Question

Water at ${20}^{\circ }C(\rho ={10}^{3}kg/m^3,c_p=4.18kJ/k{g}^{\circ }C,k=0.64W/m^{\circ }C)$ flows through a 10 mm diameter tube at a uniform velocity of $0.5m/s$. Assume that the flow at entry is fully developed. A constant heat flux of $50kW/m^2$ is imposed over the tube. If the water is to be heated from ${20}^{\circ }C$ to ${80}^{\circ }C$, then the minimum length of the tube required will be

• A. 4.27 m
• B. 3.37 m
• C. 6.27 m
• D. 2 m

Answer 1

The correct option is C 6.27 m

$Forwatertoheatfrom{20}^{\circ }Cto{80}^{\circ }C,theamountofheatrequiredis$

$Q=q{A}_s=\dot{m}{c}_p(80-20)$

$q\times \pi DL=\rho \times \frac{\pi }{4}{D}^2\times V_v\times c_p\times 60$

$4qL=\rho D\times V_{av}\times c_p\times 60$

Hence,
$L=\frac{\rho D\times V_{av}\times c_p\times 60}{4q}$

$=\frac{{10}^3\times 0.01\times 0.5\times 4180\times 60}{4\times 50\times {10}^3}$

$=6.27m$