1. CONDUCTION Problem Set Alejandro, Jan Rannel Granada, Gat Hendrix, Jamie Salcedo, Angel Frances Villaflor, Kaye

2. STEADY STATE APPLICATIONS 5. The following data was obtained in a test on a flat-walled furnace, the linings of which consist of a 11.5 cm. non-corrosive brick of unknown conductivity and the outer wall of 20 cm clay brick, also of unknown conductivity. The temperature of the inner wall (flame side) was found to be 735C and that of the outer wall 185C. This furnace was lagged with 5 cm of magnesia (k=0.07 W/mK) thermocouples inserted at various points and the ff. data taken: Calculate the percentage of heat loss that is saved by the lagging. 88C Temp of the outer surface of magnesia475CTemp at the junction of ordinary brick and magnesia700CTemp at the junction of brick layers735CTemp of inner wall (flame side) 3. STEADY STATE APPLICATIONS Magnesia 4. Solution: Basis: Am = 1m2From the data of composition with magnesia: q = T/R R = x/(km)(Am) q = T(km)(Am)/ x Series: q1 = q2 = q3 =qtotal q of magnesia(q3) =(475-88)K(0.07W/mK)( 1m2)= 541.8 W0.05m q of magnesia = q of non-corrosive brick =q of clay brick = q total = 541.2 W km = q(x)/(T)(Am) km of non-corrosive brick = (541.8W)(0.115m)/(735-700)( 1m2) = 1.7802 W/mK km of clay brick = (541.8 W)(0.20m)/(700-475) ( 1m2) = 0.4816 W/mK non-corrosive brickclay brickMagnesia735C700C475C88C 5. Using the data in the composition without the magnesia: Series: Rtotal = R1 + R2 + R3 Getting the R: R = x/(km)(Am) R non-corrosive brick = (0.115m) /(1.7802 W/mK) ( 1m2) = 0.06460 K/W R of clay brick = (0.20m)/(0.4816 W/mK)( 1m2) = 0.4153 K/W Rtotal = 0.4799K/W q total = (735-185)K/(0.4799 K/W) = 1146.07 W percentage of heat loss that is saved by the lagging: (q total without the magnesia q total with magnesia)x100% q total without magnesia (1146.07W-541,8W)x 100% = 52.73% 1146.07Wnon-corrosive brickclay brick735C185C 6. UNSTEADY STATE APPLICATIONS

13. (US)What length of time is required to heat a 2.5 cm diameter, 5 cm long copper cylinder ( k = 380 W/mK ) from 16 oC to 93 oC in a furnace whose temperature is 315 oC? The value of h at the cylinder surface is 204 KJ/hm2K andits thermal diffusivity is 0.186 m2/s.

7. UNSTEADY STATE APPLICATIONS

13. (US)What length of time is required to heat a 2.5 cm diameter, 5 cm long copper cylinder ( k = 380 W/mK ) from 16 oC to 93 oC in a furnace whose temperature is 315 oC? The value of h at the cylinder surface is 204 KJ/hm2K andits thermal diffusivity is 0.186 m2/s.

Given: X = 2.5cm To = 16 oC h = 204 kJ/hm2K L = 5cm T = 93 oC = 0.186 m2/s k = 38.0 W/mKT1 = 315 oC 8. UNSTEADY STATE APPLICATIONS Solution: h = 204_kJ_ x1000Jx1h___ = 56.67 W/m2K hm2K 1 kJ 3600 s X1 =2.5/100= 0.0125m 2 Y =315 93= 0.74 315 16 m =k__= 38.0 W/mK__________= 53.64 hX156.67 W/m2K (0.0125m) From the Heisler Chart(Cylinder): X = 10.1 =t__=(0.186) tt = 0.0085s X12 (0.0125)2 9. 10. Geankoplis Problem

4.2-3.Removal of Heat from a Bath . A cooling coil of 1.0ft of 308 stainless steel have an average thermal conductivity of 15.23 W/mK, inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40oF and is 80oF on the outside. Calculate the heat removal in BTU/s and watts. Where k is in BTU/ h ft oF and T is in oF.

11.

4.2-3.Removal of Heat from a Bath . A cooling coil of 1.0ft of 308 stainless steel have an average thermal conductivity of 15.23 W/mK, inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40oF and is 80oF on the outside. Calculate the heat removal in BTU/s and watts. Where k is in BTU/ h ft oF and T is in oF.

Given : L = 10 ft = 0.3048m k = 15.23 W/mK Di = 0.25 in. = 6.35x10-3m Do = 0.40 in. = 0.01016m T1 = 40oF = 4.44oC + 273K = 277.44K T2 = 80oF = 26.67oC + 273K = 299.67K 12.

4.2-3.Removal of Heat from a Bath . A cooling coil of 1.0ft of 308 stainless steel have an average thermal conductivity of 15.23 W/mK, inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40oF and is 80oF on the outside. Calculate the heat removal in BTU/s and watts. Where k is in BTU/ h ft oF and T is in oF.

Solution: X = r2 - r1 = (0.01016/2) (6.35x10-3/2) =1.905x10-3m Dlm =0.01016-6.35 x10-3___= 8.11x10-3 ln (0.1016/6.35x10-3) Am = lDlm q =T1 - T2=(299.67 277.44)K______ X__ 1.905x10-3m________________= 1380.16wattsor J/s kmAm 15.23W/mK[()(0.3048m)(8.11x10-3)]m2 q = 1380.16wattsorJx1BTU= 1.31 BTU/s s1055J 13. Geankoplis Problem

5.3-5.Cooling a Slab of Meat.A slab of meat 25.4 mm thick originally at a uniform temperature of 10cC is to be cooked from both sides until the center reaches 121cC in an oven at 177cC. The convection coefficient can be assumed constant at 25.6 W/m2-K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10-4m2/h. Use the Heisler Chart.

14.

5.3-5.Cooling a Slab of Meat.A slab of meat 25.4 mm thick originally at a uniform temperature of 10cC is to be cooked from both sides until the center reaches 121cC in an oven at 177cC. The convection coefficient can be assumed constant at 25.6 W/m2-K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10-4m2/h. Use the Heisler Chart.

Given: X = 25.4mm = 0.0254m To = 10oC T = 121 cC(X = 0,center) T1 = 177 cC h = 25.6 W/m2-K k = 0.69 W/m-K =5.85x10-4m2/h 15.

5.3-5.Cooling a Slab of Meat.A slab of meat 25.4 mm thick originally at a uniform temperature of 10cC is to be cooked from both sides until the center reaches 121cC in an oven at 177cC. The convection coefficient can be assumed constant at 25.6 W/m2-K. Neglect any latent heat changes and calculate the time required. The thermal conductivity is 0.69 W/m-K and the thermal diffusivity 5.85x10-4m2/h. Use the Heisler Chart.

Solution: X1 = 0.0254/2 = 0.0127m Y =T1 T=177 121= 0.34 T1 - To177 10 m =k__= 0.69.0 W/mK__________= 2.1 hX1(25.6 W/m2K)(0..0127m) n = 0___= 0 0.0127 From the Heisler Chart (Plate): X = 3.2 =t__=( 5.85x10-4m2/h ) tt = 0.88h x3600s= 3168s X12 ( 0.0127 )2 1h 16.