This question involves performing an energy balance around the system.  The following equation provides an example of this energy balance.

Where heat transferred, Q is a function of coolant flow rate and temperature driving force.  The relationship is derived in the textbook and summarized in the equation below. Eq (3.62)

When the energy balance is linearized, the terms become complicated and are simplified by the equations in Eq (3.65).  Don't get bogged down with the details of the linearization, for now.  The final result of the linearization is below.

Where the values of K_T and K_Fc appear in eq. (3.65) and depend only on the coolant flow rate and the temperatures.   They are independent of the flow rate, and the tank volume.

By continuing to follow the derivation performed in Example 3.7, the time constant and the gain can be evaluated.

Now we can consider the effects of changing the volume and the flow rate.

Firstly when we change the volume:

• Gain - The gain will not be affected.   The volume cancels out of the denominator leaving the gain independent of the tank volume.
• Time Constant - Due to the presence of the volume term in the numerator of the expression for the time constant, the time constant will increase with an increase in volume.  This is reasonable, for we would expect the process dynamics to slow down in a larger system.

As for a change in the flow rate into the tank:

• Gain - The gain will be reduced with an increase in flow rate.
• Time Constant - The time constant would also be reduced if the flow rate were to be increased.

Example 3.7 uses the fundamental modelling procedure and extensive algebraic manipulation.  Despite the more complicated algebra, the results are interpreted in the same way as with simpler models.