A small metal bead (radius 0.5 mm), initially at 100°C, when placed in a stream of fluid at 20°C, attains a temperature of 28°C in 4.35 seconds. The density and specific heat of the metal are 8500 kg/m³ and 400 J/kgK, respectively. If the bead is considered as lumped system, the convective heat transfer coefficient (in W/m²K) between the metal bead and the fluid stream is

The correct option is 299.8.

1. Identify the Given Data:
We are given the following parameters for the metal bead:
- Radius of the bead, $r=0.5\text{mm}=0.5\times {10}^{-3}\text{m}$
- Density of the metal, $\rho =8500{\text{kg/m}}^3$
- Specific heat of the metal, $c=400\text{J/kg·K}$
- Initial temperature of the bead, $T_i=100°\text{C}$
- Temperature of the fluid stream, $T_{\infty }=20°\text{C}$
- Temperature at a given time, $T\left(t\right)=28°\text{C}$
- Time taken to reach this temperature, $t=4.35\text{s}$

2. Governing Equation for Lumped System Analysis:
Under the lumped heat capacity assumption, the temperature variation of a body with time is given by:

$$\frac{T\left(t\right)-T_{\infty }}{T_i-T_{\infty }}=e^{-\left(\frac{h{A}_s}{\rho Vc}\right)t}$$
where:
- $h$ is the convective heat transfer coefficient.
- $A_s$ is the surface area of the bead.
- $V$ is the volume of the bead.

3. Calculate the Volume-to-Surface-Area Ratio for a Spherical Bead:
For a sphere, the ratio of volume to surface area is:

$$\frac{V}{A_s}=\frac{\frac{4}{3}\sigma r^3}{4\sigma r^2}=\frac{r}{3}$$
Substituting the radius $r=0.5\times {10}^{-3}\text{m}$:

$$\frac{V}{A_s}=\frac{0.5\times {10}^{-3}}{3}\text{m}$$

4. Step-by-Step Derivation and Calculation:
Substitute the temperature values into the left-hand side of the heat transfer equation:

$$\frac{28-20-}{100-20}=\frac{8}{80}=0.1$$
Now equate this to the exponential term:

$$0.1=e^{-\left(\frac{3h}{\rho cr}\right)t}$$
Taking the natural logarithm ($\ln$) on both sides:

$$\ln \left(0.1\right)=-\frac{3h}{\rho cr}t$$
Since $\ln \left(0.1\right)\approx -2.3026$:

$$2.3026=\frac{3ht}{\rho cr}$$
Rearranging to solve for the convective heat transfer coefficient $h$:

$$h=\frac{2.3026\times \rho \times c\times r}{3t}$$
Substituting the numeric values:
- Density ($\rho$) = 8500
- Specific heat ($c$) = 400
- Radius ($r$) = $0.5\times {10}^{-3}$
- Time ($t$) = 4.35

$$h=\frac{2.3026\times 8500\times 400\times 0.5\times {10}^{-3}}{3\times 4.35}$$
$$h=\frac{2.3026\times 1700}{13.05}$$
$$h=\frac{3914.42}{13.05}\approx 299.9{\text{W/m}}^2\text{K}$$

Rounding slightly to match the closest given choice, we obtain the correct answer of 299.8 ${\text{W/m}}^2\text{K}$.