Question

Steam enters a nozzle at \(400^{\circ} \mathrm{C}\) and \(800 \mathrm{kPa}\) with a velocity of \(10 \mathrm{m} / \mathrm{s}\), and leaves at \(300^{\circ} \mathrm{C}\) and \(200 \mathrm{kPa}\) while losing heat at a rate of \(25 \mathrm{kW}\). For an inlet area of \(800 \mathrm{cm}^{2}\) determine the velocity and the volume flow rate of the steam at the nozzle exit.

Short answer

Answer: The velocity of the steam exiting the nozzle is 223.61 m/s, and the volume flow rate at the nozzle exit is 1.305 m³/s.

Step-by-step solution

Calculate the enthalpy and specific volume of the steam entering the nozzle

First, we need to find the enthalpy, \(h_{1}\), and specific volume, \(v_{1}\), of the steam entering the nozzle using the steam tables. The entry values are: temperature, \(T_{1}=400^{\circ}\mathrm{C}\), and pressure, \(P_{1}=800\,\mathrm{kPa}\). Refer to the steam tables and interpolate the values if necessary, we have: \(h_{1} = 3230.9\,\mathrm{kJ/kg}\) \(v_{1} = 0.2405\,\mathrm{m^{3}/kg}\)

Calculate the enthalpy of the steam exiting the nozzle

Consider the heat loss at a rate of \(25\,\mathrm{kW}\), divide the heat loss by the mass flow rate, \(\dot{m}\), to find the change in enthalpy: \(\Delta h = -\cfrac{Q_{loss}}{\dot{m}} = -\cfrac{25000\,\mathrm{W}}{\dot{m}} = \cfrac{-25000}{\dot{m}}\,\mathrm{J/kg}\) Using the energy conservation equation: \(h_{2}=h_{1}+\Delta h\), we can write: \(h_{2} = 3230.9\,\mathrm{kJ/kg} - \cfrac{25000}{\dot{m}}\,\mathrm{J/kg}\)

Calculate the exit velocity

Apply the energy balance equation to the nozzle while neglecting potential energy changes and assuming steady state operation: \(h_{1}+\cfrac{V_{1}^{2}}{2} = h_{2}+\cfrac{V_{2}^{2}}{2}\) Plug in the values, we have: \(3230.9\,\mathrm{kJ/kg}+\cfrac{(10\,\mathrm{m/s})^{2}}{2\cdot1000}= \left(3230.9-\cfrac{25000}{\dot{m}}\right)\,\mathrm{kJ/kg}+\cfrac{V_{2}^{2}}{2\cdot1000}\) Rearrange the equation to solve for \(V_{2}\): \(V_{2}=\sqrt{2\cdot1000\left(3230.9+\cfrac{25000}{\dot{m}}-3230.9\right)-10^{2}}=\sqrt{2\cdot25000}\)

Calculate the volume flow rate at the nozzle exit

First, we need to find the mass flow rate \(\dot{m}\) of the steam entering the nozzle: \(\dot{m} =\cfrac{\text{velocity}\times\text{area}}{\text{specific volume}} =\cfrac{V_{1}\times A_{1}}{v_{1}} =\cfrac{10\,\mathrm{m/s} \times 800\,\mathrm{cm^{2}}}{0.2405\,\mathrm{m^{3}/kg}}\) Convert \(800\,\mathrm{cm^{2}}\) to \(\mathrm{m^{2}}\): \(800\,\mathrm{cm^{2}} = 0.08\,\mathrm{m^{2}}\) \(\dot{m} =\cfrac{10\,\mathrm{m/s} \times 0.08\,\mathrm{m^{2}}}{0.2405\,\mathrm{m^{3}/kg}}=3.326\,\mathrm{kg/s}\) Since mass flow rate is constant across the nozzle, the volume flow rate at the nozzle exit can be written as: \(\dot{V_{2}} = 3.326\,\mathrm{kg/s} \times v_{2}\) But first, we have to determine the specific volume, \(v_{2}\), at nozzle exit using the steam tables, considering the exit temperature, \(T_{2}=300^{\circ}\mathrm{C}\), and pressure, \(P_{2}=200\,\mathrm{kPa}\): \(v_{2} = 0.3926\,\mathrm{m^{3}/kg}\) Now calculate the volume flow rate at the nozzle exit: \(\dot{V_{2}}=3.326\,\mathrm{kg/s} \times 0.3926\,\mathrm{m^{3}/kg}= 1.305\,\mathrm{m^{3}/s}\) So, we have the velocity of the steam exiting the nozzle: \(V_{2}=\sqrt{2\cdot25000}=223.61\,\mathrm{m/s}\) and the volume flow rate at the nozzle exit: \(\dot{V_{2}}=1.305\,\mathrm{m^{3}/s}\).