### 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.

Lumped system analysis provides great simplification in certain classes of heat transfer problems without much sacrifice from accuracy.
When a solid body is being heated by the hotter fluid surrounding it (such as a potato being baked in a oven), heat is first convected to the body and subsequently conducted within the body. The Biot number is the ratio of the internal resistance of a body to heat conduction to its external resistance to heat convection. Therefore, a small Biot number represents small resistance to heat conduction, and thus small temperature gradients within the body.
Lumped system analysis assumes a uniform temperature distribution throughout the body, which will be the case only when the thermal resistance of the body to heat conduction (the conduction resistance) is zero. Thus, lumped system analysis is exact when Bi=0 and approximate when Bi >0. Of course, the smaller the Biot number, the more accurate the lumped system analysis.

Biot Number, $Bi = \dfrac{h L_c}{k} = \dfrac{\dfrac{L_c}{K}}{\dfrac{1}{h}} = \dfrac{\mathrm{Conductive\ resistance\ within\ the\ body}}{\mathrm{Convection\ resistance\ at\ the\ surface\ of\ the\ body}}$

It is generally accepted that lumped system analysis is applicable if Bi $<$ 0.1

### 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*}