Question

consider a thin 16-cm-long and 20-cm-wide horizontal plate suspended in air at 20C. The plate is equipped with electric resistance heating elements with a rating of 20 W. Now the heater is turned on and the plate temperature rises. Determine the temperature of the plate when steady operating conditions are reached. The plate has an emissivity of 0.90 and the surrounding surfaces are at 17C.

Answer 1

47.94° C

Step-by-step explanation:

DATA GIVEN:

length of thin horizontal plate = 16 cm long = 0.16

the width of the thin horizontal plate = 20 cm wide = 0.20 m

temperature of the outside air
`T \infty` = 20° C

Power rating of heating element Q = 20 W

Emissivity of the plate E = 0.9

Surrounding surface temperature
`T_(amb)` = 17° C

Let assume that the surface temperature of the plate is
`T_S` = 40 ° C

then the heat transfer by convection can be expressed by the formula:

`\mathbf{Q_(conv) = hA_s(T_s - T _(\infty))}`

Also;the heat transfer by radiation is:

`\mathbf{Q_(rad) = E A_2 \sigma (T_s^1 - T^1_(amb))}`

So; obtaining the properties of Air at 1 atm and film temperature
`T_f`

`T_f = (T_s + T_(\infty))/(2)`

`T_f = (40+ 20)/(2)`

`T_f = 30^0C`

Thermal conductivity k = 0.02588 W/m ° C

Prandtl number, Pr = 0.7282

Kinematic viscosity v = 1.608 × 10⁻⁵ m²/s

Volume expansion coefficient ;

`\beta = (1)/(T_f)`

`\beta =(1)/(30+273)`

`\beta = 3.3*10^(-3)K^(-1)`

The characteristic length
`L_c` of the horizontal plate is as follows:

`L_c = (A_s)/(P)`

`L_c =(L*W)/(2(L+W))`

`L_c =(0.16*0.20)/(2(0.16+0.20))`

`L_c =0.044 \ m`

Rayleigh number Ra =
`(g \beta(T_s - T_(\infty))L_c^3 )/(v^2)*Pr`

Ra =
`(9.8 *3.3*10^(-3)(40 - 20)0.044^3 )/(1.608^2)*0.7282`

Ra = 155327.931

To determine the heat transfer from the top surface; let's first find the Nusselt number :

`\mathbf{Nu = 0.54Ra^(0.25)}`

Nu =
`\mathbf{0.54(155327.931)^(0.25)}`

Nu = 10.72

Also, the Heat transfer coefficient is;

`h = (k*Nu)/(L_c)`

`h = (0.02588*10.72)/(0.044)`

`h = 6.3053 \ W/m^2 .K`

`Q_(top) = hA_s(T_s - T _(\infty))`

`Q_(top) = 6.3053*0.16*0.20(T_s - 20})`

`Q_(top) = 0.2017(T_s-20) ------- \ equation (1)`

Calculating the heat transfer from the bottom surface;

`\mathbf{Nu = 0.27Ra^(0.25)}`

Nu =
`\mathbf{0.27(155327.931)^(0.25)}`

Nu = 5.36

the Heat transfer coefficient is;

`h = (k*Nu)/(L_c)`

`h = (0.02588*5.36)/(0.044)`

`h = 3.1526 \ W/m^2 .K`

`Q_(top) = hA_s(T_s - T _(\infty))`

`Q_(top) =3.1526*0.16*0.20(T_s - 20})`

`Q_(top) = 0.1008(T_s-20) ------- \ equation (2)`

The heat transfer by radiation is :

`\mathbf{Q_(rad) =2*s* \sigma A_s(T^4_s-T^4_(amb))}`

`\mathbf{Q_(rad) =2*0.9* 5.67*10^(-8)*0.16*0.20((T_s+273)^4-(17+273)^4)}`

`\mathbf{Q_(rad) =3.265*10^(-9)((T_s+273)^4-(290)^4)} ------- \ equation(3)`

From the steady state condition:

The power rating of the heating element is:

`\mathbf{Q=Q_(rad)+Q_(top)+Q_(bottom)}`

replacing all the values from the above equations ; we have:

`20 = 3.265*10^(-9)[(T_s+273)^4-(290)^4]+(Ts-20)[0.2017-0.1008]`

`20 = 3.265*10^(-9)[(T_s+273)^4-(290)^4]+(Ts-20)0.3025`

The surface temperature
`T_S` from solving the above equation = 47.94° C