Physics

Physics · thermodynamics

Problem

In this problem, we explore how fast an iceberg can melt, through the dominant mode of forced convection. For simplicity, consider a very thin iceberg in the form of a square with side lengths

L=100 \mathrm{~m}

and a height of

1 \mathrm{~m}

, moving in the arctic ocean at a speed of

0.2 \mathrm{~m} / \mathrm{s}

with one pair of edges parallel to the direction of motion (Other than the height, these numbers are typical of an average iceberg). The temperature of the surrounding water and air is

2^{\circ} \mathrm{C}

, and the temperature of the iceberg is

0^{\circ} \mathrm{C}

. The density of ice is

917 \mathrm{~kg} / \mathrm{m}^{3}

and the latent heat of melting is

L_{w}=334 \times 10^{3} \mathrm{~J} / \mathrm{kg}

. The heat transfer rate

\dot{Q}

between a surface and the surrounding fluid is dependent on the heat transfer coefficient

h

, the surface area in contact with the fluid

A

, and the temperature difference between the surface and the fluid

\Delta T

, via

\dot{Q}=h A \Delta T

. In heat transfer, three useful quantities are the Reynold's number, the Nusselt number, and the Prandtl number. Assume they are constant through and given by (assuming laminar flow):

\operatorname{Re}=\frac{\rho v_{\infty} L}{\mu}, \quad \mathrm{Nu}=\frac{h L}{k}, \quad \operatorname{Pr}=\frac{c_{p} \mu}{k}

where: -

\rho

: density of the fluid -

v_{\infty}

: speed of the fluid with respect to the object (at a very far distance) -

L

: length of the object in the direction of motion -

\mu

: dynamic viscosity of the fluid -

k

: thermal conductivity of the fluid -

c_{p}

: the specific heat capacity of the fluid Through experiments, the relationship between the three dimensionless numbers is, for a flat plate:

\mathrm{Nu}=0.664 \operatorname{Re}^{1 / 2} \operatorname{Pr}^{1 / 3}

Use the following values for calculations:

	Air	Water
$\rho\left(\mathrm{kg} / \mathrm{m}^{3}\right)$	1.29	1000
$\mu(\mathrm{kg} /(\mathrm{m} \cdot \mathrm{s}))$	$1.729 \times 10^{-5}$	$1.792 \times 10^{-3}$
$c_{p}(\mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K}))$	1004	4220
$k(\mathrm{~W} /(\mathrm{m} \cdot \mathrm{K}))$	0.025	0.556

The initial rate of heat transfer is

\dot{Q}

. Assuming this rate is constant (this is not true, but will allow us to obtain an estimate), how long (in days) would it take for the ice to melt completely? Assume convection is only happening on the top and bottom faces. Round to the nearest day. Given constants: - Proton mass,

m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}

- Neutron mass,

m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}

- Electron mass,

m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}

- Avogadro's constant,

N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}

- Universal gas constant,

R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})

- Boltzmann's constant,

k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}

- Electron charge magnitude,

e=1.60 \cdot 10^{-19} \mathrm{C}

- 1 electron volt,

1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}

- Speed of light,

c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}

- Universal Gravitational constant,

G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}

- Solar Mass

M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}

- Acceleration due to gravity,

g=9.8 \mathrm{~m} / \mathrm{s}^{2}

- 1 unified atomic mass unit,

1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}

- Planck's constant,

h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}

- Permittivity of free space,

\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)

- Coulomb's law constant,

k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}

- Permeability of free space,

\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}

- Magnetic constant,

\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}

- 1 atmospheric pressure,

1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}

- Wien's displacement constant,

b=2.9

10^{-3} \mathrm{~m} \cdot \mathrm{K}

- Stefan-Boltzmann constant,

\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}

Related theory

thermodynamics

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