Question

Steam at \(4 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\) enters a nozzle steadily with a velocity of \(60 \mathrm{m} / \mathrm{s}\), and it leaves at \(2 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\) The inlet area of the nozzle is \(50 \mathrm{cm}^{2}\), and heat is being lost at a rate of \(75 \mathrm{kJ} / \mathrm{s}\). Determine \((a)\) the mass flow rate of the steam, ( \(b\) ) the exit velocity of the steam, and ( \(c\) ) the exit area of the nozzle.

Short answer

(a) The mass flow rate of the steam is 4.089 kg/s. (b) The exit velocity of the steam is 592.4 m/s. (c) The exit area of the nozzle is 10 cm².

Step-by-step solution

Find Specific Volume and Enthalpy

Consult the steam tables to find the specific volume (v) and specific enthalpy (h) at the given inlet (4 MPa, 400°C) and outlet conditions (2 MPa, 300°C): For Inlet: (4 MPa, 400°C) \(v_1 = 0.0734 \mathrm{m}^3 / \mathrm{kg}\) \(h_1 = 3230.9 \mathrm{kJ} / \mathrm{kg}\) For Outlet: (2 MPa, 300°C) \(v_2 = 0.14734 \mathrm{m}^3 / \mathrm{kg}\) \(h_2 = 3025.5 \mathrm{kJ} / \mathrm{kg}\)

Calculate Mass Flow Rate

Use the continuity equation for incompressible flow, which states that the mass flow rate is constant throughout the nozzle: \(\dot{m} = \rho_1 A_1 V_1\) We can find the density (\(\rho_1\)) of the steam at the inlet using the specific volume: \(\rho_1 = \frac{1}{v_1} = \frac{1}{0.0734 \mathrm{m}^3 / \mathrm{kg}} = 13.63 \mathrm{kg} / \mathrm{m}^3\) We're given the inlet area (\(A_1\) = 50 cm²) and inlet velocity (\(V_1\) = 60 m/s). Convert the area to square meters: \(A_1 = 50 \mathrm{cm}^2 * \frac{1 \mathrm{m}^2}{10000 \mathrm{cm}^2} = 0.005 \mathrm{m}^2\) Now, calculate the mass flow rate: \(\dot{m} = (13.63 \mathrm{kg} / \mathrm{m}^3)(0.005 \mathrm{m}^2)(60 \mathrm{m / s}) = 4.089 \mathrm{kg / s}\)

Calculate Exit Velocity

Use the energy equation for a nozzle with heat transfer: \(V_2^2 = V_1^2 + 2(h_1 - h_2 - q)\) We're given the heat loss rate (\(\dot{Q}\) = 75 kJ/s) and need to find the heat transfer per unit mass (q): \(q = \frac{\dot{Q}}{\dot{m}} = \frac{75 \mathrm{kJ/s}}{4.089 \mathrm{kg/s}} = 18.34 \mathrm{kJ/kg}\) Then, plug the specific enthalpies (h), heat transfer (q), and inlet velocity (\(V_1\) = 60 m/s) values into the energy equation: \(V_2^2 = (60 \mathrm{m/s})^2 + 2(3230.9 - 3025.5 - 18.34) \times 10^3\) \(V_2^2 = 3600 + 348.26 \times 10^3\) \(V_2 = \sqrt{351860} \approx 592.4 \mathrm{m /s}\)

Calculate Exit Area

Finally, use the mass flow rate and exit velocity to find the exit area. First, find the density of the steam at the outlet (\(\rho_2\)): \(\rho_2 = \frac{1}{v_2} = \frac{1}{0.14734 \mathrm{m}^3 / \mathrm{kg}}= 6.787 \mathrm{kg} / \mathrm{m}^3\) Now, use the mass flow rate equation to find the exit area: \(A_2 = \frac{\dot{m}}{\rho_2 V_2} = \frac{4.089 \mathrm{kg/s}}{(6.787 \mathrm{kg/m^3})(592.4 \mathrm{m/s})} = 0.00100 \mathrm{m}^2\) Convert this to square centimeters: \(A_2 = 0.00100 \mathrm{m}^2 * \frac{10000 \mathrm{cm}^2}{1 \mathrm{m}^2} = 10 \mathrm{cm}^2\) The results are as follows: (a) Mass flow rate: 4.089 kg/s (b) Exit velocity: 592.4 m/s (c) Exit area: 10 cm²

Key concepts

Mass Flow Rate

Understanding the mass flow rate is crucial in thermodynamics, especially when analyzing devices like nozzles. The mass flow rate, denoted as \(\frac{m}{s}\) or \(\dot{m}\), quantifies the amount of mass passing through a given area per unit of time.

For steam moving through a nozzle, it’s the consistent flow of steam mass per second despite the changes in pressure and temperature it undergoes. It ensures continuity in the system and is calculated using the density \(\rho\), cross-sectional area \(A\), and velocity \(V\) of the steam at a certain point, expressed by the equation \(\dot{m} = \rho A V\).

In practice, if we have the inlet conditions and the area, we can calculate the mass flow rate at that point. An important aspect of the mass flow rate is that it remains constant throughout the nozzle as long as there is no accumulation or depletion of mass within the system, aligning with the principle of conservation of mass.

Exit Velocity

As for exit velocity, this refers to the velocity of steam as it leaves the nozzle. The exit velocity of steam is essential for determining the nozzle's effectiveness in converting thermal energy into kinetic energy.

To find the exit velocity, the energy equation for a nozzle is employed, incorporating specific enthalpies and any heat transfer occurring. As steam travels through the nozzle, it experiences a change in enthalpy, converting heat into kinetic energy, accelerating the steam. The equation takes into account the initial velocity, changes in specific enthalpy, and any heat loss or gain.

In the given problem, the heat loss is factored in as \(q\) when applying the energy equation \(V_2^2 = V_1^2 + 2(h_1 - h_2 - q)\), where \(V_2\) is the exit velocity. The substantial increase from the inlet velocity to the exit velocity observed is attributed to the nozzle's design, which effectively harnesses the energy from the high-pressure steam.

Specific Enthalpy

Specific enthalpy, \(h\), is a property of a substance that represents the total energy per unit mass, including both internal energy and the energy required to make room for it by displacing the environment. For a working fluid like steam, specific enthalphy is a key factor in evaluating the performance of thermodynamic processes.

In a nozzle, changes in specific enthalpy between the inlet and outlet are indicative of energy conversion efficiency. The higher the drop in specific enthalpy, the more mechanical or kinetic energy is produced. Specifically, for a steam nozzle operating without any external work inputs or outputs, the drop in specific enthalpy from the inlet to the outlet, \((h_1 - h_2)\), correlates directly with the increase in kinetic energy of the steam, minus any heat lost or added to the surroundings.

The careful consideration of specific enthalpy in the exercise ensures an accurate calculation of steam conditions upon exiting the nozzle, demonstrating the nozzle’s capacity to convert thermal energy into motion.