Process and Instrumentation Diagram for Milk Pasteurization Plant and First Order Temperature Sensor Equation and Plot

This article presents a process and instrumentation diagram for a milk pasteurization plant and approximating the temperature sensor by a first-order process. It provides the equation of the first-order process and plots the measured temperature and the estimated ones based on the equation of the first-order temperature sensor.

Part 1: P&ID Diagrams (10 mks)
Process and instrumentation diagram for the simplified milk pasteurization plant. The process
variable to be controlled here is the pasteurized milk temperature.
Solution
The initial P&ID diagram for the milk pasteurization plant is demonstrated as,
The process line design has some product-related, process-related, economic, and legal
considerations. The raw material is treated and a quality end product is obtained. The raw milk is
fed to the balance tank, also referred to as the constant heat tank. It passes to the milk pump and
flows further to the controller and the result is fed into the plate heat exchanger. The plate heat
exchanger has the heating, holding, and cooling sections. The plant has a constant head tank, a
flow controller, the milk entry pump, a filter and clarifier, a homogenizer, and the plate heat
exchange. When the milk is processed it is extracted through the flow diversion valve, recording
and indicating instruments, cooling systems as well as the piping systems that link to various
components [1].

The block chart that illustrates the milk pasteurization process is
Part 2: First Order Temperature Sensor (10 mks)
Approximating the temperature sensor by a first order process
(i) Provide the equation of the first order process
(ii) Plot the measured temperature and the estimated ones based on the equation of the
first order temperature sensor.
Solution
Temperature in 0C Time in Seconds Estimated Values
(Equation)
19 0 0
57 1 49.8361
69 2 68.6267
73 3 75.7117
75 4 78.3831
77 5 79.3904

78 6 79.7701
79 7 79.9133
79.6 8 79.9673
79.8 9 79.9877
80 10 79.9954
The first order sensors have one element that stores energy while another dissipates it. In this
case, the water dissipates its energy while the sensor stores the energy. Therefore,
a1
dy
dt + a0 y ( t )=x ( t )
Using Laplace transform on the equation above,
1
a1 s+a0
= Y ( s )
X ( s )
Y ( s )
X ( s ) = k
τs+1
The sensor’s initial temperature was 190C and the fluid temperature was 800C. The heat flow
through the sensor covering,
heat flow=θF −θt
R , θF −temp of fluid ,θt−sensor temp
The rise in temperature in the sensor,

dθ ( t )
dt = θF −θ ( t )
RC
Using the Laplace transform,
sθ ( s )= θF ( s )−θ ( s )
RC
¿ ( RC+1 ) θ ( s )=θF ( s )
θ ( s )= θF ( s )
RCs +1
Taking the equation back to the time domain,
θ(t)=θF ( 1−e
−t
RC )
θF ( 1−e
−t
τ ) , τ−time constant
To obtain the thermal capacitance of the sensor and the thermal resistance of the fluid,
T =T fluid− ( T fluid−T initialSensor ) e
−t
τ
T + T fluid= ( T fluid −TinitialSensor ) e
−t
τ
e
−t
τ = T +T fluid
( T fluid−T initialSensor )
Introducing natural logarithm, ln, on both sides,
−t
τ =ln ( T +T fluid
( T fluid−T initialSensor ) )

−t
τ =ln ( T +T fluid
( T fluid−T initialSensor ) )
τ = −1
slope
Evaluating the sensor on MATLAB,
%%
Ot=19;
Of=80;
temp=[19 57 69 73 75 77 78 79 79.6 79.8 80];
time=0:1:10;
% The first order equation of the temperature sensor
a=temp-80;
b=Ot-Of;
gs=log(a/b);
figure(1)
plot(time,gs,'g-*');
grid on
vals=diff(gs)./diff(time);
slope=vals(1); % finds the gradient of the line
t_const=-1/slope;
temp_sensor=Of.*(1-exp(-time/t_const));
figure(2)
plot(time,temp,'b-*')
grid on
hold on
plot(time,temp_sensor,'r-*')
xlabel('Time (sec)')
ylabel('temperature (deg.celsius)')
title('Plot of Temperature Rise against Time')
legend('Measured','Estimated')
It is observed that there is a slight difference between the temperature sensor values in the
measured and estimated values based on the equation,