Question

A centrifugal compressor has an impeller tip speed of \(200\ \text{m/s}\). The whirl component of absolute velocity at the impeller exit is \(150\ \text{m/s}\). The whirl component of velocity at the inlet is zero. The specific heat at constant pressure is \(1\ \text{kJ/kg K}\). Using Euler's compressor principle, determine the temperature rise across the impeller.

• A. \(15\ \text{K}\)
• B. \(30\ \text{K}\)
• C. \(45\ \text{K}\)
• D. \(60\ \text{K}\)

Hint:

For centrifugal compressors, \[ \boxed{ u\left(V_{w2}-V_{w1}\right) = C_p\Delta T. } \] This relates the work input directly to the rise in stagnation temperature.

Answer 1

The Correct Option is B

Step 1: Use Euler's compressor equation. The work done per unit mass is \[ \boxed{ W=u\left(V_{w2}-V_{w1}\right), } \] where
• \(u\) = impeller tip speed,
• \(V_{w1}\) = inlet whirl velocity,
• \(V_{w2}\) = exit whirl velocity.

Step 2: Calculate the work input. Given, \[ u=200\ \text{m/s}, \] \[ V_{w2}=150\ \text{m/s}, \] \[ V_{w1}=0. \] Hence, \[ W = 200(150-0) = 30000\ \text{J/kg} = 30\ \text{kJ/kg}. \]

Step 3: Determine the temperature rise. Using \[ W=C_p\Delta T, \] where \[ C_p=1\ \text{kJ/kg K}, \] \[ 30 = 1\times\Delta T. \] Therefore, \[ \Delta T=30\ \text{K}. \] Hence, \[ \boxed{30\ \text{K}} \] is the correct answer. Thus, \[ \boxed{(B)} \] is the correct answer.