The nozzle diameter at the various pressures.

Write the expression for the nozzle diameter.

    d2=4A2π (I)

Here, the diameter of the nozzle is d2 and the area of the nozzle at exit is A2.

Write the expression for the area at the exit of the nozzle.

    A2=m˙v2V2 (II)

Here, the mass flow rate is m˙, the volume is at the exit of the nozzle V2 and the specific volume is at exit of the nozzle v2.

Write the expression for the volume of the nozzle at exit of the nozzle.

    V2=2(ho−h2) (III)

Here, the specific enthalpy at stagnation pressure is ho and the specific enthalpy at exit is h2.

Write the formula of interpolation method of two variables.

  y2=(x2−x1)(y3−y1)(x3−x1)+y1 (IV)

Here, the variables denote by x and y are pressure and specific enthalpy.

Conclusion:

Refer to table A-4 “properties of superheated water vapor” to obtain the specific enthalpy and the specific entropy at temperature T1=280°C and pressure p1=1.83 bar using interpolation as 3031.9 kJ/kg and 7.8839 kJ/kg⋅K respectively.

Refer to table A-4 “properties of superheated water vapor” to obtain the specific enthalpy at temperature T1=280°C and pressure p=1.5 bar as 3032.8 kJ/kg.

Refer to table A-4 “properties of superheated water vapor” to obtain the specific enthalpy at temperature T1=280°C and pressure p=3 bar as 3028.6 kJ/kg.

Show the pressure at 3028.6 kJ/kg and 3032.8 kJ/kg as in below Table.

Substitute 1.5 bar for x1, 1.83 bar for x2, 3 bar for x3, 3028.6 kJ/kg for y1, and 3032.8 kJ/kg for y3 in Equation (V).

  y2={[(3028.6 kJ/kg−3032.8 kJ/kg)(1.83 bar−1.5 bar)](3 bar−1.5 bar)+3032.8 kJ/kg}=(−1.386)1.5 kJ/kg+3032.8 kJ/kg=−0.924 kJ/kg+3032.8 kJ/kg≈3031.87 kJ/kg

So, the specific enthalpy is 3031.87 kJ/kg.

Refer to table A-4 “properties of superheated water vapor” to obtain the specific entropy at temperature T1=280°C and pressure p=1.5 bar as 7.9555 kJ/kg⋅K.

Refer to table A-4 “properties of superheated water vapor” to obtain the specific entropy at temperature T1=280°C and pressure p=3 bar as 7.6299 kJ/kg⋅K.

Show the pressure at 7.6299 kJ/kg⋅K and 7.9555 kJ/kg⋅K as in below Table.

Substitute 1.5 bar for x1, 1.83 bar for x2, 3 bar for x3, 7.6299 kJ/kg⋅K for y1, and 7.9555 kJ/kg⋅K for y3 in Equation (V).

  y2={[(7.6299 kJ/kg⋅K−7.9555 kJ/kg⋅K)(1.83 bar−1.5 bar)](3 bar−1.5 bar)+7.9555 kJ/kg⋅K}=(−1.074)1.5 kJ/kg⋅K+7.9555 kJ/kg⋅K=−0.071632 kJ/kg⋅K+7.9555 kJ/kg⋅K≈7.8838 kJ/kg⋅K

So, the specific entropy is 7.8838 kJ/kg⋅K.

Since the stagnation pressure and stagnation temperature is equal to the temperature and pressure at the inlet of the nozzle, so the specific entropy and specific enthalpy are also same.

Since the process is isentropic, so the entropy at state 2 is equal to the stagnation entropy.

Refer to table A-4 “properties of superheated water vapor” to obtain the specific enthalpy at specific entropy s=7.8052 kJ/kg⋅K and pressure p=1.5 bar as 2952.7 kJ/kg.

Refer to table A-4 “Properties of superheated water vapor” to obtain the specific enthalpy at specific entropy s=7.9555 kJ/kg⋅K and pressure p=3 bar as 3032.8 kJ/kg.

Show the pressure at 7.6299 kJ/kg⋅K and 7.9555 kJ/kg⋅K as in below Table.

Substitute 7.8052 kJ/kg⋅K for x1, 7.8838 kJ/kg⋅K for x2, 7.9555 kJ/kg⋅K for x3, 2952.7 kJ/kg for y1, and 3032.8 kJ/kg for y3 in Equation (V).

  y2={[(7.8838 kJ/kg⋅K−7.8052 kJ/kg⋅K)(3032.8 kJ/kg−2952.7 kJ/kg)](7.9555 kJ/kg⋅K−7