Questions: Using Steam Tables and Thermodynamic Diagrams

Question 1

Steam at 1 MPa has a quality of x = 0.75. The saturated liquid enthalpy is h_f = 763 kJ/kg and the latent heat h_fg = 2015 kJ/kg. What is the specific enthalpy of this steam?

• A. 763 kJ/kg — quality does not affect enthalpy, only temperature matters
• B. 2274 kJ/kg — use h = h_g since quality indicates significant vapor
• C. 1274 kJ/kg — calculated as h_f + x·h_fg = 763 + 0.75 × 2015
• D. 1511 kJ/kg — calculated as x·h_fg since we weight only the vapor fraction

Answer: (C)

In the two-phase region, properties are calculated by weighting saturated liquid and vapor values by quality: h = h_f + x·h_fg = 763 + 0.75 × 2015 = 763 + 1511 = 1274 kJ/kg. Quality x represents the mass fraction that is vapor. A quality of 0 is pure saturated liquid (h = h_f); quality of 1 is pure saturated vapor (h = h_g = h_f + h_fg). Intermediate qualities require this interpolation formula. Option A is wrong — enthalpy in the two-phase region depends critically on quality. Option D omits the liquid baseline h_f.

Question 2

On a T-s diagram, which of the following correctly describes an irreversible adiabatic expansion through a turbine, compared to an ideal isentropic expansion between the same inlet and outlet pressures?

• A. The irreversible process appears as a vertical line shifted left of the isentropic line
• B. The irreversible process curves to the right, ending at higher entropy than the isentropic endpoint
• C. The irreversible process ends at higher temperature than the isentropic process at the same pressure
• D. The irreversible process is indistinguishable from the isentropic process on a T-s diagram

Answer: (B)

An isentropic process is adiabatic and reversible — on a T-s diagram it appears as a vertical line (constant entropy). An irreversible adiabatic process generates entropy internally, so the final state has higher entropy than the isentropic endpoint at the same final pressure. The path bows to the right, ending at a higher entropy value. Because enthalpy drops less in the irreversible case (entropy generation absorbs some of the energy that could have done work), the turbine produces less work. Isentropic efficiency is literally the ratio of actual to ideal enthalpy drops — a geometric ratio visible on the Mollier diagram.

Question 3

On the Mollier (h-s) diagram, the work output of an ideal turbine is represented as a vertical distance between the inlet and outlet states.

• T. True
• F. False

Answer: True

For a steady-flow turbine with negligible kinetic and potential energy changes, the work output per unit mass equals the enthalpy drop: w = h_in - h_out. On the Mollier diagram, enthalpy is on the vertical axis, so the enthalpy drop appears as a vertical distance. An ideal (isentropic) process runs vertically downward at constant entropy. An irreversible expansion runs rightward as well as downward, so the vertical distance (enthalpy drop) is smaller — less work extracted. This geometric representation makes turbine efficiency visually intuitive.

Question 4

Steam at exactly 200°C is generally in the two-phase (wet steam) region and requires quality to determine its thermodynamic properties.

• T. True
• F. False

Answer: False

At 200°C, the saturation pressure is about 1.55 MPa. If the actual pressure is below 1.55 MPa, the steam is superheated — a single-phase gas described by the superheated steam tables without needing quality. If the pressure is exactly 1.55 MPa, the state is on the saturation boundary, where quality determines whether it is saturated liquid (x=0), saturated vapor (x=1), or a mixture. If pressure is above 1.55 MPa at 200°C, the state is compressed liquid. Temperature alone does not determine the phase region; both temperature and pressure are required.

Question 5

Why is the T-s diagram particularly useful for visualizing thermodynamic cycle efficiency, and what does an irreversible process look like on it compared to a reversible one?

Model answer: The T-s diagram makes irreversibility visible as geometry. For any reversible process, the area under the curve on a T-s diagram equals the heat transferred (Q = ∫T ds). A reversible adiabatic (isentropic) process is a vertical line — zero area, zero heat transfer, constant entropy. An irreversible adiabatic process generates entropy internally, so the exit state lies to the right of the isentropic exit at the same final pressure. The horizontal shift directly represents entropy generation — the larger the rightward deviation, the more irreversible the process. For a turbine, this means less enthalpy drop and less work output, making inefficiency literally visible as the deviation from vertical.

Isentropic efficiency of a turbine is defined as (actual enthalpy drop)/(isentropic enthalpy drop). On the Mollier diagram (where enthalpy is vertical), this ratio is a ratio of vertical distances. Engineers use this visual check constantly: if the turbine exit state is close to the isentropic endpoint, the turbine is efficient; if it has drifted rightward, efficiency has been lost to irreversibility. No calculation is needed to form this qualitative judgment — the diagram shows it directly.