, find the temperature in degrees Celsius of the outer surface of the shell.arrow_forwardAn electric heating system is installed in the ceiling of a room 5 m (length) × 5 m (width) ×2.5 m (height). The temperature of the ceiling is 315 K whereas under equilibrium conditions the walls are at 295 K. If the floor is non-sensitive to radiations and the emissivities of the ceiling and wall are 0.75 and 0.65, respectively. Calculate the radiant heat loss from the ceiling to the walls. The answer should be 1595 W. Please show steps in your solutionarrow_forwardGive step-by-step calculation and explanation Consider a person sitting nude on a beach in Florida. On a sunny day, visible radiation energy from the sun is absorbed by the person at a rate of 30 kcal/h or 34.9 W. The air temperature is a warm 30 °C and the individual’s skin temperature is 32 °C. The effective body surface exposed to the sun is 0.9 m². (Assume this same area for sun absorption, radiative transfer, and convective loss. Is this a good assumption?) a. Find the net energy gain or loss from thermal radiation each hour. (Assume thermal radiative gain and loss according to the equation 6.51 in Herman and an emissivity of 1.) -(4). Equalion (6.51) - (40Tin)Eskin Aşkin (Tskin – Troom) dt = (4 x 5.67 x 10¬8 w/m²–K* x (307 K)')€skin Askin (Tskin – Troom). (6.52) b. If there is a 4 m/s breeze, find the energy lost by convection each hour. (Use Eq. 6.61 with eq. 6.63.) 1 Equation he(Tskin – Tair), (6.61) A dt he 10.45 – w + 10w0.5 (6.63) - c. If the individual’s metabolic rate is…arrow_forward
- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning