Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:solve all parts completely , Nitrogen, modeled as an ideal gas, flows at a rate of
3
k
g
s
thro

Question

Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:solve all parts completely , Nitrogen, modeled as an ideal gas, flows at a rate of 
3
k
g
s
 ﻿through a well
-
insulated horizontal nozzle operating at steady state. The nitrogen enters the nozzle with a velocity of 
2
0
m
s
 ﻿at 
5
0
0
K
,
4
0
0
kPa and exits the nozzle at 
1
0
0
 ﻿kPa 
.
To achieve an exit velocity of 
3
0
0
m
s
,
 ﻿determine:
(
a
)
 ﻿the exit temperature, in K
.
(
b
)
 ﻿the exit area, 
i
n
2
.

VIPSolutions · Accepted Answer

# Nitrogen No Problem (Ideal)

We are:

- **Mass flow rate:**(\dot{m} = 3\, \text{kg/s}\)
-Inlet velocity:** $V_1 = 20\, \text{m/s}$
- **Inlet temperature:** $T_1 = 500\, \text{K}$
- **Inlet pressure:** $P_1 = 400\, \text{kPa}$
- **Exit pressure:** $P_2 = 100\, \text{kPa}$
- **Exit velocity:** $V_2 = 300\, \text{m/s}$
- **Nitrogen is an ideal gas.**

We need to find:

- (a) **Exit temperature, $T_2$ (in K)**
- (b) **Exit area, $A_2$ (in m²)**

---

## Step 1: Apply the Steady-Flow Energy Equation (SFEE)

Since the nozzle is horizontal and well-insulated (adiabatic), the SFEE simplifies to:

$$
h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}
$$

Where $h$ is specific enthalpy.

For an ideal gas:

$$
h = c_p T
$$

The specific heat at constant pressure for nitrogen ($N_2$) is approximately:

$$
c_p = 1.039\, \text{kJ/kg}\cdot\text{K} = 1039\, \text{J/kg}\cdot\text{K}
$$

---

## (a) Find the Exit Temperature $T_2$

Plug in values (keep units consistent):

$$
c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{V_2^2}{2}
$$

$$
T_2 = T_1 + \frac{V_1^2 - V_2^2}{2 c_p}
$$

Plug in the numbers:

- $T_1 = 500\, \text{K}$
- $V_1 = 20\, \text{m/s}$
- $V_2 = 300\, \text{m/s}$
- $c_p = 1039\, \text{J/kg}\cdot\text{K}$

Calculate:

$$
T_2 = 500 + \frac{20^2 - 300^2}{2 \times 1039}
$$

$$
T_2 = 500 + \frac{400 - 90000}{2078}
$$

$$
T_2 = 500 + \frac{-89600}{2078}
$$

$$
T_2 = 500 - 43.11
$$

$$
T_2 \approx 456.9\, \text{K}
$$

**Final answer (a):**

$$
\boxed{T_2 \approx 457\, \text{K}}
$$

---

## (b) Find the Exit Area $A_2$

Use the mass flow rate relationship:

$$
\dot{m} = \rho_2 V_2 A_2
$$

First, find $\rho_2$ (density at exit):

Ideal gas law:

$$
\rho_2 = \frac{P_2}{R T_2}
$$

Where $R$ for nitrogen, $R = \frac{R_u}{M}$:

- $R_u = 8314\, \text{J/kmol}\cdot\text{K}$
- $M = 28.013\, \text{kg/kmol}$

So,

$$
R = \frac{8314}{28.013} \approx 297.1\, \text{J/kg}\cdot\text{K}
$$

Plug in values:

- $P_2 = 100\, \text{kPa} = 100,000\, \text{Pa}$
- $T_2 \approx 457\, \text{K}$

$$
\rho_2 = \frac{100{,}000}{297.1 \times 457}
$$

$$
\rho_2 \approx \frac{100{,}000}{135,794}
$$

$$
\rho_2 \approx .736\, \text{kg/m}^3
$$

Now, rearrange for $A_2$:

$$
A_2 = \frac{\dot{m}}{\rho_2 V_2}
$$

$$
A_2 = \frac{3}{.736 \times 300}
$$

$$
A_2 = \frac{3}{220.8}
$$

$$
A_2 \approx .0136\, \text{m}^2
$$

**Final answer (b):**

$$
\boxed{A_2 \approx .0136\, \text{m}^2}
$$

---

## **Summary of Answers**

- **(a) Exit temperature:** $\boxed{457\, \text{K}}$
- **(b) Exit area:** $\boxed{.0136\, \text{m}^2}$

---

**Note:**  
- Calculations use rounded values for readability.  
- For more precision, carry more significant digits through calculations.  
- Assumed $c_p$ and $R$ for nitrogen are constant in this temperature range; for more accurate results, use property tables.

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