Question

Steam enters a diffuser at 20 psia and \(240^{\circ} \mathrm{F}\) with a velocity of \(900 \mathrm{ft} / \mathrm{s}\) and exits as saturated vapor at \(240^{\circ} \mathrm{F}\) and \(100 \mathrm{ft} / \mathrm{s}\). The exit area of the diffuser is \(1 \mathrm{ft}^{2}\). Determine (a) the mass flow rate of the steam and ( \(b\) ) the rate of entropy generation during this process. Assume an ambient temperature of \(77^{\circ} \mathrm{F}\).

Short answer

Answer: The mass flow rate is 0.272 lbm/s, and the rate of entropy generation is 0.0348 Btu/s·R.

Step-by-step solution

Find the enthalpy and entropy of steam at inlet and outlet conditions

We can look up the necessary properties in the steam tables. At the inlet, the steam is at 20 psia and \(240^{\circ} \mathrm{F}\). The steam is a superheated vapor at these conditions, so we look up the properties in the superheated section of the steam table, and we get: \(h_1 = 1201.1 \mathrm{Btu} / \mathrm{lbm}\) \(s_1 = 1.6969 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}\) At the outlet, the steam is saturated vapor at \(240^{\circ} \mathrm{F}\), so we look up the properties in the saturated section of the steam table, and we get: \(h_2 = 1150.7 \mathrm{Btu} / \mathrm{lbm}\) \(s_2 = 1.8287 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}\)

Calculate the mass flow rate

Using the fact that the outlet area is \(1 \mathrm{ft}^2\), and the velocities are given as 900 ft/s at the inlet and 100 ft/s at the outlet, we can apply the continuity equation (conservation of mass) to find the mass flow rate \(\dot{m}\): \(\dot{m} \rho_1 A_1 v_1 = \dot{m} \rho_2 A_2 v_2\) Here, we have only given the exit area, \(A_2 = 1 \mathrm{ft}^2\), so we can rearrange to find the mass flow rate: \(\dot{m} = \frac{\rho_2 v_2 A_2}{\rho_1 v_1}\) We can find the specific volume of the steam at the inlet and outlet conditions (\(v_1\) and \(v_2\)) from the steam tables: At the inlet: \(v_1 = 4.425 \mathrm{ft}^3 / \mathrm{lbm}\) At the outlet: \(v_2 = 4.755 \mathrm{ft}^3 / \mathrm{lbm}\) Now, we can calculate the mass flow rate: \(\dot{m} = \frac{(1/\mathrm{4.755 \, ft^3 / lbm})(100 \, \mathrm{ft}/ \mathrm{s})(1 \, \mathrm{ft}^2)}{(1/\mathrm{4.425 \, ft^3 / lbm})(900 \, \mathrm{ft}/ \mathrm{s})} = 0.272 \, \mathrm{lbm} / \mathrm{s}\)

Determine the rate of entropy generation

We can now use the mass flow rate, enthalpy, and entropy values to calculate the rate of entropy generation \(\dot{S}_{gen}\). We apply the first law of thermodynamics and the definition of the entropy rate, which can be written as: \(\dot{S}_{gen} = \dot{m}[(s_2 - s_1)+R \ln(v_2/v_1)]\) Here, R is the gas constant for water (1.986 Btu/lbm·R). Now we can calculate the rate of entropy generation: \(\dot{S}_{gen} = 0.272 \, \mathrm{lbm} / \mathrm{s} [(1.8287 - 1.6969) \, \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} + 1.986 \, \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} \cdot \ln(4.755 / 4.425)]\) \(\dot{S}_{gen} = 0.272 \, \mathrm{lbm} / \mathrm{s} \cdot 0.128 \, \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R} = 0.0348 \, \mathrm{Btu / s} \cdot \mathrm{R}\) Thus, the rate of entropy generation during this process is \(0.0348 \, \mathrm{Btu / s} \cdot \mathrm{R}\).