Question

An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of -15 degree C and a cylindrical layer of ice forms on the inner surface of the wall If the mean water temperature is 3 degree C and a convection coefficient of 2000 W/m^2 middot K is maintained at the inner surface of the ice, which is at 0 degree C, what is the thickness of the ice layer?

Answer 1

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe $T_p$ = -15° C = (-15 +273)K =258 K

Temperature of water $T_w$ = 3° C = (3 + 273)K = 276 K

Temperature of ice $T_i$ = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K  ;  R = 0.05

convection coefficient $Lh_l$ =2000 W/m².K

The energy balance can be expressed as:

$q_{conduction} =q_{convention}$

where;

$q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}$       -------------   equation (1)

$q_{convention} = \pi DLh_l(T_w-T_i)$  ------------ equation(2)

Equating both equation (1) and (2); we have;

$\frac{2\pi LK(T_i-T_p)}{In(R/r)}$ $= \pi DLh_l(T_w-T_i)$

Replacing the given data; we have:

$\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)}$ $= \pi (0.1)*2000(276-273)$

$\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96$

$In(\frac{0.05}{r})*1884.96 = 182.84$

$In(\frac{0.05}{r}) = \frac{182.84}{1884.96}$

$In(\frac{0.05}{r}) =0.0970$

$\frac {0.05}{r} =e^{0.0970}$

$\frac {0.05}{r} =1.102$

$r=\frac{0.05}{1.102}$

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