Question

For this problem we consider a radiant heat transfer system commonly found in space/room hoaters. Ihe ipu io the plani is (hcat) eey q(Wais) and he ouipui of the sysiem is is temperature (K). The ODE that describes the system is given below where, ền is the ambient temperature (27°C), b:91.6 is an input constant, m 0.1 kg is the mass, C-420 J/Kg. K is the specific heat of the heater and a-Acơ. A-0.25 m2 is the surface area of the heater, ε = 0.85 is the emittance and σ = 5.67 x 10-8 T/(sm2K4) is the Stephan-Boltzman constant. Given, the state x, input u and output y for the system is θ, q and θ respectively find the value of q at which the system reaches equilibrium with an output temperature of 800°C, (note: the equation considers SI metrics and so temperatures should be in 1. K) 2. Using the equilibrium values for 0e and ge computed above, linearize the system and write the state space equations. 3. Find the Transfer function P(s) of the plant from u to y

Answer 1

1)

a = Αεσ = 0.25 * 0.85 * 5.67 * 10-8-1 .205 * 10-8Ts_1K-4

b=91.6

300K

θ-1073K

At equilibrium $\dot_{\theta}$ = 0

From equation

$mC \dot{\theta} =-a(\theta^4-\theta_a^4)+bq$

$a(\theta^4-\theta_a^4)=bq$

$q=a(\theta^4-\theta_a^4)/b$

1.205 * 10-8(10733-3004)/91,6 q

q=173.311

2)

$\dot{\theta} =-a/mC(\theta^4-\theta_a^4)+bq/mC=f(\theta,q)$

$\partial f/\partial \theta = -a/mC*4\theta_e^3$

$\partial f/\partial q = b/mC$

On linearization:

$\dot{\theta}=\partial f/\partial \theta.\theta + \partial f/\partial q .q$

So linearized form is as follows:

$\dot{\theta}=-a/mC*4\theta_e^3.\theta + bq/mC$. (1)

State space equation is $[\dot{\theta}]=[-a/mC*4\theta_e^3].\theta + [b/mC]q$

3)

u is q and y is $\theta$

From (1)

$\dot{\theta}+a/mC*4\theta_e^3.\theta = bq/mC$

$s\theta (s)+a/mC*4\theta_e^3.\theta (s)= bq(s)/mC$

$(s+a/mC*4\theta_e^3)\theta (s)= bq(s)/mC$

$P(s)=\theta (s)/q(s)= (b/mC )/(s+a/mC*4\theta_e^3)$