Control ENG HW_Part_45

Control ENG HW_Part_45 - (a Obtain an expression for the...

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( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 2 10 5 15 ) ( 2 1 20 1 5 15 ) ( ) 2 1 )( 1 ( 20 15 ) ( 1 10 ) 2 1 )( 1 ( 5 ) ( 1 ) ( 2 1 ) 2 1 )( 1 ( ) ( 2 1 ) ( 1 ) ( 2 1 ) 1 ( ) ( 2 1 ) ( t t b a b e e t T s s s s T s s s s s T s s s s s s T s s T s s s T s T s s T s s T s T - - - - = + - + - = + + + = + - + + = + - + + = + - + = (c) T 2 ’(2) = 10.64 o F T 2 (2) = T 2 ’(2) + T 2s = 60 + 10.64 = 70.64 o F Q – 9.3. Heat transfer equipment shown in fig. consist of tow tanks, one nested inside the other. Heat is transferred by convection through the wall of inner tank. 1. Hold up volume of each tank is 1 ft 3 2. The cross sectional area for heat transfer is 1 ft 2 3. The over all heat transfer coefficient for the flow of heat between the tanks is 10 Btu/(hr)(ft 2 )( o F) 4. Heat capacity of fluid in each tank is 2 Btu/(lb)( o F) 5. Density of each fluid is 50 lb/ft 3 Initially the temp of feed stream to the outer tank and the contents of the outer tank are equal to 100 o F. Contents of inner tank are initially at 100 o F. the flow of heat to the inner tank (Q) changed according to a step change from 0 to 500 Btu/hr.

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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...
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Control ENG HW_Part_45 - (a Obtain an expression for the...

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