This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: (a) Obtain an expression for the laplace transform of the temperature of inner tank T(s). (b) Invert T(s) and obtain T for t= 0,5,10, ∞ Solution: (a) For outer tank WC(Ti – T o ) + hA (T 1 – T 2 ) WC(T 2 – T o ) = ρ C V 2 dt dT 2 (1) At steady state WC(Tis – T o ) + hA (T 1s – T 2s ) WC(T 2s – T o ) = 0  (2) (1) – (2) gives WCTi’ + hA (T 1 ’ – T 2 ’) WCT 2 ’ = ρ C V 2 dt dT ' 2 Substituting numerical values 10 Ti’ + 10 ( T 1 ’ – T 2 ’) – 10 T 2 ’ = 50 dt dT ' 2 Taking L.T. Ti(s) + T 1 (s) – 2T 2 (s) = 5 s T 2 (s) Now Ti(s) = 0, since there is no change in temp of feed stream to outer tank. Which gives s s T s T 5 2 1 ) ( ) ( 1 2 + = Q 10 lb/hr T 1 T 2...
View
Full Document
 Spring '10
 Koslov
 Thermodynamics, Heat, Heat Transfer, TI, Outer Tank

Click to edit the document details