Download Thermodynamics Problems: Heat Transfer and Temperature Profiles and more Assignments Heat and Mass Transfer in PDF only on Docsity! Problem 1 For a constant mass, pressure process p dT Q mC dt = so ( ) f f i i T Tt t t p p p 0 0 T T dT Q Qdt mC dt m C dT mR C / R dT dt = = = =∫ ∫ ∫ ∫ ( )f i T 2 t T Q mR A BT CT dT= + +∫ ( ) ( ) ( )2 2 3 3t f i f i f iB CQ mR A T T T T T T2 3 = − + − + − The molecular weight of water is 18g / mol , so 1 mol m 500 g x 27.78 mol 18 g = = , or J 1 mol R 8.314 x 0.4619 J / g K mol K 18 g = = − − o i o 3 6 f T 25 C 298.15K T 100 C 373.15K A 8.712 B 1.25 x10 C 0.18x10− − = = = = = = = − so 5tQ 1.58 x10 J = Problem 2 at thermal equilibrium, f f1 2 fT T T= = For block 1: ( )f i T 1 t,1 1 1 1 1 f 1T U Q m C dT m C T T∆ = = = −∫ (constant mass; no volume change) For block 2: ( )2 t,2 2 2 f 2U Q m C T T∆ = = − with t ,1 t ,2Q Q= − then ( ) ( )1 1 f 1 2 2 f 2m C T T m C T T− = − − ( ) ( ) ( ) 1 1 f 1 2 2 f 2 f 1 1 2 2 1 1 1 2 2 2 m C T T m C T T 0 T m C m C m C T m C T − + − = + = + 1 1 1 2 2 2f 1 1 2 2 m C T m C T T m C m C += + The final temperature is simply a weighted average of the initial temperatures, with the weights related to the product mC for each block. For ( )1 2 1 2 f 1 2 1 m m , C C T T T 2 = = = + ( ) ( )1 2 1 2 f 1 1 2 2 1 2m m , C C T m T m T / m m≠ = = + + Since t ,1 t ,2 t ,1Q Q , we choose to solve for Q= − , in which ( ) ( )1 1 2 2 11 1 1 2 2 2t,1 1 1 f 1 1 1 1 1 2 2 1 1 2 2 m C m C )Tm C T m C T Q m C T T m C m C m C m C m C + += − = − + + so ( )1 1 2 2 2 1 1 t,1 1 1 2 2 m C m C T T U Q m C m C − ∆ = = + ( )1 1 2 2 1 2 2 t,2 1 1 2 2 m C m C T T U Q m C m C − ∆ = = + where 1 2U U∆ = −∆ m1 C1 m2 C2 T1 T2 T1 f T2 f time (seconds) T To T1 a 3a 0 time (seconds) 0 Q a 3a ( ) ( )11 oo T T m A B T T a − + − ( )1 omA T T a − ( )1 2 omA T T a − − ( ) ( )11 2 o o T T m A B T T a − − + − time (seconds) Qt 0 a 3a 0 ( ) ( ) 2 1 1 2 o o B T T m A T T − − + (A,B>0) Problem 4 For one dimensional steady-state conduction, with constant cross-sectional area ( )1 2T Tq k L − = The x-axis is defined such that the + x direction is to the right. Hence, if q > 0 then heat is flowing from left to right, or in the + x direction. If q < 0, then heat is flowing from right to left, or in the –x direction. Also, dT q k dx = − Fourier’s Law of heat conduction so dT q k 25 W / m K L 0.5 m dx k = − = − = Case 1: 1 2T 400K T 300K= = so ( )( ) 225W / m K 100Kq 5000W / m 0.5m − = = dT q 200 K / m dx k = − = − or ( ) ( )2 1T T 300 400 KdT 200 K / m dx L 0.5 m − − = = = − Case 2: o1 dT T 100 C 250K / m dx = = − 2 dT q k 6250W / m dx = − = Now 1 2 qL T T 125K k − = = this unit is really a temperature difference and not an absolute temperature Note that 1 degree Kelvin = 1 degree o C so o2T 25 C= − x T 0 L 180oC 80oC direction of heat flux dT/dx>0 T2 T1 Case 3: 2 dT W K q k 25 200 5000 W / m dx m K m = − = − = − − 1 2 o o 1 o 2 qL T T 100K k 100 C with T 80 C then T 180 C − = = − = − = = Case 4: ( ) ( )( ) ( ) 2 o 1 2 0.5mqL T T 4000W / m 80K 80 C temperature difference k 25W / m K − = = = = − with o2T 5 C= − then o1T 75 C= Finally, ( )1 2 2 1T T T TdT q 160K / m dx k L L − −−= = − = = − Case 5: ( ) ( )( ) 2 1 2 0.5mqL T T 3000W / m 60K k 25W / m K − = = − = − − with o o1 2T 30 C, then T 90 C= = 2 1 T TdT 120K / m dx L −= = Case 2 Case 3 x T 0 L T1 T2 100oC -25oC direction of heat flux dT/dx<0
