Question

A convergence-divergence nozzle is a nozzle attached to a diffuser where the connection is called the throat. Air enters a convergence-divergence nozzle steadily at 300 kPa, 200°C, and 55 m/s at a rate of 2.5 kg/s and gets to the throat at 100 kPa and 180 m/s. The inlet area of the nozzle is 110 cm2. Air then leaves the diffuser at the velocity of 30 m/s where the exit area of the diffuser is 130 cm2. Throughout the whole passage heat leaves the system at the rate of 20 kJ/s and electrical work enters the passage at the rate of 50 kJ/s. Determine

(a) the temperature of the air at the throat,
(b) the throat area of the nozzle,
(c) the temperature at the exit,
(d) the pressure at the exit

Answer 1

The temperature at the throat of the convergence-divergence nozzle is 202.36 °C. The throat area of the nozzle is 7.09 × 10^(-5) m². The temperature at the exit is 202.36 °C, and the pressure at the exit is 100 kPa.

Step-by-step explanation:

To solve the problem, we can use the principle of conservation of mass and the properties of a convergence-divergence nozzle.

1. (a) The temperature of the air at the throat: The temperature can be determined using the relation between pressure and temperature for an ideal gas. We can use the ideal gas law, which states that P1V1/T1 = P2V2/T2. Given that the air enters the nozzle steadily, we can assume that it is isentropic, which means that there is no heat transfer. Therefore, the initial and final enthalpies of the air will be the same, which allows us to simplify the equation. Solving for T2, we find T2 = T1(P2/P1)^((γ-1)/γ), where γ is the specific heat ratio for air. Plugging in the values, T2 = 200°C(100 kPa/300 kPa)^((1.4-1)/1.4) = 202.36 °C.
2. (b) The throat area of the nozzle: The mass flow rate of the air through the nozzle can be calculated using the equation m_dot = ρAV, where m_dot is the mass flow rate, ρ is the density, A is the cross-sectional area, and V is the velocity. Rearranging the equation, we have A = m_dot / (ρV). Plugging in the values, A = 2.5 kg/s / (ρV) = 2.5 kg/s / (1.184 kg/m³ × 180 m/s) = 7.09 × 10^(-5) m².
3. (c) The temperature at the exit: Similar to part (a), we can use the ideal gas law to determine the temperature. Since the air is isentropic, we can use the same equation as before. Plugging in the values, T_exit = 200°C(100 kPa/300 kPa)^((1.4-1)/1.4) = 202.36 °C.
4. (d) The pressure at the exit: The pressure at the exit can be found using the ideal gas law. Using the equation P_exit = P_throat × (T_exit / T_throat)^(γ / (γ-1)), we can calculate the pressure at the exit. Plugging in the values, P_exit = 100 kPa × (202.36 °C / 202.36 °C)^(1.4 / (1.4-1)) = 100 kPa.