### Lumped System Analysis

• Assess when the spatial variation of temperature is negligible, and temperature varies nearly uniformly with time, making the simplified lumped system analysis applicable.

• Solve the transient conduction problem in large mediums using the similarity variable, and predict the variation of temperature with time and distance from the exposed surface.

### Solved Example:

#### 83-1-01

Biot number signifies the ratio of:

Solution:
The Biot number (Bi) is a dimensionless quantity used in heat transfer calculations. It gives a simple index of the ratio of the heat transfer resistances inside of and at the surface of a body. The Biot number is defined as: $\mathrm{Bi} = \frac{h L_C}{\ k_b}$ where: h = film coefficient or heat transfer coefficient or convective heat transfer coefficient
$L_C$ = characteristic length, which is commonly defined as the volume of the body divided by the surface area of the body, such that, $\mathit{L_C} = \frac{V_\mathrm{body}}{A_\mathrm{surface}}$ $k_b$ = thermal conductivity of the body

### Solved Example:

#### 83-1-02

Lumped parameter analysis for transient heat conduction is essentially valid for:

Solution:
It is generally accepted that lumped system analysis is applicable if Biot number is less than 0.1.

### Solved Example:

#### 83-1-03

According to lumped system analysis, solid possesses thermal conductivity that is:

Solution:
In lumped system analysis, thermal conductivity of solid is assumed to be infinitely large, such that the solid is considered as a lump, irrespective of its dimensions. Due to this, once the heat reaches the solid, it is instantaneously distributed within the solid.

### Solved Example:

#### 83-1-04

Which is true regarding lumped system analysis?

I. Conductive resistance = 0

II. Convective resistance = 0

III. Thermal conductivity = 0

IV. Thermal conductivity = $\infty$

Identify the correct statements.

Solution:
Solids have infinite thermal conductivities. It implies that internal conductance resistance is very low.

### Solved Example:

#### 83-1-05

Consider heat transfer between two identical hot solid bodies and the surrounding air. The first solid is cooled by a fan, while the second one is cooled by natural convection in air. For which case is the lumped capacity assumption more applicable?

### Solved Example:

#### 83-1-06

A steel ball of diameter 60 mm is initially in thermal equilibrium at 1030$^{\circ}\mathrm{C}$ in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30$^{\circ}\mathrm{C}$, with convective heat transfer coefficient h = 20 $W/m^2K$. The thermo-physical properties of steel are: Density = 7800 $kg/m^3$, conductivity k = 40 W/mK and specific heat c = 600 J/kgK. The time required in seconds to cool the steel ball in air from 1030$^{\circ}\mathrm{C}$ to 430$^{\circ}\mathrm{C}$ is:

Solution:
For sphere, characteristic length, $L_c = \dfrac{V}{A} = \dfrac{\dfrac{4}{3} \pi r^3}{4 \pi r^2} = \dfrac{r}{3} = \dfrac{D}{6} = 10\ \mathrm{mm}$ Biot Number, $Bi = \dfrac{h L_c}{K} = \dfrac{20 \times 10\times 10^{-3}}{40} = 0.005$ which is very very less than 1, so lumped mass system approximation can be done.
Time constant, $\tau = \dfrac{\rho C_p L_c}{h} = \dfrac{7800 \times 600 \times 10\times 10^{-3}}{20} = 2340\ \mathrm{sec}$ $\beta = \dfrac{1}{\tau} = \dfrac{1}{2340} = 4.27 \times 10^{-4} sec^{-1}$ \begin{align*} \dfrac{T-T_a}{T_i-T_a} &= e^{-\beta t}\\ \dfrac{430-30}{1030-30} &= e^{-4.27 \times 10^{-4}t}\\ t &= 2144\ \mathrm{sec} \end{align*}

A small copper ball of 5 mm diameter at 500 K is dropped into an oil bath whose temperature is 300 K. The thermal conductivity of copper is 400 W/ m K, its density is 9000 kg/m$^3$ and its specific heat is 385 J/kg K. If the heat transfer coefficient is 250 W/m$^2$ is and the lumped analysis is assumed to be valid, the rate of fall of temperature of the ball at the beginning of cooling will be, :
Solution: d =5 mm, $\rho$ =9000 kg/m$^{3}$, C$_{p}$ =385 J/kgK, h = 250 W/m$^{2}$K, L$_c$ = $\dfrac{d}{6}$ = $\dfrac{5}{6}$ mm \begin{align*} \mathrm{time\ constant}, \tau &= \dfrac{\rho C_p L_c}{h}\\ &= \dfrac{9000 \times 385 \times \dfrac{5}{6} \times 10^{-3}}{250}\\ &= 11.55\ sec \end{align*} $\beta = \dfrac{1}{\tau} = \dfrac{1}{11.55} = 0.087\ sec^{-1}$ \begin{align*} \dfrac {T-300}{500-300}&=e^{-\beta t}\\ \left( T-300\right) &=200e^{-\beta t}\\ T &=300+200e^{-\beta t}\\ \dfrac {dT}{dt}&=0+200e^{-\beta t}\left( -\beta \right) =-200 \beta e^{-\beta t} \end{align*} at t=0, $e^{-\beta t}=e^{-0}=1$ \begin{align*} \dfrac {dT}{dt}=-200\beta =-200\left( 0.087\right) =17.3 K/s \end{align*}