Answer to Question #232090 in Molecular Physics | Thermodynamics for Unknown346307

Question

When 0.5 kg of water per minute is passed through a tube of 20 mm

diameter, it is found to be heated from 20°C to 50°C. The heating is accomplished by condensing

steam on the surface of the tube and subsequently the surface temperature of the tube is main￾tained at 85°C. Determine the length of the tube required for developed flow.

Take the thermo-physical properties of water at 60°C as :

ρ = 983.2 kg/m2, cp = 4.178 kJ/kg K, k = 0.659 W/m°C, ν = 0.478 × 10–6 m2/s.

Answer 1

Mass flow rate $=density \times area \times velocity$

Here mass flow rate =0.5 kg/min= 0.0083 kg/sec

Area $=\frac{\pi}{4} \times (0.02)^2=3.141 \times 10^{-4} \;m^2$

Density=983.2 kg/m³

$0.00833=983.2 \times 3.141 \times 10^{-4} \times v$

Velocity (v) = 0.027 m/sec

Heat duty

$Q =(mcp) \times ( 50-20) \\ = 0.00833 \times 4.178 \times (50-20) \\ Q= 1.0445 \;kw$

Now Reynold number $=\frac{ veocity \times diameter}{kinematics \;viscosity}$

$Re=\frac{ 0.027 \times 0.02}{0.478 \times 10^{-6}}$

Re =1129.7<2100 ( laminor)

Prandtl number $= μ \times \frac{cp}{k}$

$Pr = \frac{983.2 \times 0.478 \times 10^{-6} \times 4.178 \times 10^3}{0.659} \\ Pr=2.98$

Now nusselt number for laminor

$Nu = 0.332 \times Re^{0.5} \times Pr^{0.33} \\ Nu=16.05 \\ Nu= \frac{h \times d}{k} =16.05$

Heat transfer coefficient

$h = 529 w/m^2 \cdot s \\ Now \; (∆T)lm =\frac{ (50-85)-(20-85)}{ ln (\frac{50-85}{20-85})} \\ (∆T)lm= 48.46 \; °C \\ Q =h \times A \times (∆T)lm \\ A=0.0407 \; m^2= \pi \times d \times L \\ Length \; (L) =0.648\; m$

Answer: 0.648 m