Question

(3) Figure 3 shows a system including three components: heater, thermostat, and room. Assume the initial room temperature is 0 C. The heater is turned on to heat the room to the desired temperature 25'C (i.e heater temperature is 25 C). A fan takes room air, and passes it through the heater and back to the room, with volumetric air flow rate q,. The density of air is assumed to 1.225 kg/m The space of the room is 30 m. Assume the heat loss to the ambient is ignorable. Develop a differential equation to describe the room temperature dynamics. Assume 0.2 mscc. Please compute the time constant and time needed to reach the desired room temperature (within 2%). Outside Room Heat Room roo Heat room Air Flow Thermostat Onvo Control Heater av Figure 3

Answer 1

We assume density and specific heat of air are constants.

Refer the control volume diagram below:

om

From 1st law of thermodynamics.

(Net heat gain by room)= (Increase of internal energy of air in room)

$\rho c_pV_{room}\frac{dT_{room}}{dt}=\rho c_pq_v(T_h-T_{room})$

Or

$\frac{dT_{room}}{dt}+\frac{q_v}{V_{room}}T_{room}=\frac{q_v}{V_{room}}T_h$

Substituting values

$\frac{dT_{room}}{dt}+\frac{0.2}{30}T_{room}=\frac{0.2}{30}(25)$

Or

$\frac{dT_{room}}{dt}+0.006667T_{room}=0.1667$

The solution of the above differential equation is

$T_{room}=ce^{-0.006667t}+25$

At t=0 T_room=0

Hence

c=-25

Therefore

$T_{room}=25(1-e^{-0.006667t})$

This is the required equation for room temperature.

The time constant is

$\tau=\frac{1}{0.006667}=150\: s$

For reaching within 2% of final value of 25⁰C

$T_{room}=0.98(25)=24.5^0C$

Hence

$24.5=25(1-e^{-0.006667t})$

Or

t= 587 s

It will take 587 s to reach within 2% of final temperature of 25⁰C