Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 40 W/m-K and a thickness L = 0.3 m, with no internal heat generation. T1 -T2 L Determine the heat flux, in kW/m?, and the unknown quantity for each case. Case T1(°C) T2(°C) dT/dx(K/m) q (kW/m?) 1 50 -20 -233.333 9333.32 2 -30 -10 66.667 75466.67 3 70 117.85 160 6400 4 63.85 40 -80 3200 5 -30.15 30 200 8000
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- A plane wall 15 cm thick has a thermal conductivity given by the relation k=2.0+0.0005T[W/mK] where T is in kelvin. If one surface of this wall is maintained at 150C and the other at 50C, determine the rate of heat transfer per square meter. Sketch the temperature distribution through the wall.1- Consider steady- state conduction for one-dimensional conduction in a plane wall having a thermal conductivity k=50 W/m.K and a thickness L-0.25 m, with no internal heat generation. Determine the heat flux and the unknown quantity (blanks) for each case and sketch the temperature distribution, indicating the direction of heat flux. Case TI(°C) 50 T2(°C) -20 dT/dx(K/m) 1 -30 -10 70 160 40 -80 5 30 200 234nConsider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 40 W/m-K and a thickness L = 0.3 m, with no internal heat generation. T2 L Determine the heat flux, in kW/m?, and the unknown quantity for each case. T1 (°C) T2(°C) dT/dx(K/m) 9 (kW/m?) Case 1 50 -20 i i 2 -30 -10 i 70 i 160 i 40 -80 i 5 i 30 200 i
- Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 40 W/m-K and a thickness L = 0.3 m, with no internal heat generation. -T2 L Determine the heat flux, in kW/m², and the unknown quantity for each case. Case T1(°C) T2(°C) dT/dx(K/m) q (kW/m?) 1 50 -20 i i 2 -30 -10 i i 3 70 i 160 i 4 i 40 -80 i 5 i 30 200 iConsider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 40 W/m-K and a thickness L = 0.4 m, with no internal heat generation. -T2 L Determine the heat flux, in kW/m2, and the unknown quantity for each case. Case T1(°C) T2(°C) dT/dx(K/m) 9% (kW/m²) 1 50 -20 i i -30 -10 i i 3 70 i 160 i 4 i 40 -80 i i 30 200 i LOConsider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 40 W/m•K and a thickness L = 0.3 m, with no internal heat generation. L Determine the heat flux, in kW/m2, and the unknown quantity for each case. Case T1(°C) T2(°C) dT/dx(K/m) (kW/m²) 1 50 -20 i i 2 -30 -10 i 3 70 i 160 i 4 i 40 -80 i i 30 200 i
- Consider 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.= 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 XQ: A large plane wall of thickness (L= 0.3 m), thermal conductivity (k = 40 W/m-°C), and surface area 0.4 m2. The left side of the wall is maintained at a constant temperature of 70°C while the right side loses heat to the surrounding air with heat flux q' expressed by: 1200k q = k+20L A- Find a relation to determine the variation of temperature in the wall. B- Determine the rate of heat transfer through the wall. C- Determine the temperature at the right side of the wall. wall T1= 70 °C L= 0.3m
- Conduction Heat Transfer X material (a) material (b) . both (a) and (b) have the same thermal conductivity the temperature distribution is independent of thermal conductivity it's not that simple 9. Fin efficiency is defined as: • tanh (mL) (hP/k Ac)1/2 (heat transfer with fin) / (heat transfer without fin) (actual heat transfer through fin) / (heat transfer assuming all fin is at T = Tb) (Tx=L-Tf)/(Tb-Tf) 10. For an infinite fin, the temperature distribution is given by: (T-Tf)/(Tb-Tf)= e-mx. The heat flow through the fin is therefore given by: k (Tb-Tf)/L ● zero, because the fin is infinite ● infinite because the fin is infinite ● (Tb-Tf) (hP/k Ac)1/2 ● (Tb – Tf) (h P / k Ac)1/2 tanh (mL) 11. The Biot number, Bi, is defined as: • Bi=hk/L • Bi=hL/k • Bi= k/LH • Bi=qL/k • Bi=p UL/k 12. For a plate of length L, thickness, t, and width, W, subjected to convection on the two faces of area L x W. What is the correct length scale for use in the Biot number? . L ● W ● t • t/2 • L/2 13. If Bi…1-D, steady-state conduction with uniform internal energy generation occurs in a plane wall with a thickness of 50 mm and a constant thermal conductivity of 5 W/m/K. The temperature distribution has the form T = a + bx + cx² °C. The surface at x=0 has a temperature of To = 120 °C and experiences convection with a fluid for which T.. surface at x= 50 mm is well insulated (no heat transfer). Find: (a) The volumetric energy generation rate q. (15) (b) Determine the coefficients a, b, and c. 20 °C and h 500 W/m² K. The To: = 120°C T = 20°C h = 500 W/m².K 111 Fluid T(x)- = q, k = 5 W/m.K L = 50 mmQ1. A 50 meter long cast iron pipe with outer diameter of 10 cm passes in an open space of 288 K temperature. The outer surface of the pipe temperature is 423 K and the combined heat transfer coefficient on the outer surface of the pipe is 25 W/m? K. Considering and stating the necessary assumptions determine, (a) The rate of heat loss from the pipe (b) The energy lost per year if the cost of the fuel is 0.52 $/therm ( 1 therm = 105,500 kJ) (c) The thickness of the insulation if 98% of the energy loss is planned to be saved. Consider the conduction coefficient of the insulation is 0.035 W/mK. Tair = 288 K 423 K Steam 50 m Fiberglass insulation Figure Q1.