Act like a helpful tutor and exlain me :
The boiling point of nitrogen at sea level
(
1
atm pressure
)
at atmospheric pressure is
-
1
9
6
°
C
.
Therefore, nitrogen is commonly u

Question

Act like a helpful tutor and exlain me : 
The boiling point of nitrogen at sea level 
(
1
 ﻿atm pressure
)
 ﻿at atmospheric pressure is 
-
1
9
6
°
C
.
 ﻿Therefore, nitrogen is commonly used in low
-
temperature scientific studies. The temperature of liquid nitrogen in a tank open to the atmosphere will remain constant at 
-
1
9
6
°
C until it is consumed. Any heat transfer to the tank will result in the evaporation of some liquid nitrogen, which has a heat of vaporization of 
1
9
8
 ﻿kJ
/
kg and a density of 
8
1
0
 ﻿kg
/
m
^
3
 ﻿at 
1
 ﻿atm. Consider a spherical tank 
3
 ﻿m in diameter that is initially filled with liquid nitrogen at 
1
 ﻿atm and 
-
1
9
6
°
C
.
 ﻿The tank is exposed to ambient air at 
1
5
°
C
,
 ﻿with a combined convective and radiant heat transfer coefficient of 
3
5
 ﻿W
/
m
^
2
.
K
.
 ﻿The temperature of the thin shell of the spherical tank should be nearly the same as the temperature of the nitrogen inside. Determine the evaporation rate of liquid nitrogen in the tank as a result of heat transfer from the ambient air if the tank is: a
)
 ﻿Uninsulated. b
)
 ﻿Insulated with 
5
 ﻿cm of insul

VIPSolutions · Accepted Answer

Absolutely! Let’s break this down step by step:

## **Given Data**
- **Boiling point of nitrogen**: $ -196^\circ C $ (at 1 atm)
- **Heat of vaporization ($h_{fg}$)**: $ 198 $ kJ/kg $ = 198,000 $ J/kg
- **Density ($\rho$)**: $ 810 $ kg/m³
- **Tank shape**: Sphere, **diameter** $ d = 3 $ m $\rightarrow r = 1.5$ m
- **Ambient temperature ($T_{air}$)**: $ 15^\circ C $
- **Nitrogen temperature ($T_{N_2}$)**: $ -196^\circ C $
- **Combined heat transfer coefficient ($h$)**: $ 35 $ W/m²·K
- **Insulation thickness**: $ 5 $ cm $ = 0.05 $ m (for part b)
- **Thermal conductivity of insulation ($k$)**: $ 0.035 $ W/m·K

---

## **Step 1: Calculate the Temperature Difference**

$$
\Delta T = T_{air} - T_{N_2} = 15 - (-196) = 211^\circ C = 211\,K
$$

---

## **Step 2: Find the Surface Area of the Sphere**

$$
A = 4\pi r^2 = 4\pi (1.5)^2 = 28.27\, m^2
$$

---

## **Step 3: Calculate Heat Transfer Rate ($\dot{Q}$)**

### **a) Uninsulated Tank**

The heat transfer is only through convection/radiation:

$$
\dot{Q} = hA\Delta T
$$
$$
\dot{Q} = 35 \times 28.27 \times 211 \approx 208,349\,W = 208.3\,kW
$$

---

### **b) Insulated Tank**

First, find the total thermal resistance:

#### **(i) Convective resistance (outside):**
$$
R_{conv} = \frac{1}{hA}
$$

#### **(ii) Conductive resistance (through spherical insulation):**
For a spherical shell:
$$
R_{cond} = \frac{r_2 - r_1}{4\pi k r_1 r_2}
$$
- $ r_1 = 1.5\,m $ (inner radius)
- $ r_2 = r_1 + 0.05 = 1.55\, m $

$$
R_{cond} = \frac{1.55 - 1.5}{4\pi \times 0.035 \times 1.5 \times 1.55}
= \frac{0.05}{4\pi \times 0.035 \times 2.325}
$$

Calculate denominator:
- $ 4\pi \times 0.035 \times 2.325 \approx 1.022 $

So,
$$
R_{cond} = \frac{0.05}{1.022} \approx 0.0489\, K/W
$$

#### **(iii) Convective resistance:**
$$
R_{conv} = \frac{1}{hA} = \frac{1}{35 \times 30.2} \approx 0.000941\, K/W
$$
Where $A = 4\pi r_2^2 = 4\pi (1.55)^2 \approx 30.2\,m^2$

#### **Total resistance:**
$$
R_{total} = R_{cond} + R_{conv} = 0.0489 + 0.000941 = 0.04984\,K/W
$$

#### **Heat transfer rate:**
$$
\dot{Q} = \frac{\Delta T}{R_{total}} = \frac{211}{0.04984} \approx 4,235\,W = 4.24\,kW
$$

---

## **Step 4: Evaporation Rate Calculation**

$$
\dot{m} = \frac{\dot{Q}}{h_{fg}}
$$

### **a) Uninsulated:**
$$
\dot{m}_{uninsulated} = \frac{208,349}{198,000} \approx 1.053\, kg/s
$$

### **b) Insulated:**
$$
\dot{m}_{insulated} = \frac{4,235}{198,000} \approx 0.0214\, kg/s
$$

---

## **Summary Table**

| Case           | Heat Transfer ($\dot{Q}$) | Evaporation Rate ($\dot{m}$) |
|----------------|----------------------------|------------------------------|
| Uninsulated    | 208.3 kW                   | 1.053 kg/s                   |
| Insulated (5cm)| 4.24 kW                    | 0.0214 kg/s                  |

---

## **Key Takeaways**
- **Insulation drastically reduces heat transfer** and thus the evaporation rate of liquid nitrogen.
- The **main limiting factor** is the heat transfer into the tank, which is much higher without insulation.

If you need to see all the formulas or want a step-by-step with all numbers plugged in, just ask!

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