Question 10.1: A hot water heating coil has a set-point of 35°C with a thro...

Using Eq. (10.1), the relationship between the heat rate Q, and the error in the air temperature at the coil outlet can be put in the form of:

u =K_{p}e+u_{0}           (10.1)

Q = K_{p} (T_{set-point} – T_{air}) + Q_{o}

(i) When the heat rate Q = Q_{min}= 0 kW, the coil outlet air temperature is T_{air} = T_{min} = 35°C – 5°C= 30°C.
(ii) When the heat rate Q = Q_{max}= 50 kW, the coil outlet air temperature is T_{air} = T_{max} = 35°C +5°C = 40°C.

The proportional gain K_{p} can be determined as follows:

Q_{max} − Q_{min} = K_{p}(T_{min} − T_{max})

or

K_{p} = [Q_{max} − Q_{min}]/[T_{min} − T_{max}]= −50kW/10 °C = −5 kW/°C

Similarly, the constant Q_{o} can be determined from

Q_{min} = K_{p}(T_{setpoint} − T_{min}) + Q_{o}

or

Q_{o} = Q_{min} − K_{p}(T_{set-point} − T_{min}) = 0 + 5 kW/°C * (35°C − 30°C) = +25 kW

Therefore, the relationship between the heat rate output and the air temperature for the heating coil is:

Q = −5 (T_{set-point} − T_{air}) + 25

Thus, as long as the heat rate is different from Q_{o} = 25 kW, the quantity ( T_{set-point} − T_{air}) which is the error in the proportional control equation cannot be equal to zero.