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 𝑑𝑉1𝑑𝑡

= 𝑢 − 𝑄1

Since the cross sectional area A is constant, while height (H) of water is function of time:

𝑑(𝐴1 × 𝐻1)𝑑𝑡

= 𝐴1𝑑𝐻1𝑑𝑡

= 𝑢 − 𝑄1

𝐴1𝑑𝐻1𝑑𝑡

= 𝑢 − 𝐾1�𝐻1

2

Tank 2 It can be seen in the figure 2 that the area of section PQR equals:

𝐴22

= 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟 𝑃𝑂𝑅 − 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑃𝑄𝑂

And, by Pythagorean Theorem:

𝑚𝑃𝑄���� = �𝑟2 − (𝑟 − 𝐻2)2 = �2𝑟𝐻2 − 𝐻22

Where r is the radius of the tank, and H2 is the height of liquid (liquid level) as shown in the figure 2. So:

𝐴22

=12𝜑𝑟2 −

12

(�2𝑟𝐻2 − 𝐻22 )(𝑟 − 𝐻2)

or

𝐴2 = 𝜑𝑟2 − (�2𝑟𝐻2 − 𝐻22 )(𝑟 − 𝐻2)

Further, since

𝜑 = cos−1(𝑟 − 𝐻2𝑟

) = cos−1(1 −𝐻2𝑟

)

So

𝐴2 = 𝑟2 cos−1(1 −𝐻2𝑟

) − (�2𝑟𝐻2 − 𝐻22 )(𝑟 − 𝐻2)

Now, if ‘L2’ is the length of tank 2, then volume of liquid in the tank is V2=L2A2. Since length is constant, so

𝑑𝑉2𝑑𝑡

= 𝐿2𝑑𝐴2𝑑𝑡

= 𝐿2𝑑𝑑𝑡 �

𝑟2 cos−1 �1 −𝐻2𝑟 �

− ��2𝑟𝐻2 − 𝐻22 � (𝑟 − 𝐻2)� = 𝑄1 − 𝑄2

𝐿2𝑑𝑑𝑡 �

𝑟2 cos−1 �1 −𝐻2𝑟 �

− ��2𝑟𝐻2 − 𝐻22 � (𝑟 − 𝐻2)� = 𝐾1�𝐻1 − 𝐾2�𝐻2

Figure 2: Cross-section, perpendicular to plane of page, of tank 2 in figure 1

3

Tank 3 𝑑𝑉3𝑑𝑡

= 𝑄2 − 𝑢

Since the cross sectional area A3 is constant, while height (H3) of water is function of time:

𝐴3𝑑𝐻3𝑑𝑡

= 𝐾2�𝐻2 − 𝑢

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.