Question

A metallic sphere of 1 kg mass, with a surface area of 0.0314 m², is maintained at an initial temperature of 50°C. The fluid circulating the sphere is maintained at a temperature of 10°C. Specific heat of metallic sphere is 314 J/kg-K and the heat transfer coefficient between the fluid and the sphere is 10 W/m² -K. The time is taken (in seconds) for the sphere to cool down to 20°C is___

Answer 1

Concept:

Lumped parameter analysis:

Internal/conductive resistance is very little as compared to surface convective resistance and the temperature distribution is given by

\(\frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~}=~{{e}^{\left( \frac{hA}{ρ V{{C}_{p}}} \right)t}}\)

\(\frac{hA}{ρ V{{C}_{p}}}t=\ln\left( \frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~} \right)\)

where

• Ti = Initial temperature of beat at t = 0,
• T = Temperature of body at any instant ‘t’ sec
• T∞ = Ambient fluid temperature
• h = heat transfer coefficient, ρ is the density of the metal sphere

For sphere: \(\frac{V}{A} = \frac{{Volume\;of\;body}}{{surface\;area}} = \frac{{\frac{4}{3}π {r^3}}}{{4π {r^2}}} = \frac{r}{3}\) 

V/A = r/3

Calculation:

Given:

Cp = 314 J/kgK, h = 10 W/m²K, m = 1 kg, A = 0.0314 m2

Ti = 50°C, T∞ = 10°C, T = 20°C

surface area of sphere, A = 4 π r² = 0.0314 m²

⇒ r = 0.05 m

density of the metal sphere,

\(\rho = {mass\over volume \;of \;sphere} = {m\over{\frac{4}{3}\pi r^3}}\)

\(={ 1\over{\frac{4}{3}\pi\;0.05^3 }}\) =1909.86 kg/m³

Using the equation below and putting the given values: 

\(\frac{hA}{ρ V{{C}_{p}}}t=\ln\left( \frac{{{T}_{i}}-{{T}_{\infty }}}{T-{{T}_{\infty }}~} \right)\)

\(\ln\left( {\frac{{50 - 10}}{{20 - 10}}} \right) = \left( {\frac{{10 × 3 }}{{0.05× 1909.86 × 314}}} \right) × t\)

\(⇒ t= \frac{{1.386 × \;0.05 × 1909.86 × 314}}{{10 × 3}} \)

⇒ t =1385.29 seconds

∴  Time is taken (in seconds) for the sphere to cool down to 20°C is = 1385.29 sec