Solutions to Quiz 12 – Radiation Heat Transfer

Solutions to Quiz Twelve –Radiation Heat Transfer Consider a triangular enclosure that has the two rectangular surfaces shown at the right. These two surfaces are perpendicular to 0.8 each other with a common edge which is 1.6 m long. The horizontal surface is 0.8 m wide and the vertical surface is 1.2 m high. The horizontal surface has an emissivity of 0.75 and is maintained at 400 K. The vertical surface is maintained at 550 K and has an emissivity of 0.8. The back sides of the surfaces are insulated. The remaining three sides of the triangular enclosure (one flat surface that joins the outer edges of the horizontal and vertical surfaces shown and two triangular surfaces at the ends) are at 290 K and can be considered to have an emissivity of 0.85. Determine the net rate of radiation heat transfers between the two surfaces, and between the horizontal surface and the surroundings. (Problem and figure P13-45 from Çengel, Heat and Mass Transfer.) The equations for a three surface enclosure are taken from slides 29 and 30 of the May 2 lecture presentation. ⎛ 1 − ε1 ⎞ 1 − ε1 1 − ε1 1 − ε1 ⎜⎜1 + F12 + F13 ⎟⎟ J1 − F12 J 2 − F13 J 3 = σT14 ε ε ε ε 1 1 1 1 ⎝ ⎠ − ⎛ 1 − ε2 ⎞ 1 − ε2 1− ε2 1 − ε2 F21 J1 + ⎜⎜1 + F21 + F23 ⎟⎟ J 2 − F23 J 3 = σT24 ε2 ε ε ε 2 2 2 ⎝ ⎠ − ⎛ 1 − ε3 ⎞ 1 − ε3 1 − ε3 1 − ε3 F31 J1 − F32 J 2 + ⎜⎜1 + F31 + F32 ⎟⎟ J 3 = σT34 ε3 ε3 ε3 ε3 ⎝ ⎠ We can start by computing the black body emissive power for the three surfaces that appear on the right-hand sides of our equations. Eb1 = σT14 = 5.670 x10 −8 W m2 ⋅ K 4 (400 K )4 = 14522W m E b3 = σT34 = Eb 2 = σT24 = 5.670 x10 −8 W m ⋅K 2 4 5.670 x10 −8 W m2 ⋅ K 4 (550 K )4 = 51882W m (290 K )4 = 401.02 W m The remainder of the coefficients will involve view factors and areas. The areas of the two flat surfaces in the enclosure are A1 = L1W = (0.8 m) (1.6 m) = 1.28 m2 and A2 = L2W = (1.2 m) (1.6 m) = 1.92 m2; the area of the third radiation surface is the area of a single diagonal surface, W(L12 + L22)1/2 plus the area two triangles enclosing the ends: 2[(1/2)L1L2]. This third surface area is thus (1.6 m)[(0.8 m)2 +(1.2 m)2]1/2 + (0.8 m)(1.2 m) = 3.268 m2. Quiz twelve solutions Page 2 We have to find six view factors F12, F13, F21, F23, F31, and F32. We can find F12 from the view factor graph in Figure 13-6. For this problem we have L1/W = (0.8 m) / (1.6 m) = 0.5 and L2/W = (1.2 m) / (1.6 m) = 0.75; for these values I read a view factor of F12 = 0.28 from Figure 13-6. We can find F13 from the summation rule: F11 + F12 + F13 = 1, where F11 = 0 because surface 1 is a flat surface. This gives F13 = 1 – 0 – 0.2749 = 0.72. We can use the reciprocity rule to find F21 and F31: A2F21 = A1F12; so, F21 = A1F12/A2 = (1.28 m2)(0.28)/ (1.92 m2) = 0.1867; A3F31 = A1F13; so, F31 = A1F13/A3 = (1.28 m2)(0.72)/ (3.268 m2) = 0.2820. Since surface 2 is a flat surface, F22 = 0, and we can use the summation rule, F21 + F22 + F23 = 1 to find F23 = 1 – F21 – F22 = 1 – 0.1867 – 0 = 0.8133. Finally, we can use the reciprocity rule to find F32 = A2F23/A3 = (1.92 m2)(0.8133)/ (3.268 m2) = 0.4799. The emissivity factors in the equations that we have to solve are (1 – ε1)/ε1 = (1 – 0.75)/0.75 = 1/3, (1 – ε2)/ε2 = (1 – 0.8)/0.8 = 1/4, and (1 – ε3)/ε3 = (1 – 0.85)/0.85 = 0.1765. We can now compute the coefficients and right-hand-side terms in our system of two equations. We now have all the data required to compute the coefficients on the left-hand sides of the three equations. Note that the calculations of diagonal terms use the results of the previous calculations for the other terms in the same equation. 1 − ε1 1 (0.28) = −0.09333 − 1 − ε1 F13 = 1 (0.72) = −0.24 ε1 ε1 3 3 1 − ε1 1 − ε1 1− F12 − = 1 + 0.09333 + 0.24 = 1.3333 − F12 ε1 − 1− ε2 ε2 ε1 1 (0.1967) = −0.04667 − 1 − ε 2 F23 = 1 (0.8133) = −0.2033 ε2 4 4 1− ε2 1− ε2 F21 + F23 = 1 + 0.04667 + 0.20333 = 1.25 1+ F21 = ε2 − 1 − ε3 ε3 ε2 F31 = (0.1765)(0.2820) = −0.049773 − 1+ 1 − ε3 ε3 F31 + 1 − ε3 ε3 1 − ε3 ε3 F32 = (0.1765)(0.4779) = −0.08434 F32 = 1 + 0.049773 + 0.08434 = 1.1341 We have the following system of equations to solve. 1452 W m2 5188 W − 0.046667 J1 + 1.253 J 2 − 0.20333 J 3 = m2 401.0 W − 0.049773J1 − 0.084337 J1 − 1.1341J 3 = m2 1.3333 J1 − 0.093333 J 2 − 0.24 J 3 = We see that the left-hand side coefficients are dimensionless and the right-hand side terms have units of W/m2 so our results for radiosity will have units of W/m2. Solving these simultaneous equations gives J1 = 1525 W/m2, J2 = 4238 W/m2 and J3 = 742.4 W/m2. The heat transfer from the vertical surface (2) to the horizontal surface (1) is ( ) ⎛ 4328 W 1525 W ⎞ 2 − Q& 21 = A2 F21 ( J 2 − J1 ) = 1.92 m 2 (0.1867 )⎜ ⎟ = 1005 W/m 2 2 m m ⎝ ⎠