Coatings applied to long metallic strips are cured by installing the strips along the walls of a long oven of square cross section.
Thermal conditions in the oven are maintained by a long silicon carbide rod (heating element), which is of diameter
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Fundamentals of Heat and Mass Transfer
- 2.30 An electrical heater capable of generating 10,000 W is to be designed. The heating element is to be a stainless steel wire having an electrical resistivity of ohm-centimeter. The operating temperature of the stainless steel is to be no more than 1260°C. The heat transfer coefficient at the outer surface is expected to be no less than in a medium whose maximum temperature is 93°C. A transformer capable of delivering current at 9 and 12 V is available. Determine a suitable size for the wire, the current required, and discuss what effect a reduction in the heat transfer coefficient would have. (Hint: Demonstrate first that the temperature drop between the center and the surface of the wire is independent of the wire diameter, and determine its value.)arrow_forwardQuestion 5: A copper bar 35 cm long, with square cross-section 2 cm x 2 cm is fitted with a resistive heater at one end, and a large heat sink at the other. The bar itself is ideally thermally lagged. (i) (ii) Sketch how you think the two thermometers would behave when the heater is switched on; (ii) Calculate the thermal conductivity if in the steady-state, T1=64.7 deg C and T2=40.0 deg C. Hint: rate of flow of heat = k*A*AT/X V=6 V, I=2.5A heater T1 25 cm T2 Heat sinkarrow_forwardIn order to reduce the heat loss through a large furnace wall, the decision has been made to add external insulation. Calculate the thickness of insulation required to reduce the heat loss by 75%. Before the change is made, no outer steel shell is used.Data: Refractory brick and wall brick:k = 0.87 W m-1 K-1Insulation: k = 0.090 W m-1 K-1•Steel: k = 43 W m-1 K-1h = 55 W m-1 K-1 (inside furnace).h = 11 W m-1 K-1 (outside furnace).arrow_forward
- The Diamond Ring Solution. The processing chip on the computer that controls the navigation equipment on your spacecraft is overheating. Unless you fix the problem, the chip will be damaged and the navigation system will shut down. You open the panel and find that the small copper disk that was supposed to bridge the gap between the smooth top of the chip and the cooling plate is missing, leaving a 2.0 mm gap between them. In this configuration, the heat cannot escape the chip at the required rate. You notice by the thin smudge of thermal grease (a highly thermally conductive material used to promote good thermal contact between surfaces) that the missing copper disk was 2.0 mm thick and had a díameter of 1.2 cm. You know that the chip is designed to run below 65 °C, and the copper cooling plate is held at a constant 5.0 °C. (a) What was the rate of heat flow from the chip to the copper plate when the original copper disk was in place and the chip was at its maximum operating…arrow_forwardThe cross-sectional area of a conical piece made of pure aluminum has a diameter D = a.x ^ 1/2 and a = 0.5m ^ 1/2.The left edge surface of the part is at x1 = 25 mm, and the right edge surface is x2 = 125 mm. Edge temperatures are T1 = 600K and T2 = 400K and the side surface is completely insulated.a) Assuming that the heat conduction is one dimensional, write a relation for the temperature distribution T (x).b) Calculate the heat transfer.arrow_forwardAn industrial company manufactures tungsten cubes that need to be sintered to a temperature of 800K after processing. Hot combustion gas at 100OK is used to heat the tungsten cubes with a side length of 0.3m. The cube is initially at a temperature of 298K. Calculate the time that is needed for the cube to reach 800K. Be sure to show that you have verified your method is valid. 0.3m T: 1000K h: 100 W/m?/Karrow_forward
- The Diamond Ring Solution. The processing chip on the computer that controls the navigation equipment on your spacecraft is overheating. Unless you fix the problem, the chip will be damaged and the navigation system will shut down. You open the panel and find that the small copper disk that was supposed to bridge the gap between the smooth top of the chip and the cooling plate is missing, leaving a 2.0 mm gap between them. In this configuration, the heat cannot escape the chip at the required rate. You notice by the thin smudge of thermal grease (a highly thermally conductive material used to promote good thermal contact between surfaces) that the missing copper disk was 2.0 mm thick and had a diameter of 1.0 cm. You know that the chip is designed to run below 70 °C, and the copper cooling plate is held at a constant 5.0 °C. (a) What was the rate of heat flow from the chip to the copper plate when the original copper disk was in place and the chip was at its maximum operating…arrow_forwardA coil-shaped cooling pipe is made of SS-304 material. This pipe is 1 ft long, 0.4 inch outside diameter, and inch inside diameter. This coil cooling pipe is used to cool the water in the bath. The temperature of the inner coil pipe is 40oF while the outer coil in contact with water is 80oF. The thermal conductivity of SS-304 is a function of temperature where k(T) = 7.75 + (7.78 x 10-3).T where k is in Btu/h.ft.oF and T is in oF. Calculate the rate of heat dissipation in watts! (1287.7)arrow_forward4. The section of a vertical wall is made up of fiberglass insulation slabs separated by wooden studs. The thermal conductivity of fiberglass and wood are 0.04 W/m-K and 0.18 W/m-K, respectively. The thickness of the wall is 160 mm. The temperatures of the inner and outer surfaces of the wall section are 22 °C and 4 °C, respectively. The ratio of the insulation area to the total wall area is 0.8. Calculate the total heat flow rate (in W/m2) through the wall per unit area. Wooden Stud Fiberglass Insulation Follow-up question to Question 4, calculate the rate of heat flow (in W/m2) through unit area of studs.arrow_forward
- Calculate the thermal resistance of a copper tube with diameter1=0.153 mm diameter2= 0.157 mm and thermal conductivity of 310 W/m.K insulated with a layer of Styrofoam diameter = 0.09 mmarrow_forwardBuild a spreadsheet to do the step by step method for estimating the temperature of unprotected steel work exposed to the ASTM E119 Standard Fire. Assume a convective heat transfer coefficient of 25 W/m2K and an emissivity of 0.5. Assume the specific heat to remain constant at 600 J/kgK. Use a time step of 30 seconds and an ambient temperature of 20°C. Provide an Excel plot showing the curves of the fire and a steel member through a time of 60 minutes with F/V ratios of: a)25 m-1 b)100 m-1 c)200 m-1 d)300 m-1 Note your plot should contain 5 curves. One for the fire and one for each of the F/V values. Also on the spreadsheet, highlight in green, the cells identifying the temperature of the unprotected steel member with an F/V of 200m-1 at the following times: a)3 minutes b)10 minutes c)20 minutes d)50 minutesarrow_forward1. For a steam pipe with a given diameter of 10 cm covered by two (2) layers of insulation. The first insulation has a thickness of 4 cm and a coefficient of thermal conductivity of 0.08 W/m.K. and the second insulation has a thickness of 3 cm and a thermal conductivity of 0.15 W/m.K. The steam main conveys steam at a pressure of 1.70 MPa with 25°C superheat. Outside temperature is 27°C. The pipe is 30 meters in length. (tsat @ 1.70 MPa = 204°C). Determine the following: a) The heat loss in KW b) Explain the concepts/principles that were considered and the factors that affected the condition of the above mentioned items (a & b).arrow_forward
- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning