CSU MECH 344 - HW2 - Solutions (9 pages)

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HW2 - Solutions



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HW2 - Solutions

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Pages:
9
School:
Colorado State University- Fort Collins
Course:
Mech 344 - Heat and Mass Transfer
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PROBLEM 2 14 KNOWN Two dimensional body with specified thermal conductivity and two isothermal surfaces of prescribed temperatures one surface A has a prescribed temperature gradient FIND Temperature gradients T x and T y at the surface B SCHEMATIC ASSUMPTIONS 1 Two dimensional conduction 2 Steady state conditions 3 No heat generation 4 Constant properties ANALYSIS At the surface A the temperature gradient in the x direction must be zero That is T x A 0 This follows from the requirement that the heat flux vector must be normal to an isothermal surface The heat rate at the surface A is given by Fourier s law written as T W K 10 2m 30 600W m y A m K m q y A k w A On the surface B it follows that T y B 0 in order to satisfy the requirement that the heat flux vector be normal to the isothermal surface B Using the conservation of energy requirement Eq 1 12c on the body find q y A q x B 0 or q x B q y A Note that q x B k w B T x B and hence T x B q y A k wB 600 W m 10 W m K 1m 60 K m COMMENTS Note that in using the conservation requirement q in q y A and q out q x B PROBLEM 2 17 KNOWN Electrical heater sandwiched between two identical cylindrical 30 mm dia 60 mm length samples whose opposite ends contact plates maintained at To FIND a Thermal conductivity of SS316 samples for the prescribed conditions A and their average temperature b Thermal conductivity of Armco iron sample for the prescribed conditions B c Comment on advantages of experimental arrangement lateral heat losses and conditions for which T1 T2 SCHEMATIC ASSUMPTIONS 1 One dimensional heat transfer in samples 2 Steady state conditions 3 Negligible contact resistance between materials PROPERTIES Table A 2 Stainless steel 316 T 400 K k ss 15 2 W m K Armco iron T 380 K k iron 67 2 W m K ANALYSIS a For Case A recognize that half the heater power will pass through each of the samples which are presumed identical Apply Fourier s law to a sample q kA c k T x 0 5 100V 0 353A 0 015 m q x 15 0 W m K 2 A c T 0 030 m 4 25 0 C The total temperature drop across the length of the sample is T1 L x 25 C 60 mm 15 mm 100 C Hence the heater temperature is Th 177 C Thus the average temperature of the sample is T To Th 2 127 C 400 K We compare the calculated value of k with the tabulated value see above at 400 K and note the good agreement b For Case B we assume that the thermal conductivity of the SS316 sample is the same as that found in Part a The heat rate through the Armco iron sample is Continued PROBLEM 2 17 Cont 0 030 m 15 0 C qiron q heater qss 100V 0 601A 15 0 W m K 4 0 015 m qiron 60 1 10 6 W 49 5 W 2 where q ss k ssA c T2 x 2 Applying Fourier s law to the iron sample q x 49 5 W 0 015 m k iron iron 2 70 0 W m K 2 Ac T2 0 030 m 4 15 0 C The total drop across the iron sample is 15 C 60 15 60 C the heater temperature is 77 60 C 137 C Hence the average temperature of the iron sample is T 137 77 C 2 107 C 380 K We compare the computed value of k with the tabulated value see above at 380 K and note the good agreement c The principal advantage of having two identical samples is the assurance that all the electrical power dissipated in the heater will appear as equivalent heat flows through the samples With only one sample heat can flow from the backside of the heater even though insulated Heat leakage out the lateral surfaces of the cylindrically shaped samples will become significant when the sample thermal conductivity is comparable to that of the insulating material Hence the method is suitable for metallics but must be used with caution on nonmetallic materials For any combination of materials in the upper and lower position we expect T1 T2 However if the insulation were improperly applied along the lateral surfaces it is possible that heat leakage will occur causing T1 T2 PROBLEM 2 34 KNOWN Steady state conduction with uniform internal energy generation in a plane wall temperature distribution has quadratic form Surface at x 0 is prescribed and boundary at x L is insulated FIND a Calculate the internal energy generation rate q by applying an overall energy balance to the wall b Determine the coefficients a b and c by applying the boundary conditions to the prescribed form of the temperature distribution plot the temperature distribution and label as Case 1 c Determine new values for a b and c for conditions when the convection coefficient is halved and the generation rate remains unchanged plot the temperature distribution and label as Case 2 d Determine new values for a b and c for conditions when the generation rate is doubled and the 2 convection coefficient remains unchanged h 500 W m K plot the temperature distribution and label as Case 3 SCHEMATIC T x a bx cx2 k 5 W m K q T 0 To 120oC Insulated boundary o Too 20 C h 500 W m2 K Fluid x L 50 mm ASSUMPTIONS 1 Steady state conditions 2 One dimensional conduction with constant properties and uniform internal generation and 3 Boundary at x L is adiabatic ANALYSIS a The internal energy generation rate can be calculated from an overall energy balance on the wall as shown in the schematic below E in E out E gen 0 q conv E in where h T To q L 0 1 q h T To L 500 W m 2 K 20 120 C 0 050 m 1 0 106 W m3 To q conv k q To q conv Too h Egen L x a Overall energy balance qx 0 qx L x L x b Surface energy balances 2 b The coefficients of the temperature distribution T x a bx cx can be evaluated by applying the boundary conditions at x 0 and x L See Table 2 2 for representation of the boundary conditions and the schematic above for the relevant surface energy balances Boundary condition at x 0 convection surface condition E in E out q conv q x 0 0 where q x 0 k dT dx x 0 h T To k 0 b 2cx x 0 0 Continued PROBLEM 2 34 Cont b h T To k 500 W m 2 K 20 120 C 5 W m K 1 0 104 K m 2 Boundary condition at x L adiabatic or …


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