Question

In a guarded hot plate, the dimension of the square test plate is 15 cm $\times$ 15 cm. Keeping the temperatures of the test plate and the air at 35$^\circ$C and 20$^\circ$C, respectively, the power losses from the test plate with and without fabric specimen are 16 W and 40 W, respectively. The intrinsic transmittance [W/(m$^2$ K)] of the fabric (rounded off to 2 decimal places) is ...................

Hint:

For guarded hot-plate problems, treat “with specimen” as series resistances: $R_{\text{tot}} = R_{\text{background}} + R_{\text{specimen}}$. Intrinsic transmittance is simply $1/R_{\text{specimen}}$.

Answer 1

Step 1: Area and temperature difference
Plate area, $A = 0.15 \times 0.15 = 0.0225\ \text{m}^2$; \ $\Delta T = 35-20 = 15\ \text{K}$.
Step 2: Overall conductance without fabric
$P_0 = h_0 A \Delta T ⇒ h_0 = \dfrac{40}{0.0225 \times 15} = 118.52\ \text{W/(m}^2\text{K)}$.
Background resistance: $R_0 = 1/h_0 = 0.00844\ \text{m}^2\text{K/W}$.
Step 3: Overall conductance with fabric
$P = h A \Delta T ⇒ h = \dfrac{16}{0.0225 \times 15} = 47.41\ \text{W/(m}^2\text{K)}$.
Total resistance with fabric: $R_{\text{tot}}=1/h=0.02110\ \text{m}^2\text{K/W}$.
Step 4: Intrinsic resistance and transmittance of fabric
$R_{\text{fabric}} = R_{\text{tot}} - R_0 = 0.02110 - 0.00844 = 0.01266\ \text{m}^2\text{K/W}$.
Intrinsic transmittance: $T_i = 1/R_{\text{fabric}} = 78.98 \approx 79.00\ \text{W/(m}^2\text{K)}$.