Interpretation:
The exit velocity and cross-sectional area at the nozzle exit under the given conditions needs to be calculated
Concept Introduction:
- The energy balance equation that relates the inlet and outlet enthalpies and velocities is given based on the first law of
thermodynamics as:
- For a process that takes place at constant entropy i.e. isentropic, the change in entropy is zero. In other words, the entropy in the final state (S2) is equal to that in the initial state (S1). The change in entropy is given as:
The rate of mass flow
Exit velocity = 534.4 m/s
Cross-sectional area at the nozzle exit =
Given Information:
Inlet pressure of steam, P1 = 800 kPa
Inlet Temperature of steam T1=
Outlet pressure P2 = 525 kPa
Explanation:
Since this is a constant enthalpy process, S2 = S1
From the final state entropy, the exit temperature and the exit enthalpy (H2) can be deduced. The change in enthalpy can be used to calculate the exit velocity based on equation (1). Finally, from the calculated value of exit velocity, the exit cross-sectional area can be calculated using equation (3).
Calculation:
Step 1:
Calculate the final state entropy, S2
Based on the steam tables for inlet conditions, P1 = 800 kPa and T1=
Specific enthalpy of vapor, Hg = H1= 3014.5 kJ/kg
Specific enthalpy of vapor, Sg = S1= 7.1593 kJ/kg-K
Since, S2 = S1
We have, S2= 7.1593 kJ/kg-K
Step 2:
Calculate the final state enthalpy, H2
The exit temperature, T2corresponding to the exit pressure P2 = 525 kPa and S2= 7.1593 kJ/kg-K, can be calculated by interpolation:
At T =
At T =
Similarly, H2 at T2 =
Step 3:
Calculate the exit velocity, u2
Based on equation (3) we have:
Step 3:
Calculate the exit cross-sectional area, A
Based on equation (3) we have:
Where:
Thus, exit velocity = 534.4 m/s
The cross-sectional area at the exit =
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