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- Q4) A long steel bar 5 by 10 cm is initially maintained at a uniform temperature of 250 "C. It is suddenly subjected to a change such that the environment temperature is lowered to 35 °C. Assuming a heat-transfer coefficient of 23 W/m² °C, use a numerical method to estimate the time required for the center temperature to reach 105 °C. Check this result with a calculation using the Heisler charts.A copper rod has an initial length 5.0 cm and a thermal expansion coefficient of 16 x 106 /K. It is heated from room temperature (22 °C) to a final temperature T, and its final length is 5.0064 cm. What is the final temperature in °C? In your answer, use the relation (L final – Linitial)/Linitial a(Tfinal – Tinitial) where Linitial is the initial length, L final is the final length, Tinitial is the initial temperature, Tfinal is the final temperature, and a is the thermal expansion coefficient. (This question has only one correct answer) a. 133 b. 102 c. 107 d. 1412-88 A 1.5-mm-diameter stainless-steel rod [k = 19 W/m.°C] protrudes from a wall maintained at 45°C. The rod is 12 mm long, and the convection coefficient is 500 W/m² . °C. The environment temperature is 20°C. Calculate the temperature of the tip of the rod. Repeat the calculation for h = 200 and 1500 W/m² . °C.
- = Consider a large plane wall of thickness L=0.3 m, thermal conductivity k = 2.5 W/m.K, and surface area A = 12 m². The left side of the wall at x=0 is subjected to a net heat flux of ɖo = 700 W/m² while the temperature at that surface is measured to be T₁ = 80°C. Assuming constant thermal conductivity and no heat generation in the wall, (a) express the differential equation and the boundary equations for steady one- dimensional heat conduction through the wall, (b) obtain a relation for the variation of the temperature in the wall by solving the differential equation, and (c) evaluate the temperature of the right surface of the wall at x=L. Ti до L XThe two-dimensional solid shown in Figure P3-64 generates heat internally at the rate of 90 MW/m3. Using the numerical method calculate the steady-state nodal temper- atures for k= 20 W/m.°C. %3D Figure P3-64 h = 100 W/m2. °C !3! T = 20°C 2 T 100°C 5 7 8 9. 10 11 12 Insulated Ax = Ay = 1 cm k= 20 W/m • °C 4 = 90 MW/m3 %3D Insulated3- A large plane wall has a thickness L=50 cm and thermal conductivity k=25 W/m.K. On the left surface (x=0), it is subjected to a uniform heat flux go, while the surface temperature To is constant. On the right surface, it experiences convection and radiation heat transfer while the surface temperature is TL = 225°C and surrounding temperature is 25°C. The emissivity and the convection heat transfer coefficient on the right surface are 0.7 and 15 W/m²K, respectively. (a) Derive the temperature distribution equation for the wall parametrically (based on x, T₁, qo, L, and k). Plane wall Tsurr = 25°C T% = 25°C k = 25 W/m.K h = 15 W/m²K 90 ε = 0.7 T₁ = 225°C (b) Determine the temperature of the left surface of the wall at x=0? L x
- 2-1. Temperature Response in Cooling a Wire. A small copper wire with a diameter of 0.792 mm and initially at 366.5 K is suddenly immersed in a liquid held constant at 311 K. The convection coefficient h 85.2 W/m K. The physical properties can be assumed constant and are k =374 W/m K, c, = 0.389 kJ/kg K, and p = 8890 kg/m2. (a) Determine the time in seconds for the average temperature of the wire to drop to 338.8 K (one-half the initial temperature difference). (b) Do the same but for h 11.36 W/m2 K (c) For part (b), calculate the total amount of heat removed for a wire 1.0 m р u (A long. (a) t 5.66 s Ans. the Smou hire is lons asbume Nhere rraolius , here radlius yinder hitu x Chapter 14 Principles of Unsteady-State Heat Transfer4x F2 # 3 E 4, F3 54 $ R F4 Ac = 1m² ▬ H DII x= 1 m (4) Consider a wall (as shown above) of thickness L-1 m and thermal conductivity k-1 W/m-K. The left (x=0) and the right (x=1 m) surfaces of the wall are subject to convection with a convectional heat transfer coefficient h= 1 W/m²K and an ambient temperature T. 1 K. There is no heat generation inside the wall. You may assume 1-D heat transfer, steady state condition, and neglect any thermal contact resistance. Find T(x). % To,1 = 1 K h₁ = 1 W/m²K 5 Q Search F5 T T₁ A 6 x=0 F6 à = 0 W/m³ k= 1W/mK L=1m Y 994 F7 & 7 T₂ U Ton2 = 1 K h₂ = 1 W/m²K1 PrtScn F8 Page of 7 ) 0 PgUp F11 PConsider a copper plate that has dimensions of 3 cm x 3 cm x 7 cm (length, width, and thickness, respectively). As shown in the following figure, the copper plate is exposed to a thermal energy source that puts out 126 J every second. The density of copper is 8,900 kg/m³. Assume there is no heat loss to the surrounding block. 126 J Copper Insulation Ⓡ What is the specific heat of copper (in J/(kg K))? J/(kg. K) What is the mass of the copper plate (in kg)? kg How much energy (in J) will be consumed during 11 seconds? J Determine the temperature rise (in K) in the plate after 11 seconds.
- (Q4) A 4m x 6m wall consists of 4 glass windows of 2m x 1.5m dimensions. The wall has thickness of 0.13m and a thermal conductivity of 0.5 W/m.K, while the glass windows are 6 mm thick with a thermal conductivity of 1.228 W/m.K. The values of intemal and external surface conductance for the wall (including glass) are 7.8 W/m? K and 34.4 W/m².K, respectively. The intemal and extemal temperatures are 22° C and 42°C, respectively. Calculate the total heat transfer rate through the wall. What percentage of this heat transfer is through the windows?A large 100-mm thick steel plate is initially at uniform temperature T = 180°C at time t = 0. Both sides of the plate are exposed to 30°C oil with convection coefficient h = 1500 W/m².K. The thermal conductivity, heat capacity, and density of the steel are k = 38 W/m-K, cp = 485 J/kg-K, and p = 7800 kg/m³. a) Find the temperature at the center of the slab when t = 125 s. b) Find the temperature at the surface of the slab when t=125 s.You are asked to estimate the maximum human body temperature if the metabolic heat produced in your body could escape only by tissue conduction and later on the surface by convection. Simplify the human body as a cylinder of L=1.8 m in height and ro= 0.15 m in radius. Further, simplify the heat transfer process inside the human body as a 1-D situation when the temperature only depends on the radial coordinater from the centerline. The governing dT +q""=0 dr equation is written as 1 d k- r dr r = 0, dT dr =0 dT r=ro -k -=h(T-T) dr (k-0.5 W/m°C), ro is the radius of the cylinder (0.15 m), h is the convection coefficient at the skin surface (15 W/m² °C), Tair is the air temperature (30°C). q" is the average volumetric heat generation rate in the body (W/m³) and is defined as heat generated per unit volume per second. The 1-D (radial) temperature distribution can be derived as: T(r) = q"¹'r² qr qr. + 4k 2h + 4k +T , where k is thermal conductivity of tissue air (A) q" can be calculated…