A rectangular slab (k= 10 W/m-K) of thickness 15cm and inside temperature of 4000C is insulated by a materials of thickness 10cm (k= 30 W/m-k). The ambient air is at 280C and the outside convective heat transfer coefficient is 15 W/m2K. Determine the steady state heat transfer per unit surface area and the temperature of outside surface of the slab and the insulation.

Given,

δ1=15cm=0.15m $\hspace{1cm}$ (Thickness of slab)

K=10W/mK $\hspace{1.8cm}$ (Thermal conductivity of slab material)

t1=400℃=673K $\hspace{1.2cm}$ (Inside temperature of slab)

δ2=10cm=0.1m $\hspace{1.2cm}$ (Thickness of insulation)

Kins=30W/mK $\hspace{1.4cm}$ (Thermal conductivity of insulation material)

t4=28℃=301K $\hspace{1.4cm}$ (Ambient air temperature)

h=15W/m2K $\hspace{1.7cm}$ (Convective heat transfer coefficient of ambient air)

A=1m2 $\hspace{2.5cm}$ (cross sectional area of slab)

Find:

(1) Q (Heat transfer rate)

(2) t2 (Outside surface temperature of slab)

(3) t3 (Outside surface temperature of insulation)

Solution:

Calculation for thermal resistance

Thermal resistance due to rectangular slab

R1 = δ1K1A1=0.1510×1

R1 = 0.015 K/W

Thermal resistance due to insulation

R2 = δ2K2A2=0.130×1

R2 = 3.333×10-3 K/W

Thermal resistance due to convective layer

R3 = 1h3A3=115×1

R3 = 0.066 K/W

Total thermal resistance

R = R1+R2+R3=0.015+3.333×10-3+0.066

R = 0.0843 K/W

Heat transfer rate

Q = dtR=t1-t4R=673-3010.0843

Q = 4412.811 W

also,

Q = t1-t2R1

4412.811 = 673-t20.015

t2 = 606.807K=333.807℃

Q = t2 - t3R2

4412.811 = 606.807 - t33.333 × 10-3

t3 = 592.099K = 319.099℃