Question

Steam at \(3 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\) enters an adiabatic nozzle steadily with a velocity of \(40 \mathrm{m} / \mathrm{s}\) and leaves at \(2.5 \mathrm{MPa}\) and \(300 \mathrm{m} / \mathrm{s}\). Determine \((a)\) the exit temperature and ( \(b\) ) the ratio of the inlet to exit area \(A_{1} / A_{2}\).

Short answer

Inlet conditions: - Pressure: \(P_1 = 3 \mathrm{MPa}\) - Temperature: \(T_1 = 400^{\circ} \mathrm{C}\) - Velocity: \(V_1 = 40 \mathrm{m} / \mathrm{s}\) Outlet conditions: - Pressure: \(P_2 = 2.5 \mathrm{MPa}\) - Velocity: \(V_2 = 300 \mathrm{m} / \mathrm{s}\) Answer: The exit temperature is \(165.19^{\circ} \mathrm{C}\), and the area ratio (inlet to exit) is \(2.097\).

Step-by-step solution

Identify the Given Information

We are given the following conditions for the adiabatic nozzle: Inlet conditions: - Pressure: \(P_1 = 3 \mathrm{MPa}\) - Temperature: \(T_1 = 400^{\circ} \mathrm{C}\) - Velocity: \(V_1 = 40 \mathrm{m} / \mathrm{s}\) Outlet conditions: - Pressure: \(P_2 = 2.5 \mathrm{MPa}\) - Velocity: \(V_2 = 300 \mathrm{m} / \mathrm{s}\) We need to find: - Exit temperature: \(T_2\) - Ratio of inlet to exit area: \(A_1/A_2\) Assumption: The process is steady and adiabatic (no heat transfer).

Apply the First Law of Thermodynamics

For a steady, adiabatic process, the energy equation is given by: \(h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}\) Here, \(h_1\) and \(h_2\) are the specific enthalpies at the inlet and outlet, respectively. To find the enthalpies, we can use steam tables. However, we will use the Ideal Gas Law for this problem, with the assumption that the steam behaves like an ideal gas. Using the Ideal Gas Law, specific enthalpy can be expressed as: \(h = C_pT\) where \(C_p\) is the specific heat at constant pressure. For steam, we can use the value of \(C_p = 2.01 \mathrm{kJ} / \mathrm{kg \cdot K}\).

Calculate the Exit Temperature

Substitute the expressions for enthalpy into the energy equation: \(C_pT_1 + \frac{V_1^2}{2} = C_pT_2 + \frac{V_2^2}{2}\) Now, solve for the exit temperature \(T_2\): \(T_2 = \frac{C_pT_1 + \frac{V_1^2}{2} - \frac{V_2^2}{2}}{C_p}\) Plug in the values given: \(T_2 = \frac{2.01 (400) + \frac{40^2}{2} - \frac{300^2}{2}}{2.01}\) Calculate \(T_2\): \(T_2 = 165.19^{\circ} \mathrm{C}\) \((a)\) The exit temperature is \(165.19^{\circ} \mathrm{C}\).

Apply the Continuity Equation

For a steady flow through the nozzle, the mass flow rate is constant. The continuity equation is given by: \(\rho_1 V_1 A_1 = \rho_2 V_2 A_2\) Assuming ideal gas behavior, we can rewrite this equation as: \(\frac{P_1}{RT_1} V_1 A_1 = \frac{P_2}{RT_2} V_2 A_2\) The area ratio \(A_1/A_2\) can be found by rearranging this equation: \(\frac{A_1}{A_2} = \frac{P_2V_2T_1}{P_1V_1T_2}\) Now, plug in the given values and the calculated exit temperature: \(\frac{A_1}{A_2} = \frac{(2.5\times10^6)(300)(400+273.15)}{(3\times10^6)(40)(165.19+273.15)}\) Calculate the area ratio: \(\frac{A_1}{A_2} = 2.097\) \((b)\) The ratio of the inlet to exit area is \(A_1/A_2 = 2.097\).