The water level h(t) in a tank is controlled by an open-loop system as shown below. An armature-controlled DC motor turns a shaft that opens/closes a valve. The inductance of the armature circuit isnegligible: Lo=0. Also, the rotational friction of the motor shaft and the valve is negligible. The height ofthe water in the tank is given by: h(t) = S[1.60(t) – h(t)]dt. The motor torque constant K=10 andthe inertia of the motor-valve assembly is J=6x10³ kg-m?.(a) Determine the differential equation for h(t) and v(t).(b) Determine the transfer function H(s)/V(s).Amplifier10 0k- 50MotorTLValve

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Description: The water level h(t) in a tank is controlled by an open-loop system as shown below. An armature-controlled DC motor turns a shaft that opens/closes a valve. The inductance of the armature circuit isnegligible: Lo=0. Also, the rotational friction of

First of all we have (a) The height of the water tank is h(t) =∫[1.6 θ(t) -h(t) ]…

Answered: The water level h(t) in a tank is…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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The water level h(t) in a tank is controlled by an open-loop system as shown below. An armature-
controlled DC motor turns a shaft that opens/closes a valve. The inductance of the armature circuit is
negligible: Lo=0. Also, the rotational friction of the motor shaft and the valve is negligible. The height of
the water in the tank is given by: h(t) = S[1.60(t) – h(t)]dt. The motor torque constant K=10 and
the inertia of the motor-valve assembly is J=6x10³ kg-m?.
(a) Determine the differential equation for h(t) and v(t).
(b) Determine the transfer function H(s)/V(s).
Amplifier
10 0
k- 50
Motor
TL
Valve

Expert Solution

Step 1

First of all we have

(a) The height of the water tank is

$h (t) = \int_{} [1.6 θ (t) - h (t)] d t . . . . . . . . . . . . . . . . . . . . . . . . . . (1) ω (t) = \frac{d θ (t)}{d t} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) w e k n o w t h a t t h e f o r c e i s d i r e c t l y p r o p o r t i o n a l t o t h e a r m a t u r e c u r r e n t p r o p o r t i o n a l i t y c o n s \tan t X_{m} = 10 I = \frac{d ω (t)}{d t} = k_{m} i_{a} (t) = 10 i_{a} (t) T h e o u t p u t v o l t a g e (V_{a} (t) o f t h e a m p l i f i e r i s g i v e n b y v_{a} (t) = 50 v (t) a n d . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4) v_{a} (t) = 10 i_{a} (t) + v_{b} (t) . . . . . . . . . . . . . . . . . . . . . . . . . . (5) T h e v o l t a g e (v_{0} (t)) i s v o l t a g e a c r o s s m o t o r g i v e n b y \frac{d θ}{d t} = x v_{b} (t) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (6) u \sin g t h e r e l a t i o n s o f t h e e q u a t i o n s (1) t o (6) t o g e t t h e d i f f e r e n t i a l e q u a t i o n s o f v (t) ξ h (t) \frac{d^{2} h (t)}{d t^{2}} = 1.6 \frac{d θ (t)}{d t} - \frac{d h (t)}{d t} \frac{d θ (t)}{d t} = 0.625 (\frac{d^{2} h (t)}{d t^{2}} + \frac{d h (t)}{d t}) \frac{d^{2} θ (t)}{d t^{2}} = 0.625 \frac{d^{3} h (t)}{d t^{3}} + \frac{d^{2} h (t)}{d t^{2}}$

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