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Unformatted text preview: K C (1 + τ D s )(1 + τ m s )
R τ 1τ m s + (τ 1 + τ m + K Cτ D ) s + ( K C + 1) = τ2 = τ 1τ m
KC + 1 τ 1τ m τ= KC + 1 ξ= 1 τ 1 + τ m + k Cτ D
2 τ 1τ m (k C + 1) b) τ 1 = 1 min; τ m = 10s;ξ = 0.7 τD = 0
⇒ 0.7 = ×
600(k C + 1) kc=3.167 τ D = 3s
2) ⇒ 0.7 = 70 + 3k C
600(k C + 1) k C = 5.255 k
c)for R = ; c(∞) = C
kC + 1 For τ D = 0; C (∞) = 0.76; offset = 0.24
for τ D = 3s; C (∞) = 0.84; offset = 0.16
4.167 = 105.57 = period
1− ξ 2 2π ×
for τ D = 0; period = 600
6.255 = 86.17 s = period
for τ D = 3s; period =
1 − ( 0.7) 2
2π Comments: Advantage of adding derivative mode is lesser offset lesser period
13.6The thermal system shown in fig P 13.6 is controlled by PD controller.
Data ; w = 250 lb/min; ρ = 62.5 lb/ft3;
V1 = 4 ft3,V2=5 ft3; V3=6ft3;
C = 1 Btu/(lb)(°F)
Change of 1 psi from the controller changes the flow rate of heat of by 500
Btu/min. the temperature of the inlet stream may vary. There is no lag in the
(a) Draw a block diagram of the control system with the appropriate transfer
function in each block.Each transfer function should contain a numerical values of
(b) From the block diagram, determine the overall transfer function relating the
temperature in tank 3 to a change in set point.
(c ) Find the offset for a unit steo change in inlet temperature if the controller
gain KC is 3psi/°F of temperature error and the derivative time is 0.5 min. ...
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This note was uploaded on 11/13/2011 for the course COP 4355 taught by Professor Koslov during the Spring '10 term at University of Florida.
- Spring '10