The heat flow equations I got are RATE OF TEMPERATURE CHANGE

The heat flow equations I got are:

RATE OF TEMPERATURE CHANGE OF A CONDUCTOR:

HEAT FLOW RATE:

don\'t know how to solve them

I. Consider a single room building modelled as follows, where T(t) is the inside temperature and TA(t) is the outside temperature. T(t) TA(t) Let o denote the total internal floor, wall and ceiling surface area of the building. Insulation is used that has a thermal resistance per unit area y, so the overall thermal resistance of the building is on. The thermal capacitance of the building is given by 5. If we denote the average outside temperature by TR, then the temperature of the room Tt) can be written relative to this reference by TCt) TT(t) TR. Similarly, the ambient outside temperature TA can be written relative to its average value by defining TA(t) TA(t) TR

Solution

solution:

here heat flow in capacitor by current analogy as

V=q/C

V=Ic.dt/C

hence capacitor current is

Ic=V\' *C

where current through resistor is

Ir=V/RA

Ir=Ic

V/RA=V\'*C

V\'=V/RCA

now in heat system

V\'=T\'

V=dT=Ta-T

C=b(beta)

R=y(gamma)

surface area A=a(sigma)

now we get that

V\'=V/RCA

in heat system we get,

T\'=dT/b*y*a

T\'(b*y*a)+T=Ta

this is ODE governing heat flow

and on applying laplace transform as

[s(b*y*a)+1]T(s)=Ta(s)

so transfer function is

T(s)/Ta(s)=[1/b*y*a*s+1]

hence equation is proved

on applying reverse laplace transform we get that

T(s)/Ta(s)=[1/b*y*a*s+1]=(1/b*y*a)/[s+(1/b*y*a)]

which standard form of

1/s+a=e^-at

so we get that

T(t)/Asinwt=(1/b*y*a)*[e^(-1/b*y*a)t]

T(t)=(A/b*y*a)*[e^(-1/b*y*a)t]*sinwt

when we put T(t)=(T(t)-Tr)=dTt

dTt=(A/b*y*a)*[e^(-1/b*y*a)t]*sinwt

dTt<A

we can write

y>=((A/dTt)/b*a)*[e^(-1/b*y*a)t]*sinwt

we can write its magnitude as

y>=[(A/dTt)^2-1]^.5/b*y*w

in this way control equation are develop and utilizes for close control of mechanism of heat flow