A nozzle converts enthalpy into kinetic energy. Air enters the nozzle with a pressure of $27.2$ bar, a velocity of $32.8$ m/s, and an enthalpy of $559$ kJ/kg. At the nozzle exit, the pressure is $6.80$ bar, and the enthalpy is $378$ kJ/kg. The mass flow rate is $273$ kg/h, and the thermal losses are $5$ kJ/kg. Calculate the air velocity at the nozzle exit under two conditions: (a) considering the given conditions and (b) assuming the nozzle is well-insulated.

1. Problem Description

A nozzle converts enthalpy into kinetic energy. Air enters the nozzle with a pressure of $27.2$ bar, a velocity of $32.8$ m/s, and an enthalpy of $559$ kJ/kg. At the nozzle exit, the pressure is $6.80$ bar, and the enthalpy is $378$ kJ/kg. The mass flow rate is $273$ kg/h, and the thermal losses are $5$ kJ/kg. Calculate the air velocity at the nozzle exit under two conditions: (a) considering the given conditions and (b) assuming the nozzle is well-insulated.

2. Solution Steps

First, convert the mass flow rate from kg/h to kg/s:

$\dot{m} = 273 \frac{\text{kg}}{\text{h}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 0.07583 \frac{\text{kg}}{\text{s}}$

(a) Considering the given conditions (with heat losses):

We will use the steady-flow energy equation, which, neglecting potential energy changes, can be written as:

$\dot{m} (h_1 + \frac{V_1^2}{2}) = \dot{m} (h_2 + \frac{V_2^2}{2}) + \dot{Q}$

where:

$\dot{m}$ is the mass flow rate,

$h_1$ is the inlet enthalpy,

$h_2$ is the outlet enthalpy,

$V_1$ is the inlet velocity,

$V_2$ is the outlet velocity,

$\dot{Q}$ is the rate of heat loss.

The heat loss rate is calculated as:

$\dot{Q} = \dot{m}q = 0.07583 \frac{\text{kg}}{\text{s}} \times 5 \frac{\text{kJ}}{\text{kg}} = 0.37915 \text{ kW} = 379.15 \text{ W}$

$h_1 = 559 \frac{\text{kJ}}{\text{kg}} = 559000 \frac{\text{J}}{\text{kg}}$

$h_2 = 378 \frac{\text{kJ}}{\text{kg}} = 378000 \frac{\text{J}}{\text{kg}}$

$V_1 = 32.8 \frac{\text{m}}{\text{s}}$

Substitute the values into the energy equation:

$0.07583 (559000 + \frac{32.8^2}{2}) = 0.07583 (378000 + \frac{V_2^2}{2}) + 379.15$

$0.07583(559000 + 537.92) = 0.07583(378000 + \frac{V_2^2}{2}) + 379.15$

$0.07583(559537.92) = 0.07583(378000 + \frac{V_2^2}{2}) + 379.15$

$42436.71 = 28664.74 + 0.037915 V_2^2 + 379.15$

$42436.71 - 28664.74 - 379.15 = 0.037915 V_2^2$

$13392.82 = 0.037915 V_2^2$

$V_2^2 = \frac{13392.82}{0.037915} = 353219.03$

$V_2 = \sqrt{353219.03} = 594.32 \frac{\text{m}}{\text{s}}$

(b) Assuming the nozzle is well-insulated (no heat losses, $\dot{Q} = 0$):

$\dot{m} (h_1 + \frac{V_1^2}{2}) = \dot{m} (h_2 + \frac{V_2^2}{2})$

$0.07583 (559000 + \frac{32.8^2}{2}) = 0.07583 (378000 + \frac{V_2^2}{2})$

$559000 + 537.92 = 378000 + \frac{V_2^2}{2}$

$559537.92 - 378000 = \frac{V_2^2}{2}$

$181537.92 = \frac{V_2^2}{2}$

$V_2^2 = 2 \times 181537.92 = 363075.84$

$V_2 = \sqrt{363075.84} = 602.56 \frac{\text{m}}{\text{s}}$

3. Final Answer

(a) The air velocity at the nozzle exit, considering the given conditions, is $594.32$ m/s.

(b) The air velocity at the nozzle exit, assuming the nozzle is well-insulated, is $602.56$ m/s.