Superheated steam originally at P1 and T1 expands through a nozzle to an exhaust pressure P2. Assuming the process is reversible and adiabatic, determine the downstream state of the steam and ΔH for P1 = 1000 kPa, t1 = 250°C, and P2 = 200 kPa.

Question

Superheated steam originally at [latex] P_1 and T_1[/latex] expands through a nozzle to an exhaust pressure [latex] P_2[/latex] . Assuming the process is reversible and adiabatic, determine the downstream state of the steam and ΔH for [latex] P_1 = 1000 kPa, t_1 = 250°C, and P_2 = 200 kPa.[/latex]

Accepted Answer

Because the process is both reversible and adiabatic, there is no change in the entropy of the steam. For the initial temperature of 250°C at 1000 kPa, no entries appear in the tables for superheated steam. Interpolation between values for 240°C and 260°C yields, at 1000 kPa,

[latex] H_1=2942.9 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \quad S_1=6.9252 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1} [/latex]

For the final state at 200 kPa,

[latex] S_2=S_1=6.9252 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1} [/latex]

Because the entropy of saturated vapor at 200 Pa is greater than [latex] S_2 [/latex] the final state must lie in the two-phase liquid/vapor region. Thus [latex] t_2[/latex] is the saturation temperature at 200 kPa, given in the superheat tables as [latex] t_2 = 120.23°C. [/latex]Equation (6.96a) applied to the entropy becomes:

[latex] M=\left(1-x^v\right) M^l+x^v M^v [/latex] (6.96a)

[latex] S_2=\left(1-x_2^v\right) S_2^I+x_2^v S_2^v [/latex]

Numerically,

[latex] 6.9252=1.5301\left(1-x_2^v\right)+7.1268 x_2^v [/latex]

where the values 1.5301 and 7.1268 are entropies of saturated liquid and saturated vapor at 200 kPa. Solving,

[latex] x_2^v=0.9640 [/latex]

The mixture is 96.40% vapor and 3.60% liquid. Its enthalpy is obtained by further application of Eq. (6.96a):

[latex] M=\left(1-x^v\right) M^l+x^v M^v [/latex] (6.96a)

[latex] H_2 = ( 0.0360 ) ( 504.7 ) + ( 0.9640 ) ( 2706.3 ) = 2627.0 kJ ⋅kg ^−1 [/latex]

Finally,

[latex]ΔH = H_2 − H_1 = 2627.0 − 2942.9 = − 315.9 kJ ⋅kg^ −1[/latex]

For a nozzle, under the stated assumptions the steady-flow energy balance, Eq. (2.31), becomes:

[latex] \Delta H+\frac{\Delta u^2}{2}+g \Delta z=Q+W_s [/latex] (2.31)

[latex] \Delta H+\frac{1}{2} \Delta u^2=0 [/latex]

Thus the decrease in enthalpy is exactly compensated by an increase in kinetic energy of the fluid. In other words, the velocity of a fluid increases as it flows through a nozzle, which is its usual purpose. Nozzles are treated further in Sec. 7.1.

Question 6.10: Superheated steam originally at P1 and T1 expands through a ...

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