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Assume steady-state, one-dimensional heat conduction through the symmetric shape shown.
Assuming that there is no internal heat generation, derive an expression for the thermal conductivity
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Introduction to Heat Transfer
- 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.arrow_forwardFind the two-dimensional temperature distribution T(x,y) and midplane temperature T(B/2,W/2) under steady state condition. The density, conductivity and specific heat of the material are ρ =1200 kg/m 3, k=400 W/m.K, and cp=2500 J/kg.K, respectively. A uniform heat flux q =1000 W/m 2 is applied to the upper surface. The right and left surfaces are also kept at 0oC. Bottom surface is insulated.arrow_forwardAssume steady-state, one-dimensional heat conduction through the symmetric shape shown in Figure 1.Assuming that there is no internal heat generation, derive an expression for the thermal conductivity k(x) for these conditions: A(x) = (1 -x), T(x) = 300(1 - 2x -3x),and q = 6000 W, where A is in square meters, T in Kelvin’s, and x in meters. Consider x= 0 and 1.arrow_forward
- Assume steady-state, one-dimensional heat conduction through the symmetric shape shown in Figure 1. Assuming that there is no internal heat generation, derive an expression for the thermal conductivity k(x) for these conditions: A(x) = (1 - x), T(x) = 300(1 - 2x - x3), and q = 6000 W, where A is in square meters, T in kelvins, and x in meters. Consider x= 0 and 1.arrow_forwardAssume steady-state, one-dimensional heat conduction through the symmetric shape shown in Figure 1. Assuming that there is no internal heat generation, derive an expression for the thermal conductivity k(x) for these conditions: A(x) = (1 - x), T(x) = 300(1 - 2x - x3), and q = 6000 W, where A is in square meters, T in kelvins, and x in meters. Consider x= 0 and 1.arrow_forwardWhich formula is used to calculate the heat conduction in the AXIAL direction in a vertically located pipe segment whose inner and outer surfaces are perfectly insulated. Here r, is inner radius, r, outer radius, Tri pipe inner surface temperature, Tro pipe outer surface temperature, L is the length of the pipe, T the temperature on the lower surface, Ty the temperature on upper surface. Tu r; Tro rarrow_forward
- under steady-state conditions. If you are given T1 = 200 °C and T2 = 164 °C, determine: a) the conduction heat flux, q,.cond, in m2 W from x = 0 to x = L b) if the dimensions of the triangle ares 15 mm and h 13 mm, calculate the heat transfer due to convection, q,y, in W at x = L Finsulation T2 T T = 20°C h = 500 W/m2.K Triangular Prism x L x 0 L= 50 mm k = 100 W/m-Karrow_forward1-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 mmarrow_forwardYou 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…arrow_forward
- Q. 5: Assume steady-state, one-dimensional heat conduction through the symmetric shape shown in Figure 1. Assuming that there is no internal heat generation, derive an expression for the thermal conductivity k(x) for these conditions: A(x) = (1 - x), T(x) = 300(1 - 2x - x3), and q = 6000 W, where A is in square meters, T in kelvins, and x in meters. Consider x= 0 and 1.arrow_forwardDo fast i will give you good ratearrow_forwardThe initial temperature distribution of a 5 cm long stick is given by the following function. The circumference of the rod in question is completely insulated, but both ends are kept at a temperature of 0 °C. Obtain the heat conduction along the rod as a function of time and position ? (x = 1.752 cm²/s for the bar in question) 100 A) T(x1) = 1 Sin ().e(-1,752 (³¹)+(sin().e (-1,752 (²) ₁ + 1 3π TC3 .....) 100 t + ··· ....... 13) T(x,t) = 200 Sin ().e(-1,752 (²t) + (sin (3). e (-1,752 (7) ²) t B) 3/3 t + …............) C) T(x.t) = 200 Sin ().e(-1,752 (²t) (sin().e(-1,752 (7) ²) t – D) T(x,t) = 200 Sin ().e(-1,752 (²)-(sin().e (-1,752 (²7) ²) t E) T(x.t)=(Sin().e(-1,752 (²t)-(sin().e(-1,752 (²) t+ t + ··· .........) t +.... t + ··· .........) …..)arrow_forward
- Principles of Heat Transfer (Activate Learning wi...Mechanical EngineeringISBN:9781305387102Author:Kreith, Frank; Manglik, Raj M.Publisher:Cengage Learning