Air at 312 K (k = 1.4, R = 287 J/kg·K) flows at 498 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 312) = 354.1 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 498/354.1 = 1.407$$

5. Classification

   $$M = 1.41 \to supersonic$$

Why the other options are there

• c = 299.2 m/s (k omitted)
• M = 0.711 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 286 K (k = 1.4, R = 287 J/kg·K) flows at 569 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 286) = 339.0 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 569/339.0 = 1.679$$

5. Classification

   $$M = 1.68 \to supersonic$$

Why the other options are there

• c = 286.5 m/s (k omitted)
• M = 0.596 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 267 K (k = 1.4, R = 287 J/kg·K) flows at 201 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 267) = 327.5 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 201/327.5 = 0.614$$

5. Classification

   $$M = 0.61 \to subsonic compressible$$

Why the other options are there

• c = 276.8 m/s (k omitted)
• M = 1.630 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 311 K (k = 1.4, R = 287 J/kg·K) flows at 305 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 311) = 353.5 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 305/353.5 = 0.863$$

5. Classification

   $$M = 0.86 \to subsonic compressible$$

Why the other options are there

• c = 298.8 m/s (k omitted)
• M = 1.159 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 260 K (k = 1.4, R = 287 J/kg·K) flows at 597 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 260) = 323.2 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 597/323.2 = 1.847$$

5. Classification

   $$M = 1.85 \to supersonic$$

Why the other options are there

• c = 273.2 m/s (k omitted)
• M = 0.541 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 301 K (k = 1.4, R = 287 J/kg·K) flows at 182 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 301) = 347.8 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 182/347.8 = 0.523$$

5. Classification

   $$M = 0.52 \to subsonic compressible$$

Why the other options are there

• c = 293.9 m/s (k omitted)
• M = 1.911 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 248 K (k = 1.4, R = 287 J/kg·K) flows at 656 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 248) = 315.7 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 656/315.7 = 2.078$$

5. Classification

   $$M = 2.08 \to supersonic$$

Why the other options are there

• c = 266.8 m/s (k omitted)
• M = 0.481 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 277 K (k = 1.4, R = 287 J/kg·K) flows at 432 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 277) = 333.6 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 432/333.6 = 1.295$$

5. Classification

   $$M = 1.29 \to supersonic$$

Why the other options are there

• c = 282.0 m/s (k omitted)
• M = 0.772 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 271 K (k = 1.4, R = 287 J/kg·K) flows at 496 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 271) = 330.0 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 496/330.0 = 1.503$$

5. Classification

   $$M = 1.50 \to supersonic$$

Why the other options are there

• c = 278.9 m/s (k omitted)
• M = 0.665 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number

Air at 260 K (k = 1.4, R = 287 J/kg·K) flows at 517 m/s. Compute the local speed of sound and the Mach number, and classify the flow regime.

Start with the thinking

• The speed of sound depends only on temperature for an ideal gas — pressure does not appear.
• M < 0.3 lets you treat the flow as incompressible; M > 1 is supersonic.

Step-by-step solution

1. Formula

   $$c = \sqrt{kRT}$$

2. Substituting

   $$c = \sqrt(1.4 \times 287 \times 260) = 323.2 m/s$$

3. Formula

   $$M = V/c$$

4. Substituting

   $$M = 517/323.2 = 1.600$$

5. Classification

   $$M = 1.60 \to supersonic$$

Why the other options are there

• c = 273.2 m/s (k omitted)
• M = 0.625 (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Mach Number