Steam expands isentropically through a converging nozzle operating at steady state from a large tank at 1.83 bar1.83\ \text{bar}, 280C280^{\circ} \mathrm{C}. The mass flow rate is 2 kg/s2 \mathrm{~kg} / \mathrm{s}, the flow is choked, and the exit plane pressure is 1 bar1\ \text{bar}. Determine the diameter of the nozzle, in cm\mathrm{cm}, at locations where the pressure is 1.5 bar1.5\ \text{bar}, and 1 bar1\ \text{bar}, respectively.


Answered 1 year ago
Answered 1 year ago
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In this problem we need to calculate the diameter the exit nozzle for two different exit pressures. For the calculation we will need the given temperature T0=280°CT_0=280\,\text{\textdegree}\text{C} and pressure p0=183kPap_0=183\,\text{kPa} to determine the enthalpy h0h_0 and entropy s0=s2s_0=s_2.

h0=3030kJkgs0=s2=7.86kJkg K\begin{align*} h_0&=3030\,\frac{\text{kJ}}{\text{kg}} \\ s_0&=s_2=7.86\,\frac{\text{kJ}}{\text{kg K}} \\ \end{align*}

The first pressure given is exit pressure p2=150kPap_2=150\,\text{kPa}. We can use that with the calculated entropy s2s_2 to determine the enthalpy h2h_2 and specific volume V2V_2.

h2=2980kJkgV2=1.61m3kg\begin{align*} h_2&=2980\,\frac{\text{kJ}}{\text{kg}} \\ V_2&=1.61\,\frac{\text{m}^3}{\text{kg}} \\ \end{align*}

We can now calculate the exit velocity v2v_2 using the energy balance equation.

v222=h0h2v222=3030000Jkg2980000Jkgv2=316.2ms\begin{align*} \frac{v_2^2}{2}&=h_0-h_2 \\ \frac{v_2^2}{2}&=3030000\,\frac{\text{J}}{\text{kg}} -2980000\,\frac{\text{J}}{\text{kg}}\\ v_2&=316.2\,\frac{\text{m}}{\text{s}} \\ \end{align*}

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