Isentropic Flow Relationships

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

16 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

In an ideal gas for an isentropic process, the following relationships exist between static properties at any two points in the flow.
The stagnation temperature, T0, at a point in the flow is related to the static temperature as follows:
Energy relation between two points:
The relationship between the static and stagnation properties (T0, P0, and ρ0) at any point in the flow can be expressed as a
function of the Mach number as follows:
Compressible flows are often accelerated or decelerated through a nozzle or diffuser. For subsonic flows, the velocity decreases
as the flow cross-sectional area increases and vice versa. For supersonic flows, the velocity increases as the flow cross-sectional
area increases and decreases as the flow cross-sectional area decreases. The point at which the Mach number is sonic is called
the throat and its area is represented by the variable, A*. The following area ratio holds for any Mach number.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Isentropic flow relationships — solve for stagnation temperature — Isentropic Flow Relationships

isentropic flow relationships through a converging nozzle Given static temperature (T) = 297.0 K; specific heat ratio (k) = 1.3800; Mach number (M) = 2.0000, determine the stagnation temperature (T_0) in K.

Given

$static temperature (T) = 297.0 K$
$specific heat ratio (k) = 1.3800$
$Mach number (M) = 2.0000$

Find

stagnation temperature (T_0), in K

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except T_0 is given, so isolate T_0 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that T_0 stands alone on the left-hand side.
Step 3 — List the givens: static temperature (T) = 297.0 K, specific heat ratio (k) = 1.3800, Mach number (M) = 2.0000.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T_{0} = 522.7\ \text{K}$
Step 6 — Check: returning T_0 = 522.7 K to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

T_{0} = 522.7\ \text{K}

Why the other options are there

1,045 — kept a factor of two that cancels in the correct rearrangement.
261.4 — dropped that same factor in the other direction.
575.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 2

Isentropic flow relationships — solve for static temperature — Isentropic Flow Relationships (2)

isentropic flow relationships in a compressor inlet duct Given specific heat ratio (k) = 1.3800; Mach number (M) = 1.0000; stagnation temperature (T_0) = 309.0 K, determine the static temperature (T) in K.

Given

$specific heat ratio (k) = 1.3800$
$Mach number (M) = 1.0000$
$stagnation temperature (T_0) = 309.0 K$

Find

static temperature (T), in K

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that T stands alone on the left-hand side.
Step 3 — List the givens: specific heat ratio (k) = 1.3800, Mach number (M) = 1.0000, stagnation temperature (T_0) = 309.0 K.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T = 259.7\ \text{K}$
Step 6 — Check: returning T = 259.7 K to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

T = 259.7\ \text{K}

Why the other options are there

519.3 — kept a factor of two that cancels in the correct rearrangement.
129.8 — dropped that same factor in the other direction.
285.6 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 3

Isentropic flow relationships — solve for Mach number — Isentropic Flow Relationships (3)

isentropic flow relationships for air accelerating through a venturi Given static temperature (T) = 269.0 K; specific heat ratio (k) = 1.3600; stagnation temperature (T_0) = 400.0 K, determine the Mach number (M).

Given

$static temperature (T) = 269.0 K$
$specific heat ratio (k) = 1.3600$
$stagnation temperature (T_0) = 400.0 K$

Find

Mach number (M)

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except M is given, so isolate M symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that M stands alone on the left-hand side.
Step 3 — List the givens: static temperature (T) = 269.0 K, specific heat ratio (k) = 1.3600, stagnation temperature (T_0) = 400.0 K.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 1.6448$
Step 6 — Check: returning M = 1.6448 to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 1.6448

Why the other options are there

3.2897 — kept a factor of two that cancels in the correct rearrangement.
0.8224 — dropped that same factor in the other direction.
1.8093 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 4

Isentropic flow relationships — solve for stagnation temperature (case 2) — Isentropic Flow Relationships (4)

isentropic flow relationships through a converging nozzle Given static temperature (T) = 399.0 K; specific heat ratio (k) = 1.3500; Mach number (M) = 1.1500, determine the stagnation temperature (T_0) in K.

Given

$static temperature (T) = 399.0 K$
$specific heat ratio (k) = 1.3500$
$Mach number (M) = 1.1500$

Find

stagnation temperature (T_0), in K

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except T_0 is given, so isolate T_0 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that T_0 stands alone on the left-hand side.
Step 3 — List the givens: static temperature (T) = 399.0 K, specific heat ratio (k) = 1.3500, Mach number (M) = 1.1500.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T_{0} = 491.3\ \text{K}$
Step 6 — Check: returning T_0 = 491.3 K to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

T_{0} = 491.3\ \text{K}

Why the other options are there

982.7 — kept a factor of two that cancels in the correct rearrangement.
245.7 — dropped that same factor in the other direction.
540.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 5

Isentropic flow relationships — solve for static temperature (case 2) — Isentropic Flow Relationships (5)

isentropic flow relationships in a compressor inlet duct Given specific heat ratio (k) = 1.3500; Mach number (M) = 1.3500; stagnation temperature (T_0) = 539.0 K, determine the static temperature (T) in K.

Given

$specific heat ratio (k) = 1.3500$
$Mach number (M) = 1.3500$
$stagnation temperature (T_0) = 539.0 K$

Find

static temperature (T), in K

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that T stands alone on the left-hand side.
Step 3 — List the givens: specific heat ratio (k) = 1.3500, Mach number (M) = 1.3500, stagnation temperature (T_0) = 539.0 K.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T = 408.7\ \text{K}$
Step 6 — Check: returning T = 408.7 K to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

T = 408.7\ \text{K}

Why the other options are there

817.3 — kept a factor of two that cancels in the correct rearrangement.
204.3 — dropped that same factor in the other direction.
449.5 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 6

Isentropic flow relationships — solve for Mach number (case 2) — Isentropic Flow Relationships (6)

isentropic flow relationships for air accelerating through a venturi Given static temperature (T) = 266.0 K; specific heat ratio (k) = 1.3000; stagnation temperature (T_0) = 682.0 K, determine the Mach number (M).

Given

$static temperature (T) = 266.0 K$
$specific heat ratio (k) = 1.3000$
$stagnation temperature (T_0) = 682.0 K$

Find

Mach number (M)

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except M is given, so isolate M symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that M stands alone on the left-hand side.
Step 3 — List the givens: static temperature (T) = 266.0 K, specific heat ratio (k) = 1.3000, stagnation temperature (T_0) = 682.0 K.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 3.2289$
Step 6 — Check: returning M = 3.2289 to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 3.2289

Why the other options are there

6.4579 — kept a factor of two that cancels in the correct rearrangement.
1.6145 — dropped that same factor in the other direction.
3.5518 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 7

Isentropic flow relationships — solve for stagnation temperature (case 3) — Isentropic Flow Relationships (7)

isentropic flow relationships through a converging nozzle Given static temperature (T) = 293.0 K; specific heat ratio (k) = 1.3300; Mach number (M) = 0.5500, determine the stagnation temperature (T_0) in K.

Given

$static temperature (T) = 293.0 K$
$specific heat ratio (k) = 1.3300$
$Mach number (M) = 0.5500$

Find

stagnation temperature (T_0), in K

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except T_0 is given, so isolate T_0 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that T_0 stands alone on the left-hand side.
Step 3 — List the givens: static temperature (T) = 293.0 K, specific heat ratio (k) = 1.3300, Mach number (M) = 0.5500.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T_{0} = 307.6\ \text{K}$
Step 6 — Check: returning T_0 = 307.6 K to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

T_{0} = 307.6\ \text{K}

Why the other options are there

615.2 — kept a factor of two that cancels in the correct rearrangement.
153.8 — dropped that same factor in the other direction.
338.4 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 8

Isentropic flow relationships — solve for static temperature (case 3) — Isentropic Flow Relationships (8)

isentropic flow relationships in a compressor inlet duct Given specific heat ratio (k) = 1.3500; Mach number (M) = 1.0500; stagnation temperature (T_0) = 466.0 K, determine the static temperature (T) in K.

Given

$specific heat ratio (k) = 1.3500$
$Mach number (M) = 1.0500$
$stagnation temperature (T_0) = 466.0 K$

Find

static temperature (T), in K

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except T is given, so isolate T symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that T stands alone on the left-hand side.
Step 3 — List the givens: specific heat ratio (k) = 1.3500, Mach number (M) = 1.0500, stagnation temperature (T_0) = 466.0 K.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T = 390.6\ \text{K}$
Step 6 — Check: returning T = 390.6 K to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

T = 390.6\ \text{K}

Why the other options are there

781.3 — kept a factor of two that cancels in the correct rearrangement.
195.3 — dropped that same factor in the other direction.
429.7 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 9

Isentropic flow relationships — solve for Mach number (case 3) — Isentropic Flow Relationships (9)

isentropic flow relationships for air accelerating through a venturi Given static temperature (T) = 259.0 K; specific heat ratio (k) = 1.3600; stagnation temperature (T_0) = 385.0 K, determine the Mach number (M).

Given

$static temperature (T) = 259.0 K$
$specific heat ratio (k) = 1.3600$
$stagnation temperature (T_0) = 385.0 K$

Find

Mach number (M)

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except M is given, so isolate M symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that M stands alone on the left-hand side.
Step 3 — List the givens: static temperature (T) = 259.0 K, specific heat ratio (k) = 1.3600, stagnation temperature (T_0) = 385.0 K.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$M = 1.6440$
Step 6 — Check: returning M = 1.6440 to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

M = 1.6440

Why the other options are there

3.2880 — kept a factor of two that cancels in the correct rearrangement.
0.8220 — dropped that same factor in the other direction.
1.8084 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships

Example 10

Isentropic flow relationships — solve for stagnation temperature (case 4) — Isentropic Flow Relationships (10)

isentropic flow relationships through a converging nozzle Given static temperature (T) = 251.0 K; specific heat ratio (k) = 1.3400; Mach number (M) = 0.2500, determine the stagnation temperature (T_0) in K.

Given

$static temperature (T) = 251.0 K$
$specific heat ratio (k) = 1.3400$
$Mach number (M) = 0.2500$

Find

stagnation temperature (T_0), in K

Start with the thinking

The governing relation printed in this handbook section is Isentropic flow relationships.
Everything except T_0 is given, so isolate T_0 symbolically first — never rearrange after the numbers are in.
Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
Isentropic flow relationships connect static and stagnation properties for compressible flow at a given Mach number.

Step-by-step solution

Step 1 — State the governing relation:
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
Step 2 — Rearrange the relation so that T_0 stands alone on the left-hand side.
Step 3 — List the givens: static temperature (T) = 251.0 K, specific heat ratio (k) = 1.3400, Mach number (M) = 0.2500.
Step 4 — Substitute the given values into the rearranged relation.
Step 5 — Evaluate:
$T_{0} = 253.7\ \text{K}$
Step 6 — Check: returning T_0 = 253.7 K to
$\dfrac{T_0}{T} = 1 + \dfrac{k-1}{2} M^2$
reproduces the given quantities, and both sides carry the same units.

Answer:

T_{0} = 253.7\ \text{K}

Why the other options are there

507.3 — kept a factor of two that cancels in the correct rearrangement.
126.8 — dropped that same factor in the other direction.
279.0 — rounded an intermediate value before the final step.

Reference: FE Handbook — Isentropic Flow Relationships