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Contains practicals related to ipdc lab BVCOE
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Contents
1 Description 2 Specifications 3 Installation requirements 4 Installation Commissioning
5 Troubleshooting 6 Components used 7 Packing slip 8 Warranty
9 Theory 10 Experiments
Apex Innovations
The set up is designed to study dynamic response of single and multi capacity processes when connected in interacting and non- interacting mode. It is combined to study
in interacting and non-interacting mode. The components are assembled on frame to form tabletop mounting.
Rotameter
R2 R
R
Tank 1
Tank 2 Tank 3
Pump
Product Interacting and Non interacting system Product code 328 Rotameter 10-100 LPH Process tank Acrylic, Cylindrical, Inside Diameter 92mm With graduated scale in mm. (3 Nos) Supply tank SS Pump Fractional horse power, type submersible Overall dimensions 410Wx350Dx705H mm
Shipping details Gross volume 0.24m^3 , Gross weight 60kg, Net weight 26kg
Electric supply Provide 230 +/- 10 VAC, 50 Hz, single phase electric supply with proper earthing. (Neutral – Earth voltage less than 5 VAC)
Specifications
Description
Installation requirements
This product is warranted for a period of 12 months from the date of supply against manufacturing defects. You shall inform us in writing any defect in the system noticed during the warranty period. On receipt of your written notice, Apex at its option either repairs or replaces the product if proved to be defective as stated above. You shall not return any part of the system to us before receiving our confirmation to this effect. The foregoing warranty shall not apply to defects resulting from: Buyer/ User shall not have subjected the system to unauthorized alterations/ additions/ modifications. Unauthorized use of external software/ interfacing. Unauthorized maintenance by third party not authorized by Apex. Improper site utilities and/or maintenance. We do not take any responsibility for accidental injuries caused while working with the set up.
Apex Innovations Pvt. Ltd. E9/1, MIDC, Kupwad, Sangli-416436 (Maharashtra) India Telefax:0233-2644098, 2644398 Email: [email protected] Web: www.apexinnovations-ind.com
Warranty
Step function: Mathematically, the step function of magnitude A can be expressed as X (t) = A u (t) where u (t) is a unit step function. It can be graphically represented as
To study the transient response for step function, consider the system consisting of a tank of uniform cross sectional area A1 and outlet flow resistance R such as a valve. q (^) o ,volumetric^ flow^ rate^ through^ the^ resistance,^ is^ related^ to^ head^ h^ by^ a^ linear relationship qo = h/R -------(1) Writing a transient mass balance around the tank: Mass flow in - Mass flow out = rate of accumulation of mass in the tank. 〈q (t) - 〈 qo (t) = d(〈Ah)/dt q(t) - qo (t) = A1 dh /dt ------(2) Combining equation (1) and (2) to eliminate qo (t) gives the following linear differential equation: q - h/R = A1 dh/dt -------(3) Initially the process is operating at steady state, which means that dh/dt = 0. Therefore equation (3) becomes as qs - h (^) s /R= 0 -----------(4) Where, the subscript s indicates the steady state value of the variable. Subtracting equation (4) from (3) (q - q (^) s) = 1/R ( h - hs ) + A1 d(h - hs ) / dt ----(5) Defining deviation variable
Theory
In non-interacting system we assume the tanks have uniform cross sectional area and the flow resistance is linear. To find out the transfer function of the system that relates h 2 to q, writing a mass balance around the tanks, we proceed as follows
We can write mass balance at tank1 as q-q 1 = A 1 (dh 1 / dt) ………………………… (1)
A mass balance at tank 2 is given as q 1 -q 2 = A 2 (dh 2 / dt) ……………………….. (2)
The flow head relationships for the two linear resistances in non-interacting system are given by the expressions q 1 =(h 1 / R 1 ) ………………………………. (3)
q 2 =(h 2 / R 2 ) ……………………………… (4)
From (1) and (3) Q 1 (s) 1 ----------- = ------------ ……………………… (5) Q(s) τ 1. S + 1
Where Q 1 = q 1 - q1s, Q = q - qs and τ 1 = A1 R
From (2) and (4)
H 2 (s) R 2 ----------- =------------ ……………………….. (6) Q1(s) τ2. S + 1 Where H 2 = h 2 - h2s and τ 2 = A 2 R 2
Overall transfer function can be calculated as follows H 2 (s) R 2 ----------- =------------------------ ………………. ( 7) Q(s) (τ1. S + 1) ( τ2. S + 1 ) For a step change of magnitude A Q(t) = A u(t) So, Q(s) = A / s AxR H2(s) = -------------------------------------------------- ….( 8) s x (τ1. S + 1) X ( τ2. S + 1 )
H 2 at time t is given by (τ 1 x τ 2 ) 1 1
H 2 (t)= A R2 [1 - -------- {--- e -^ t/^ τ^1 -^ ----- e -t/^ τ^2 }] ……………(9) ( τ 1 - τ 2 ) τ 2 τ 1
Mathematically, the impulse function of magnitude A is defined as X(t) = A (t) Where (t) is the unit impulse function. Graphically it can be described as
Overall transfer function of the system as described in previous experiment H 2 (s) R 2 ----------- = ------------------------ ……………… (1) Q(s) (τ1. S + 1) ( τ2. S + 1 ) For a impulse change of magnitude V (volume added to the system) Q(t) = V (t) So, Q (s) = V VxR 2 H 2 (s) = -------------------------------------------------- … (2) (τ 1. S + 1) ( τ2. S + 1) For impulse change H 2 at time t is given by
e -^ t/τ^1 - e -^ t/τ^2 H 2 (t)= V R 2 [--------------------------] ………………… (3) ( τ 1 -^ τ 2 )
Considering non-linear resistance at outlet valve of the tank R2 can be calculated as R 2 = 2dH 2 /dQ Where dH 2 is change in level of tank2 and dQ is change of flow from initial to final state. Put the values in equation (3) to find out H (t) (^) Predicted and plot the graph of H (t)
Predicted and H (t)Observed Vs time.
As mentioned in theory part of experiment 3, impulse function is described as X (t)= A (t) Overall transfer function of the system as described in previous experiment H 2 (s) R 2 -------- = ----------------------------------------- ………(1) Q(s) τ 1 τ 2 s^2 + (τ 1 + τ 2 + A 1 R 2 ) s +
For a impulse change of magnitude V (volume added to the system) Q(t) = V (t) So, Q (s) = V V R 2 H 2 (s) = -------------------------------------------------- ---(2) τ 1 τ 2 s^2 + (τ 1 + τ 2 + A 1 R 2 ) s +
For impulse change H 2 at time t is given by
V R 2 H 2 (t)= ---------------- [e (αt)^ – e (βt)] -----(3) τ 1 τ 2 (α-β)
(For α, β refer theory part of experiment No. 4) Considering non- linear valve resistance, the resistance at outlet of both tanks can be calculated as R 1 = 2 dH1/dQ ------------(4) R 2 = 2 dH2/dQ ------------(5)
1 Step response of single capacity system
Procedure
Observations Diameter of tank mm: ID 92 mm Initial flow rate (LPH): Initial steady state tank level (mm): Final flow rate (LPH): Final steady state tank level (mm): (Fill up columns H(t) observed and H(t) predicted after calculations) Sr. No.
Time (sec)
Level (mm)
H(t) observed (mm)
H(t) predicted (mm)
Calculations H (t) (^) observed = (Level at time t - level at time 0) x 10 –3^ m H(t) (^) Predicted = AR { (1- e -t/τ) } Where H (t)Predicted is level predicted at time t in m. A = magnitude of step change = Flow after step input - Initial flow rate in m 3 /sec. R = Outlet valve resistance in sec/m 2 Considering non linear resistance at outlet, it can calculated as R = dH /dQ Where dH is change in level (Final steady state level - Initial steady state level) and dQ is change flow (Final flow rate after step change - Initial flow rate). τ = time constant in sec. =A1 x R Where A1 is area of tank in m 2 and R is resistance of outlet
Experiments
2 Step response of first order systems arranged in non-interacting mode
Procedure
Observations Diameter of tanks: ID 92mm Initial flow rate (LPH): Initial steady state level of Tank 1 (mm): Initial steady state level of Tank 2 (mm): Final flow rate (LPH): Final steady state level of Tank 1 (mm): Final steady state level of Tank 2 (mm): (Fill up columns H(t) observed and H(t) predicted after calculations) Sr. No. Time (sec)
Level of tank 2 (mm)
H(t) observed (mm)
H(t) predicted (mm) 1 0 2 30 3 60 4 --
Calculations H (t) (^) observed = (Level at time t - level at time 0) x 10 - (τ 1 x τ 2 ) 1 1
H (t) (^) Predicted = A R 2 [1 - -------- { --- e-^ t/^ τ^1 - ----- e -t/^ τ^2 }] ------(1) ( τ 1 - τ 2 ) τ 2 τ 1
Where H (t)Predicted is level in Tank2 predicted at time t in m. A = magnitude of step change = Flow after step input - Initial flow rate in m 3 /sec. τ 1 = A 1 x R 1 τ 2 = A 2 x R 2 Where τ 1 is time constant of tank1, A 1 is area of tank1 and R 1 is resistance of outlet valve of tank1. τ 2 is time constant of tank2, A 2 is area of tank2 and R 2 is resistance of outlet valve of tank Area of tank 1 = ◊/4 (d 12 ) in m 2 Area of tank 2 = ◊/4 (d 22 ) in m 2 Considering non-linear resistance at outlet valve of both tanks, it can be calculated as
R 1 = dH 1 /dQ R 2 = dH 2 /dQ Where dH 1 is change in level of tank1 and dQ is change flow of from initial to final state and dH 2 is change in level of tank2 at initial and final state. Put the values in equation (1) to find out H (t) (^) Predicted and plot the graph of H (t) Predicted and H (t)Observed Vs time. Sample calculations & results Refer MS Excel program for calculation and graph plotting. Comments Observed response fairly tallies with theoretically calculated response. Deviations observed may be due to following factors:
4 Step response of first order systems arranged in interacting mode
Procedure
Diameter of tanks: ID 92mm Initial flow rate (LPH): Initial steady state level of Tank 3 (mm): Initial steady state level of Tank 2 (mm): Final flow rate (LPH): Final steady state level Tank 3 (mm): Final steady state level Tank 2 (mm): (Fill up columns H(t) observed and H(t) predicted after calculations) Sr. No. Time (sec)
Level of tank 2 (mm)
H(t) observed (mm)
H(t) predicted (mm) 1 0 2 30 3 60 4 --
Calculations H (t) (^) Observed = (Level at time t - level at time 0 ) x 10 -3^ m [(1/α) exp (αt)] – [(1/β) exp ( βt)] (H 2 ) t (^) Predicted = AR 2 {1 - ---------------------------------------------------} -----(1) [1/α - 1/β] Where A = magnitude of step change = Flow after step input - Initial flow rate in m 3 /sec τ1 = A 1 x R 1 τ 2 = A 2 x R 2 Where τ1 is time constant of tank1, A 1 is area of tank1 and R 1 is resistance of outlet valve of tank1. τ 2 is time constant of tank2, A 2 is area of tank2 and R 2 is resistance of outlet valve of tank Considering non linear resistance at outlet valve of both tanks, it can calculated as R 1 = dH 1 /dQ and R 2 = dH 2 /dQ Where dH is change in tank height for change in flow dQ. Calculate values of b, α and β from equations given in theory part. Put the values in equation (1) to find out H (t) (^) Predicted and plot the graph of H (t)
Predicted and H (t)Observed Vs time
Sample calculations & results Refer MS Excel program for calculation and graph plotting. Comments
Observed response fairly tallies with theoretically calculated response. Deviations observed may be due to following factors:
