Pressure coefficient definition check for turbomachinery A statement claims: “The ratio of the isentropic work to the Euler work in a compressor is called the pressure coefficient.” Decide whether this terminology is correct.

Introduction / Context:Non-dimensional parameters help compare turbomachines. However, consistent terminology is important. The “pressure coefficient” commonly relates pressure rise to dynamic head or to blade speed squared, not to a ratio of two different works as defined here.

Given Data / Assumptions:

• Isentropic work = ideal thermodynamic head rise for a given inlet-to-exit state change.
• Euler work = ideal mechanical energy transfer per unit mass, U * delta V_theta.
• No losses or only conceptual comparison of definitions.

Concept / Approach:Standard dimensionless groups include head coefficient (psi = delta h / U^2), flow coefficient (phi = V_ax / U), and pressure coefficient (often delta p / (0.5 * rho * V^2)). Isentropic efficiency is defined as isentropic work divided by actual shaft work, not by Euler work. Therefore, naming the ratio (isentropic work)/(Euler work) as “pressure coefficient” is incorrect usage.

Step-by-Step Solution:Identify accepted definitions: head coefficient and pressure coefficient relate pressure or head to velocity scales.Recognize efficiency definition: eta_isentropic = isentropic work / actual work input.Euler work is a kinematic ideal; its ratio with isentropic work is not termed “pressure coefficient”.Hence, the statement is not correct.

Verification / Alternative check:Textbook glossaries consistently separate “coefficients” (psi, phi) from “efficiencies.”

Why Other Options Are Wrong:

• Conditional answers (low Mach, solidity dependence) do not fix the terminology error.
• “Only for pumps” is still incorrect; pumps use similar non-dimensional groups.

Common Pitfalls:Mixing up head/pressure coefficients with various efficiency definitions; Euler head is not the denominator in isentropic efficiency.

Final Answer:No