ME 315 Final Examination Solution 8:00-10:00 AM Friday, May 8, 2009 CIRCLE YOUR DIVISION

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1 ME 315 Final Examination Solution 8:00-10:00 AM Friday, May 8, 009 This is a closed-book, closed-notes examination. There is a formula sheet at the back. You must turn off all communications devices before starting this exam, and leave them off for the entire exam. Please write legibly and show all work for your own benefit. Please show your final answers in the boxes provided. State all assumptions. Please arrange all your sheets in the correct order. Make sure they are all included. Name: Last First CIRCLE YOUR DIVISION Div. 1 (9:30 am) Prof. Murthy Div. (1:30 pm) Prof. Choi Problem 1 (30 Points) Score (35 Points) 3 (40 Points) 4 (45 Points) Total (150 Points)

2 1. (30 points) A small, diffuse plate is initially at 300 K and has radiative properties shown below. The plate is suddenly inserted in a large furnace at T f = 000 K. (a) Sketch the spectral absorptivity of the plate, α λ. (b) Calculate total hemispherical absorptivity and emissivity. α = ε =

3 (c) Is this plate approximately gray? Circle one answer and explain your choice. Gray vs. non-gray Reason: (d) After a very long time in the furnace, the radiative properties of this plate could be changed. What do you expect in relative magnitudes of absorptivity and emissivity? Circle one answer and explain your choice. (i) α > ε (ii) α = ε (iii) α < ε (iv) There will be no change in α or ε. Reason: 3

4 Solution (a) 1 (m) Therefore, (m) 4

5 (b) Diffuse surface T surface = 300 K T surr = 000 K 1 T surr = 5 m 000 K = m K : F (0 1) = T surr = 10 m 000 K = 0000 m K : F (0 ) = = 1 F (0 1) + [F (0 ) - F (0 1) ] + 3 [1 - F (0 ) ] = (0.6) ( ) (0.) ( ) = = T surface = 5 m 300 K = 1500 m K : F (0 1) = T surface = 10 m 300 K = 3000 m K : F (0 ) = = 1 F (0 1) + [F (0 ) - F (0 1) ] + 3 [1 - F (0 ) ] = (0.6) ( ) (0.) ( ) = = (c) Non-gray surface (d) As t, T surface T surr.therefore, since both the source and emitter temperature become the same, and therefore the same part of the spectrum determines both. = 5

6 . (35 points) An engineer seeks to dry a contoured surface using only a hot dry air stream, as shown below. A liquid film of water exists on the contoured surface, and dry air at 350 K and a velocity of 10 m/s is passed over it in order to dry it. The temperature of the contoured surface is measured to be 300 K. Furthermore, the engineer also makes measurements of the average Nusselt number and finds that it is of the form: Nu L = 0.5Re L 0.6 Pr 0.4 All properties in the correlation are to be found at the film temperature. Furthermore, the surface area is given to be m and the characteristic length L is given to be 1 m. You are given the following properties: Air properties at 350K Value ρ kg/m 3 μ 08.x10-7 Ns/m C p kj/kgk k 0.03 W/mK Air properties at 300 K ρ kg/m 3 μ 184.6x10-7 Ns/m C p kj/kgk k W/mK Diffusivity of water vapor in mixture, D AB D AB 0.6x10-4 m /s The properties of saturated water are given in the formula sheet. (i) Find the rate of water vapor evaporated from the surface in kg/s. m evap = kg/s 6

7 (ii) Find the required rate of convective heat transfer to the surface in Watts in order for this rate of water vapor evaporation to occur. q conv = W 7

8 Solution (i) Assume linear interpolations are valid for air properties. T film T T s K air,35k 300k 350k kg / m 3 air,35k 300k 350k N s / m c p,air,35k c p,air,300k c p,air,350k kj / kg K k air,35k k 300k k 350k W / m K Re L U L c ; L c 1 m U 10 m / s Re L (1.078 kg / m3 ) (10 m / s) (1 m) N s / m Pr v k c p v Pr c p k c p k ( N s / m )( J / kg K) W / m K Nu L 0.5 Re L 0.6 Pr ( ) 0.6 (0.7037) h L c k air h Nu k L air (106.57) ( W / m K) W / m K L c (1 m) 8

9 h kair h D Le m AB n ; n 0.4 k W / m K 1 Le D c D (1.078 kg / m ) ( J / kg K) ( m / s) AB p AB h W / m K W / m K h h ( m / s) ( ) m m h m m / s m h A[ ] ; A m 0 evap m A, s A, A, kg / m v m / kg A, s water vapor,300k 3 water vapor,300k 3 3 mevap ( m / s) ( m ) ( kg / m ) kg / s m evap kg / s (ii) q m h ; h 438 kj / kg evap evap fg fg, water,300k 3 qevap ( ) ( J / kg) W W is required 9

10 3. (40 points) Consider two inclined diffuse-gray plates of area A 1 and A respectively, as shown below. The plates are infinitely long into the page and are at an angle of 60 o, as shown. The two plates are exposed to a large surroundings at a uniform temperature of 300 K and an emissivity ε =0.8. Surface 1 is insulated, and has an emissivity of ε 1 =0.8. Surface is at a temperature T =1000 K, and has an emissivity of ε =0.5. (i) Find the view factor F 1 F 1 = (ii) Draw a radiation circuit describing the radiation exchange between the surfaces, showing the symbolic form of all relevant resistances and potentials. Surroundings Surface 1 Surface 10

11 (iii) Find the heat transfer rate q leaving surface in W/m. q = W/m (iv) Find the irradiation G on surface in W/m. G = W/m (v) Find the net radiation heat transfer rate from surfaces 1 and to the surroundings in W/m. Heat transfer rate = W/m 11

12 Solution (i) Surface 3 Surface F 11 + F 1 + F 13 = 1 F 11 = 0 By symmetry, F 1 = F 13 F 1 = F 13 = 0.5 Surface 1 (ii) q surr Surrounding E b,surr 1 surr A surr surr J surr 1 A 1 F 13 1 A F 3 Surface 1 Surface q 1 = 0 q E b,1 J 1 J E b, A 1 1 A 1 F 1 A 1

13 (iii) Surrounding is large J surr = E b,surr q 1 A E b, E b,surr 1 1 A F surr 1 A surr surr A 1 F 1 A 1 F 13 q ' 1 L E b, E b,surr 1 1 L F 3 L F 1 1 L F 13 1 ( W / m K 4 ) ( )K (0.1 m) (0.5) 1 1 (0.1 m) (0.5) (0.1 m) (0.5) 1 (0.1 m) (0.5) W / m q ' W / m (iv) q E b, J 1 A q ' E b, J 1 L ( W / m K 4 )(1000 K) 4 J (0.1 m)(0.5) W / m J W / m 13

14 q A (J G ) q ' L (J G ) W / m (0.1 m)( w / m G ) G W / m (v) ' q 1&surr q ' ' as q 1 0 ' q 1&surr W / m 14

15 4. (45 points) A vertically-oriented large nuclear fuel solid with a thickness of t = 0.1 m generates uniform volumetric heat at q (W/m 3 ). One of its surfaces is perfectly insulated, while the other is facing a thin plate with the medium between them evacuated. Both materials are diffuse-gray with ε 1 =0.8 and ε =0.01, and are placed in a large room at 90 K. The other side of the thin plate is exposed to free convection by air at T = 300 K as shown below. The plate temperature is measured to be T = 400 K. You are given air properties evaluated at 350 K: ν = m /s, α = m /s, β = K -1, and k air = W/mK. You may assume that both sides of plate have the same emissivity. T sur = 90 K Nuclear fuel ( ) evacuated Air T air = 300 K 1 (a) Determine net heat flux between surfaces 1 and. You may assume a characteristic length, L=1 m, for natural convection. q 1 = W/m (b) Find the temperature of surface 1. T 1 = K (c) Determine the volumetric heat generation rate. q = W/m 3 15

16 Solution (a) Nu L Ra L 1/6 1 (0.49 / Pr) 9/16 8/7 Ra g (T T ) L 3 s c v (9.81 m / s) ( K 1 )( )K (1 m) 3 ( m / s)( m / s) Pr v m / s m / s Nu L ( ) 1/6 1 (0.49 / ) 9/16 8/ h L c k air ; k air,350k W / m K h Nu L k air (19.644)( W / m K) W / m K L c 1 m q 1 q conv q rad q conv hconv (T T ) ( W / m K)( )K W / m Surrounding is large J surr = E b,surr A 1 1 A 1 F 1 A A A F surr 16

17 q rad E b, E b,surr 1 1 ; F surr 1 F surr q rad (T 4 4 T surr ) 1 1 F surr ( W / m K 4 )( )K W / m For a small surface in a large surrounding (based upon Kirchhoff s law), or q rad (T 4 T 4 surr ) (0.01)( W / m K 4 )( )K W / m q 1 q conv q rad W / m W / m W / m q W / m (b) q 1 E b,1 E b, ; F F 1 q 1 (T 4 1 T 4 ) ( W / m K 4 )(T )K F W / m T K 17

18 (c) q q t ; t 0.1 m W / m q (0.1 m) q W / m 3 18