Answer to Question #231326 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

$\frac mT=\frac{0.5}{60}~\frac{kg}s,$ $d=0.02~m,$

$t_1=20°C,~t_2=50°C,~t_3=85°C,$

$\rho=983.2~\frac{kg}{m^3},~c=4178~\frac J{kg°C},$

$k=0.659~\frac W{m°C},$

$\nu=0.478\cdot 10^{-6}\frac{m^2}s,$

$t'=\frac{t_1+t_2}2=35°C,$

$t=\frac{t_3+t'}2=60°C,$

$\frac mT=\rho Av=\frac{\pi d^2}4\rho v,$

$v=\frac{4m}{\pi d^2\rho T}=0.0269~\frac ms,$

$Re=\frac{vd}{\nu}=1126<Re_0=2000\to laminar,$

$Nu=3.65,$

$Nu=\frac{hd}k,\implies h=\frac{kNu}d=120.26~\frac W{m^2°C},$

$A_s=\pi dl,$

$Q=A_sh(t_3-t')=c\frac mT(t_2-t_1),$

$l=\frac{cm(t_2-t_1)}{T\pi dh(t_3-t')}=2.76~m.$