School of Process, Environmental and Materials Engineering

PEME2200Process Engineering Systems

Part III: Principles of Electricity, Flow Measurement and Auto Control

Student Name Raja Aneece AhmedStudent ID 200564028F.A.O Professor M. WangDate of Submission 26/04/2013

Assignment 2

Problem StatementThe principle of differential flow meters can be described using Bernoulli’s ideal flow equation that states the sum of the static energy (pressure head), the kinetic energy (velocity head), and the potential energy (elevation head) of the fluid are approximately conserved in the flow across a constriction in a pipe.

Pρ . g

+V2

2 g+Z=consta nt

Equation 1.0 - Bernoulli's ideal flow equationWhere ρ is the fluid density (kg/m3), g denotes gravitational constant (m/s2), P is the pressure

(Pa), V is the fluid flow velocity (m/s) and Z the elevation head (m).

Considering the principle of the flow continuity for an incompressible fluid through a horizontal pipe as shown in Figure 2

Figure 1 - Venturi pipe

a) Prove that the flow speed V2 at A2 can be described by Equation 2 (assuming the cross section areas A1, A2 of the pipe and the pressures P1 and P2 at A1 and A2 to beknown)

V 2=1

√1−m2 √ 2. ΔPρ

Equation 1.1 - Flow speed equationWhere m= (A2/A1), ΔP = P1-P2 and ρ is the density of the flow

b) Design a flow rate measurement device using a Venturi and a U-tube manometer for a horizontal liquid flow measurement.

c) Calculate the fluid flow velocity at A2 for a pressure drop of 1 meter measured from the manometer (assuming the gravitational acceleration is 9.8m/s2 ,m=0.8 and the liquid density in the manometer is the same as that of the liquid that flows thorough the venturi).

Answer to part A

This part of the task involves rearranging equation 1.0 to equation 1.1 which relates to the calculations to work out parameters in the Venturi pipe shown in figure 2. The subsequent method shown can be applied for this to occur

Equation 1 given in the problem statement showing the Bernoulli’s equation can be applied at points 1 and 2 in the Venturi Pipe by re-writing it in the following format:

P1

ρ .g+V 1

2

2g+z1=

P2

ρ. g+V 2

2

2 g+ z2

Equation 2.0

As the diagram depicts a Venturi Pipe shown in figure 2 in the problem statement, it can be deduced that there is no elevation as the pipe is shown to be linear in the horizontal direction; therefore z1 and z2 have the same value and thus cancel out. Consequently, equation 2.1 simplifies to the following equation:

P1

ρ .g+V 1

2

2g=P2

ρ .g+V 2

2

2gEquation 2.1

Further simplification can be applied to the above equation by the multiplication of g of all terms, to give the following equations:

P1

ρ+V 1

2

2=P2

ρ+V 2

2

2Equation 2.2

Subsequently, multiplying ρ to the above equation gives:

P1+ρV 1

2

2=P2+

ρV 22

2Equation 2.3

p2can then be moved to the right hand side of the equation

(P¿¿1−P2)+ρV 1

2

2=ρV 2

2

2¿

Equation 2.4

This equation can know be rearranged to give the following:

(P¿¿1−P2)=ρV 2

2

2−ρV 1

2

2¿

Equation 2.5

The fraction shown on the right hand side of the equation has a common factor which is ρ /2 so it can be factorised. Also,(P1−P2) is the same as writing∆ P. Applying this to equation 2.5 would give:

∆ P= ρ2

(V 22−V 1

2)Equation 2.6

The equation representing the flow rate is shown below:Q=AV

Equation 2.7

The flow rates of points 1 and 2 can be calculated using the continuity equation which states:A1V 1=A2V 2

Equation 2.8

This equation can then be rearranged to give the following relationship. This rearrangement is achieved by dividing A1and V 2 to both sides of the equation as shown below:

V 1

V 2

=A2

A1

Equation 2.9

‘m’ can then be substituted A2/ A1 so the equation know becomes:V 1

V 2

=m

Equation 3.0

Going back to the rearranged and transformed Bernoulli’s equation shown by equation 2.6 it can be further changed by multiplying the entire equation by 2 and then dividing it by the ρ which gives the following equation:

2∆Pρ

=(V 22−V 1

2)

Equation 3.1

Taking V 2 as a factor would change the equation to:

2∆Pρ

=V 22(1−

V 12

V 22 )

Equation 3.2

Now it is possible to recognize a relationship emerging between equations 3.2 and 3.0. When substituting equation 3.0 into equation 3.2 the following equation emerges:

2∆Pρ

=V 22(1−m2)

Equation 3.3

By dividing (1−m2 ) to both sides of the equation the following equation appears:

1

(1−m2)2∆Pρ

=V 22

Equation 3.4

Finally, the square root of both sides can be taken to obtain equation 1.1 as required:

V 2=1

√1−m2 √ 2∆ Pρ

Equation 1.1

Answer to part B

This task involves designing flow rate measurement device using a Venturi and a U-tube manometer for a horizontal liquid flow measurement. A design of this is represented by the figure below:

Figure 2: Venturi Pipe with the addition of a flow measurement device

Answer to part C

This part of the assignment involved calculating the fluid flow velocity at A2 for a pressure drop of 1 meter measured from the manometer (assuming the gravitational acceleration is 9.8 m/s2 , m=0.8 and the liquid density in the manometer is the same as that of the liquid flowing through the Venturi).

Solution:The equation used to calculate the pressure when using a manometer is as follows:

∆ P=ρghEquation 3.5

Substituting equation 3.5 into equation 1.1 the following equation emerges:

V 2=1

√1−m2√2gh

Equation 3.6

When substituting the values given for each parameter into the above equation the answer can be determined:

V 2=1

√1−0.82√2×9.81×1=7.9m /s

Assignment 3Considering a fired charge heater given in below Figure 4, please answer below questions, assuming a feedback control loop to be applied for the control of the output temperature.

Figure 3 – Fired charge heater

a) Please identify a manipulated, controlled variables, also possible disturbancesFor this system, the flow rate can be considered as the manipulated control variable and the output temperature as the controlled variable. A possible disturbance that may occur during the operation is heat loss from the system. This may occur due to low temperature weather conditions or lack of insulation surrounding the plant.

b) Draw a physical layout based on your control strategy

Figure 4: Physical layout strategyc) Draw a block diagram with text definition to reflect your strategy

Figure 5: Block Diagram reflecting the control strategy and the fire exchange system

ControlledValve

d) Derive the transfer function of the feedback control system

Where: Gs – Required Temperature, Gc – Temperature Control, Gt – Transmission System Dynamics, Gv – Fuel Valve Dynamics, Gr – Output Temperature Dynamics, Hs –

Temperature Sensor Dynamics, Ht – Transmission System DynamicsThe G terms can be grouped together: GcGtGvGr = G

The H terms can be grouped together: HsHt = HUsing this terminology, the two relationships can be derived as:

Y = EGEquation 3.7

And E = GsX – HYEquation 3.8

When substituting equation 3.8 into equation 3.7 the following equation is obtained:Y = (GsX-HY)G

Equation 3.9

Multiplying out the above equation gives the following:Y = GsXG-HYG

Equation 4.0

Rearranging equation 4.0 gives the following equation:Y + HYG = GsXG

Equation 4.1Y(1+HG) = GsXG

Equarion 4.1Y(1+HG)/X = GsG

Equation 4.2Y/X = GsG/(1+HG)

Equation 4.1

Y

Figure 6: Mathematical representation of the flow diagram

When substituting equation 4.1 with the original terminology without grouping the symbolts together, the following equation is derived:

Y/X = GsGcGtGvGr/(1+HsHtGcGtGvGr

Equation 4.2

Assume a cascade control is to be applied for the temperature control

c) Please identify the secondary controlled variable and justify the selection.The secondary control variable is the internal temperature. The reason to why this has been identified as a secondary control variable is that it allows the room temperature to be controlled. An example can be explained when referring to the temperature of the furnace relative to the output temperature. When the internal temperature is measured and identified, this sends a signal to the fuel valve. Depending on the temperature measurements made and the set temperature, if the temperature is measured to be lower than the set amount then fuel valve would be opened further to allow more fuel to enter the system increasing the flow rate of the fuel. However if the internal temperature is measured to be greater than the set amount; this results in the flow rate of the fuel entering the system decreasing as the valve slightly closes.

d) Re-draw the physical layout based on your control strategy

Figure 5: Showing the actual diagram of the system with the cascade control physical layout

Figure 7: Mathematical Layout of figure 6

Where the symbols mean:

Figure 6: Block Diagram also showing the cascade control system

Gs – Required TemperatureGc – Temperature ControlGcc – Temperature ControlGt – Transmission System DynamicsGv – Fuel Valve DynamicsGf – Furnace Temperature DynamicsGr – Output Temperature DynamicsHs – Temperature Sensor DynamicsHt – Transmission System DynamicsIs – Temperature Sensor DynamicsIt – Transmission System DynamicsLet G = GccGtGvGf Equation 4.3Let H = HsHt Equation 4.4Let I = IsIt Equation 4.5

The secondary loop can be condensed using the algebraic law changing the block diagram to the following:

It is also possible to condense the primary loop so that it is in the following format:Figure 8: Condensed mathematical layout of the secondary loop

Because all errors have been eliminated the transfer function of the cascade control is derived to be the following:

Transfer Function=GsGcGccGtG vGf Gr

1+GccGtGvG f H tH s+G cGccGtGvGf Gr I s I t

Equation 4.6 Cascade control transfer function

Figure 9: Secondary loop condensed mathematical layout