SME
Compressible Flow · Problem type
Choked mass flow through a throat
Use this for “Air from a tank at 500 kPa and 500 K flows through a 0.001 m² throat at M = 1. Find ṁ.”
Diagram
The duct ends at its throat, where M = 1: the flow is choked, and lowering the back pressure adds no flow.
This is an example. Type your own values to start a new problem.
ṁ (kg/s)
=
0.6847
×
p₀ (kPa)
×
A∗ (m²)
R
×
T₀ (K)
Formulas
ṁ
=
0.6847
×
p₀
×
A∗
÷
√
287 ×
T₀
0.9037
=
0.6847
×
500,000
×
0.001
÷
√
287 × 500
Formulas are worked in Pa. The step-by-step shows the conversions.
Assumptions
•
The throat is at M = 1 (choked), so lowering the back pressure further adds no flow.
•
Isentropic flow of air from rest in the tank, which holds p₀ and T₀.
•
Air: γ = 1.4 gives the factor 0.6847, and R = 287 J/(kg·K).
Step-by-step
WE KNOW
p₀ = 500 kPa, T₀ = 500 K, A∗ = 0.001 m²
FIND
mass flow (ṁ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to Pa).
p₀ = 500 kPa = 500,000 Pa (1 kPa = 1000 Pa)
Step 2 · Find mass flow (ṁ)
Use ṁ = 0.6847 × p₀ × A∗ ÷ √(287 × T₀)
At
M
=
1
the
flow
per
unit
area
is
the
most
the
throat
can
pass:
0.6847
p₀
÷
√
RT₀
.
ṁ
=
0.6847
×
500,000
×
0.001
÷
√
287 × 500
ṁ
=
0.6847
×
500,000
×
0.001
÷
√
143
,500
ṁ
=
0.6847
×
500,000
×
0.001
378.8
ṁ
=
342,350
×
0.001
378.8
ṁ
=
342.35
378.8
ṁ
=
0.9037
kg/s
Check
Put the numbers back in. Both sides should match.
0.9037
=
0.6847
×
500,000
×
0.001
÷
√
287 × 500
✓
Topic check
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