Answer:
4.6 mm
Explanation:
Given data includes:
thin-walled pipe diameter = 100-mm =0.1 m
Temperature of pipe [tex]T_p[/tex] = -15° C = (-15 +273)K =258 K
Temperature of water [tex]T_w[/tex] = 3° C = (3 + 273)K = 276 K
Temperature of ice [tex]T_i[/tex] = 0° C = (0 +273)K =273 K
Thermal conductivity (k) from the ice table = 1.94 W/m.K ; R = 0.05
convection coefficient [tex]Lh_l[/tex] =2000 W/m².K
The energy balance can be expressed as:
[tex]q_{conduction} =q_{convention}[/tex]
where;
[tex]q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}[/tex] ------------- equation (1)
[tex]q_{convention} = \pi DLh_l(T_w-T_i)[/tex] ------------ equation(2)
Equating both equation (1) and (2); we have;
[tex]\frac{2\pi LK(T_i-T_p)}{In(R/r)}[/tex] [tex]= \pi DLh_l(T_w-T_i)[/tex]
Replacing the given data; we have:
[tex]\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)}[/tex] [tex]= \pi (0.1)*2000(276-273)[/tex]
[tex]\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96[/tex]
[tex]In(\frac{0.05}{r})*1884.96 = 182.84[/tex]
[tex]In(\frac{0.05}{r}) = \frac{182.84}{1884.96}[/tex]
[tex]In(\frac{0.05}{r}) =0.0970[/tex]
[tex]\frac {0.05}{r} =e^{0.0970}[/tex]
[tex]\frac {0.05}{r} =1.102[/tex]
[tex]r=\frac{0.05}{1.102}[/tex]
r = 0.0454
The thickness (t) of the ice layer can now be calculated as:
t = (R - r)
t = (0.05 - 0.0454)
t = 0.0046 m
t = 4.6 mm
