Question

A steam condensor condensing steam at ${70}^{0}C$ to have a capacity 100 kW. Water at ${20}^{0}C$ is used and outlet temperature is limited to ${45}^{0}C$. If the overall heat transfer coefficient is $3100W/m^{2}K$. The surface area required is (Assume water heat capacity is minimum and use NTU method)

• A. $0.652{\text{m}}^2$
• B. $0.894{\text{m}}^2$
• C. $0.234{\text{m}}^2$
• D. $0.789{\text{m}}^2$

Answer 1

The correct option is B $0.894{\text{m}}^2$

${\text{T}}_h={70}^0\text{C},\text{Q}=100KW$

${\text{T}}_{c_i}={20}^{0}C,{\text{T}}_{c_e}={45}^{0}C,\text{U}=3100\frac{\text{W}}{m^{2}K}$

$ϵ=\frac{45-20}{70-20}=0.5$

$ϵ=1-e^{-NTU}\Rightarrow \text{NTU}=0.6931$

$Q={\text{C}}_{min}({\text{T}}_{c_e}-{\text{T}}_{c_i})$

${\text{C}}_{min}=\frac{100\times {10}^3}{45-20}=4000\frac{W}{K}$

$\text{NTU}=\frac{\text{UA}}{{\text{C}}_{min}}$

$\Rightarrow A=\frac{\text{NTU}\times {\text{C}}_{min}}{\text{U}}$

$=\frac{0.6931\times 4000}{3100}=0.894m^2$