$U$: Overall heat transfer coefficient.

Ratio of actual heat transfer rate to the maximum possible heat transfer rate. $$\varepsilon = \frac{\text{Actual heat transfer}}{\text{Maximum possible heat transfer}} = \frac{\dot{Q}}{\dot{Q}_{max}}$$
Fluid Capacity Rate ($C$): $$C_h = \dot{m}_h C_{ph} \quad (\text{Hot fluid})$$ $$C_c = \dot{m}_c C_{pc} \quad (\text{Cold fluid})$$ $C_{min}$ is the minimum of $C_h$ and $C_c$. $C_{max}$ is the maximum of $C_h$ and $C_c$.
Maximum Possible Heat Transfer: $$\dot{Q}_{max} = C_{min} (T_{h,in} - T_{c,in})$$
Actual Heat Transfer using Effectiveness: $$\dot{Q} = \varepsilon C_{min} (T_{h,in} - T_{c,in})$$

Definition: Deposition of ash, dirt, scale, and fluid impurities on heat exchanger tube surfaces during operation, leading to reduced heat transfer efficiency.

Roof dimensions: 6m long, 8m wide, 0.25m thick.
Material: Concrete, $k = 0.8 \text{ W/(m}\cdot\text{K)}$.
Temperatures: Inner $15^\circ \text{C}$, Outer $4^\circ \text{C}$.
Calculate:
1. Rate of heat loss through the roof.
2. Cost of heat lost for 10 hours if electricity costs Rs 5/kWh.

Plate: $600\text{mm} \times 900\text{mm} \times 25\text{mm}$. $k = 45 \text{ W/(m}\cdot^\circ\text{C)}$.
Temperature: Maintained at $310^\circ \text{C}$. Air at $15^\circ \text{C}$ blows over.
Convection coefficient: $h = 22 \text{ W/(m}^2\cdot^\circ\text{C)}$.
Radiation heat loss: $250 \text{ W}$.
Calculate: Inside plate temperature. (Answer: $313.86^\circ \text{C}$).

Surface temperature: $250^\circ \text{C}$. Surroundings: $110^\circ \text{C}$.
Convection coefficient: $h = 75 \text{ W/(m}^2\cdot^\circ\text{C)}$. Emissivity: $\varepsilon = 1$.
Heat conducted to surface through solid: $k = 10 \text{ W/(m}\cdot^\circ\text{C)}$.
Calculate: Temperature gradient at the surface in the solid. (Answer: $dt/dx = -1352.21^\circ \text{C/m}$).

Wall: $320\text{mm}$ thick. Inner layer fire brick ($k = 0.84 \text{ W/(m}\cdot^\circ\text{C)}$), outer layer insulation ($k = 0.16 \text{ W/(m}\cdot^\circ\text{C)}$).
Reactor operates at $1325^\circ \text{C}$, ambient $25^\circ \text{C}$.
Calculate:
1. Thickness of fire brick and insulation for minimum heat loss.
2. Heat loss assuming insulation max temp is $1200^\circ \text{C}$.

Wall: $0.1\text{m}$ common brick ($k = 0.7 \text{ W/(m}\cdot^\circ\text{C)}$), $0.04\text{m}$ gypsum plaster ($k = 0.48 \text{ W/(m}\cdot^\circ\text{C)}$).
Calculate: Thickness of rock wool insulation ($k = 0.065 \text{ W/(m}\cdot^\circ\text{C)}$) to reduce heat loss/gain by 80%. (Answer: $58.8\text{mm}$).