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Control ENG HW_Part_42 - WC(Ti – T o WC(T 1 – T o = ρ...

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Q 9.1. Two tank heating process shown in fig. consist of two identical, well stirred tank in series. A flow of heat can enter tank2. At t = 0 , the flow rate of heat to tank2 suddenly increased according to a step function to 1000 Btu/min. and the temp of inlet Ti drops from 60 o F to 52 o F according to a step function. These changes in heat flow and inlet temp occurs simultaneously. (a) Develop a block diagram that relates the outlet temp of tank2 to inlet temp of tank1 and flow rate to tank2. (b) Obtain an expression for T 2 ’(s) (c) Determine T 2 (2) and T 2 (∞) (d) Sketch the response T 2 ’(t) Vs t. Initially Ti = T 1 = T 2 = 60 o F and q=0 W = 250 lb/min Hold up volume of each tank = 5 ft 3 Density of the fluid = 50 lb/ft 3 Heat Capacity = 1 Btu/lb ( o F) Solution: (a) For tank 1 w Ti T 1 T 2 w q
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Input – output = accumulation
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Unformatted text preview: WC(Ti – T o ) - WC(T 1 – T o ) = ρ C V dt dT 1-------------------------- (1) At steady state WC(Tis – T o ) - WC(T 1 s – T o ) = 0 ------------------------------------(2) (1) – (2) gives WC(Ti – Tis) - WC(T 1 – T 1s ) = ρ C V dt dT 1 ' WTi ’- WT 1 ’ = ρ V dt dT 1 ' Taking Laplace transform WTi(s) = WT 1 (s) + ρ V s T 1 (s) s s Ti s T τ + = 1 1 ) ( ) ( 1 , where τ = ρ V / W. From tank 2 q + WC(T 1 – T o ) - WC(T 2 – T o ) = ρ C V dt dT 2-------------------------- (3) At steady state q s + WC(T 1s – T o ) - WC(T 2s – T o ) = 0 ------------------------------------(4) (3) – (4) gives Q ‘ + WC(T 1 – T 1s ) - WC(T 2 – T 2s ) = ρ C V dt dT 2 ' Q ‘ + WCT 1 ’- WCT 2 ’ = ρ C V dt dT 2 ' Taking Laplace transform...
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