modelling of liquid-level system.pdf
TRANSCRIPT
Modelling of Liquid-Level System
LCS Lab Report
By:
H. M. Asim Qayyum
Haseeb-ul-Hassan
Abdus Samad
Azmat Saifullah Khan
H. M. Waqar Sharif
Sikandar Ali
Submitted to:
Mr. M. Abubakar DEE, PIEAS
Submission Date:
24 March 2015
1
In this lab study we modelled the height of liquid in a 3-tank system as a function of time.
Description of the System The system consisted of three cylindrical tanks; two of which had circular bases, while the third one was lying along its lateral curved surface, as shown in figure 1. Liquid could flow through valves from tank 1 to tank 2, and from tank 2 to tank 3. A pump was utilized to pump water from tank 3 to tank 1.
Figure 1: Schematic diagram of the system
Modelling Suppose that Q1 be the flow rate in the pipe joining tank 1 and tank 2, Q2 be the flow rate in the pipe joining tank 2 and tank 3, and Q3 = u be the flow rate in the pipe joining tank 3 and tank 1 through pump; and that H1, H2, and H3 be the heights of liquid in the tank 1, tank 2, and tank 3 respectively. The flow in the system is taken to be turbulent. Then the differential equations governing the heights of liquid in the tanks can be derived as follows.
Tank 1 ππ1ππ‘
= π’ β π1
Since the cross sectional area A is constant, while height (H) of water is function of time:
π(π΄1 Γ π»1)ππ‘
= π΄1ππ»1ππ‘
= π’ β π1
π΄1ππ»1ππ‘
= π’ β πΎ1οΏ½π»1
2
Tank 2 It can be seen in the figure 2 that the area of section PQR equals:
π΄22
= π΄πππ ππ π πππ‘ππ πππ β π΄πππ ππ π‘πππππππ πππ
And, by Pythagorean Theorem:
ππποΏ½οΏ½οΏ½οΏ½ = οΏ½π2 β (π β π»2)2 = οΏ½2ππ»2 β π»22
Where r is the radius of the tank, and H2 is the height of liquid (liquid level) as shown in the figure 2. So:
π΄22
=12ππ2 β
12
(οΏ½2ππ»2 β π»22 )(π β π»2)
or
π΄2 = ππ2 β (οΏ½2ππ»2 β π»22 )(π β π»2)
Further, since
π = cosβ1(π β π»2π
) = cosβ1(1 βπ»2π
)
So
π΄2 = π2 cosβ1(1 βπ»2π
) β (οΏ½2ππ»2 β π»22 )(π β π»2)
Now, if βL2β is the length of tank 2, then volume of liquid in the tank is V2=L2A2. Since length is constant, so
ππ2ππ‘
= πΏ2ππ΄2ππ‘
= πΏ2πππ‘ οΏ½
π2 cosβ1 οΏ½1 βπ»2π οΏ½
β οΏ½οΏ½2ππ»2 β π»22 οΏ½ (π β π»2)οΏ½ = π1 β π2
πΏ2πππ‘ οΏ½
π2 cosβ1 οΏ½1 βπ»2π οΏ½
β οΏ½οΏ½2ππ»2 β π»22 οΏ½ (π β π»2)οΏ½ = πΎ1οΏ½π»1 β πΎ2οΏ½π»2
Figure 2: Cross-section, perpendicular to plane of page, of tank 2 in figure 1
3
Tank 3 ππ3ππ‘
= π2 β π’
Since the cross sectional area A3 is constant, while height (H3) of water is function of time:
π΄3ππ»3ππ‘
= πΎ2οΏ½π»2 β π’
Remarks It is a first-order non-linear system. If the flow rate that the pump creates is known, the three equations derived above can be solved by using numerical methods and the height of liquid in each tank can be calculated as a function of time.