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To be properly sterilized, all bacteria must be heated for a required duration. Although individu

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Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:
To be properly sterilized, all bacteria must be heated for a required duration. Although individual bacterial sizes are small, when present in large clumps, some bacteria located deep inside the clump may require a significant amount of time to reach the medium 
(
air or water
)
 ﻿temperature. Thus, they cannot be assumed to be at the medium temperature all the time. Consider bacterial clumps of different sizes heated in air. The shape of the clump can be assumed spherical with a radius R
,
 ﻿which is related to n
,
 ﻿the number of bacteria in a clump, and r
,
 ﻿the radius of individual bacteria, by the equation R 
=
 ﻿r
[
n
/
(
1
-
E
)
]
^
(
1
/
3
)
 ﻿where E 
=
 
0
.
4
0
 ﻿is porosity. The surface heat transfer coefficient for air flow over the clump is given by h 
=
 
4
 ﻿x 
1
0
^
-
4
kf
/
R where kf is the air thermal conductivity and is equal to 
0
.
0
3
1
7
 ﻿W
/
m
.
K
.
 ﻿Consider an individual bacterial radius of 
1
um and the specific heat, density, and thermal conductivity of the bacterial clump of 
3
.
6
 ﻿kJ
/
kg
.
K
,
 
1
0
8
0
 ﻿kg
/
m
^
3
,
 ﻿and 
0
.
4
9
 ﻿W
/
m
.

VIPSolutions · Accepted Answer

# Step-by-Step Solution

Given Data

- Number of bacteria in clump: $ n = 10^, 105, 10^6 $  
- Porosity: $ E = .40 $
- Individual bacterial radius: $ r = 1\,\mu\text{m} = 1 \times 10^{-6}\,\text{m} $
- Air thermal conductivity: $ k_f = .0317\,\text{W}/\text{m}\cdot\text{K} $
- Specific heat of clump: $ c_p = 3.6\,\text{kJ}/\text{kg}\cdot\text{K} = 360\,\text{J}/\text{kg}\cdot\text{K} $
- Density of clump: $ \rho = 108\,\text{kg}/\text{m}^3 $
- Clump thermal conductivity: $ k = .49\,\text{W}/\text{m}\cdot\text{K} $
- Heat transfer coefficient: $ h = 4 \times 10^{-4} \dfrac{k_f}{R} $

---

## 1. **Determine Clump Radius $ R $ and Heat Transfer Coefficient $ h $**

### **Clump Radius Formula:**
$$
R = r \left[ \frac{n}{1-E} \right]^{1/3}
$$

#### **For each $ n $:**

### **(a) $ n = 10^2 $**
$$
R = 1 \times 10^{-6} \left[ \frac{10^2}{1-.4} \right]^{1/3}
= 1 \times 10^{-6} \left[ \frac{100}{.6} \right]^{1/3}
$$
$$
= 1 \times 10^{-6} (166.67)^{1/3}
$$
$$
166.67^{1/3} \approx 5.50
$$
$$
R \approx 5.50 \times 10^{-6}\,\text{m}
$$

### **(b) $ n = 10^5 $**
$$
R = 1 \times 10^{-6} \left[ \frac{10^5}{.6} \right]^{1/3}
= 1 \times 10^{-6} (166666.67)^{1/3}
$$
$$
166666.67^{1/3} \approx 55.1
$$
$$
R \approx 5.51 \times 10^{-5}\,\text{m}
$$

### **(c) $ n = 10^6 $**
$$
R = 1 \times 10^{-6} \left[ \frac{10^6}{.6} \right]^{1/3}
= 1 \times 10^{-6} (1666666.67)^{1/3}
$$
$$
1666666.67^{1/3} \approx 118
$$
$$
R \approx 1.18 \times 10^{-4}\,\text{m}
$$

---

### **Heat Transfer Coefficient $ h $:**
$$
h = 4 \times 10^{-4} \frac{k_f}{R}
$$
Where $ k_f = .0317\,\text{W}/\text{m}\cdot\text{K} $

#### **(a) $ R = 5.50 \times 10^{-6}\,\text{m} $**
$$
h = 4 \times 10^{-4} \frac{.0317}{5.50 \times 10^{-6}}
$$
$$
= 4 \times 10^{-4} \times 5763.64
$$
$$
= 2.31\,\text{W}/\text{m}^2\cdot\text{K}
$$

#### **(b) $ R = 5.51 \times 10^{-5}\,\text{m} $**
$$
h = 4 \times 10^{-4} \frac{.0317}{5.51 \times 10^{-5}}
$$
$$
= 4 \times 10^{-4} \times 575.3
$$
$$
= .230\,\text{W}/\text{m}^2\cdot\text{K}
$$

#### **(c) $ R = 1.18 \times 10^{-4}\,\text{m} $**
$$
h = 4 \times 10^{-4} \frac{.0317}{1.18 \times 10^{-4}}
$$
$$
= 4 \times 10^{-4} \times 268.6
$$
$$
= .107\,\text{W}/\text{m}^2\cdot\text{K}
$$

---

### **Summary Table**

| $ n $     | $ R $ (m)         | $ h $ ($\text{W}/\text{m}^2\cdot\text{K}$) |
|-------------|---------------------|-----------------------------------------------|
| $ 10^2 $  | $ 5.50 \times 10^{-6} $  | 2.31                                     |
| $ 10^5 $  | $ 5.51 \times 10^{-5} $  | .230                                    |
| $ 10^6 $  | $ 1.18 \times 10^{-4} $  | .107                                    |

---

## 2. **Show that the Lumped Parameter Approach is Valid**

The lumped parameter (lumped capacitance) method is valid if the Biot number $ Bi $ is less than .1:

$$
Bi = \frac{hR}{k}
$$

Where $ k = .49\,\text{W}/\text{m}\cdot\text{K} $

Calculate for each clump:

### **(a) $ n = 10^2 $**
$$
Bi = \frac{2.31 \times 5.50 \times 10^{-6}}{.49} = \frac{1.2705 \times 10^{-5}}{.49} \approx 2.59 \times 10^{-5}
$$

### **(b) $ n = 10^5 $**
$$
Bi = \frac{.230 \times 5.51 \times 10^{-5}}{.49} = \frac{1.2673 \times 10^{-5}}{.49} \approx 2.59 \times 10^{-5}
$$

### **(c) $ n = 10^6 $**
$$
Bi = \frac{.107 \times 1.18 \times 10^{-4}}{.49} = \frac{1.2626 \times 10^{-5}}{.49} \approx 2.58 \times 10^{-5}
$$

**All Biot numbers are much less than .1**, so the lumped parameter approach is valid for all cases.

---

## 3. **Calculate Time to Reach 99% of Total Temperature Change**

Lumped capacitance method:

$$
\frac{T(t) - T_{\infty}}{T_i - T_{\infty}} = \exp\left(-\frac{hA}{\rho c_p V} t\right)
$$

For 99% change:
$$
\left| \frac{T(t) - T_{\infty}}{T_i - T_{\infty}} \right| = .01
$$
$$
.01 = \exp\left(-\frac{hA}{\rho c_p V} t\right)
$$
$$
\ln(.01) = -\frac{hA}{\rho c_p V} t
$$
$$
t = -\frac{\ln(.01) \cdot \rho c_p V}{hA}
$$
$$
\ln(.01) = -4.605
$$
$$
t = \frac{4.605 \cdot \rho c_p V}{hA}
$$

For a sphere:
- Volume: $ V = \frac{4}{3}\pi R^3 $
- Surface area: $ A = 4\pi R^2 $

So,
$$
\frac{V}{A} = \frac{\frac{4}{3}\pi R^3}{4\pi R^2} = \frac{R}{3}
$$
$$
t = \frac{4.605 \cdot \rho c_p (R/3)}{h}
$$

Plug in values ($ \rho = 108\,\text{kg/m}^3, c_p = 360\,\text{J}/\text{kg}\cdot\text{K} $):

$$
t = \frac{4.605 \times 108 \times 360 \times R/3}{h}
$$
$$
= \frac{4.605 \times 108 \times 360}{3} \frac{R}{h}
$$
$$
= \frac{4.605 \times 388800}{3} \frac{R}{h}
$$
$$
= 596160 \frac{R}{h}
$$

Now, substitute $ R $ and $ h $:

### **(a) $ n = 10^2, R = 5.50 \times 10^{-6}\,\text{m}, h = 2.31 $**
$$
t = 596160 \times \frac{5.50 \times 10^{-6}}{2.31}
$$
$$
= 596160 \times 2.38 \times 10^{-6}
$$
$$
= 14.2\,\text{s}
$$

### **(b) $ n = 10^5, R = 5.51 \times 10^{-5}\,\text{m}, h = .230 $**
$$
t = 596160 \times \frac{5.51 \times 10^{-5}}{.230}
$$
$$
= 596160 \times 2.396 \times 10^{-4}
$$
$$
= 1428\,\text{s}
$$

### **(c) \( n = 10^6, R = 1.18 \times 10^{-4}\,\text{m}, h = .

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