Isentropic flow in nozzles

The 600 mm transmission main between Northside and the elevated reservoir has eleven air valves at its high points. When the main is filled after a repair, air must leave through them as fast as water enters, or the trapped air compresses, slows the filling and can cause pressure surges when it finally escapes. When the main is in service, small amounts of air come out of solution and collect at the high points, reducing the pipe's capacity until the valves release them. And when the main is drained, the valves must let air in to prevent a vacuum that could collapse the pipe or draw in contaminated water. Two of the valves failed last winter, and Hana has asked Jun to check whether the remaining valves have the right orifice sizes. The calculation is a compressible-flow problem: air flowing through a small opening under a pressure difference that can be large enough to make the flow sonic.

Isentropic flow with area change

Lesson 8 showed that for steady one-dimensional isentropic flow, \(dA/A = (M^2 - 1)\,dV/V\). A subsonic flow accelerates in a converging passage, a supersonic flow accelerates in a diverging one, and the sonic condition M = 1 can occur only at a minimum area, a throat. To work with finite changes it is convenient to use the sonic state as a reference. Denote by A* the area at which the flow would be exactly sonic; for a given mass flow and stagnation state, conservation of mass then gives the area-Mach relation:

For every area ratio greater than one there are two solutions, one subsonic and one supersonic. The ratio has a minimum of exactly 1 at M = 1. Combined with the stagnation relations of Lesson 8, it gives the pressure, temperature and density anywhere in the passage. The values at the sonic point, called critical values, are fixed fractions of the stagnation values. For air with k = 1.4:

# Isentropic flow: area-Mach relation, choking, and sizing an air-release orifice
import math
k, R = 1.4, 287.0
def A_ratio(M):                      # A/A* for isentropic flow
    return (1 / M) * ((2 / (k + 1)) * (1 + (k - 1) / 2 * M**2)) ** ((k + 1) / (2 * (k - 1)))
def M_from_A(Ar, supersonic):        # invert A/A* by bisection
    lo, hi = (1.0, 10.0) if supersonic else (1e-6, 1.0)
    for _ in range(200):
        mid = (lo + hi) / 2
        if (A_ratio(mid) > Ar) == supersonic: hi = mid
        else: lo = mid
    return (lo + hi) / 2
pr = lambda M: (1 + (k - 1) / 2 * M**2) ** (-k / (k - 1))   # p/p0
Tr = lambda M: 1 / (1 + (k - 1) / 2 * M**2)                 # T/T0
crit = pr(1.0)
print(f"Critical pressure ratio p*/p0 = {crit:.4f}; T*/T0 = {Tr(1.0):.4f}")
print(" M     A/A*    p/p0    T/T0")
for M in (0.2, 0.5, 0.8, 1.0, 1.5, 2.0, 2.5, 3.0):
    print(f" {M:3.1f}  {A_ratio(M):6.3f}  {pr(M):6.4f}  {Tr(M):6.4f}")

# Laboratory blowdown nozzle: exit area 1.6875 times the throat
Ae_At = 1.6875
for sup in (False, True):
    Me = M_from_A(Ae_At, sup)
    print(f"Exit Mach {'supersonic' if sup else 'subsonic'} branch: {Me:.3f}, p_e/p0 = {pr(Me):.4f}")

# Choked mass flow through an orifice: mdot = Cd A p0 sqrt(k/(R T0)) (2/(k+1))^((k+1)/(2(k-1)))
def mdot_orifice(d, p0, T0, pb, Cd=0.62):
    A = math.pi * d**2 / 4
    if pb / p0 <= crit:              # choked: independent of back pressure
        return Cd * A * p0 * math.sqrt(k / (R * T0)) * (2 / (k + 1)) ** ((k + 1) / (2 * (k - 1))), True
    M = math.sqrt(2 / (k - 1) * ((p0 / pb) ** ((k - 1) / k) - 1))
    T = T0 * Tr(M); rho = pb / (R * T)
    return Cd * A * rho * M * math.sqrt(k * R * T), False
patm = 97.4e3; T0 = 283.15
# Air-release valve on the 600 mm main: small orifice venting accumulated air under pressure
for pg in (50e3, 300e3, 600e3):
    m, ch = mdot_orifice(0.005, patm + pg, T0, patm)
    rho_n = patm / (R * T0)
    print(f"5 mm orifice at {pg/1000:.0f} kPa gauge: {m*1000:.2f} g/s ({m/rho_n*1000:.1f} L/s of free air), {'choked' if ch else 'not choked'}")
# Large orifice during pipe filling: air must leave as fast as water enters
Vfill, Dmain = 0.30, 0.600
Qair = Vfill * math.pi * Dmain**2 / 4
for d in (0.025, 0.050, 0.075):
    # find the internal pressure needed to vent Qair (free air) through orifice d
    lo, hi = patm * 1.0001, patm * 3
    for _ in range(100):
        mid = (lo + hi) / 2
        m, ch = mdot_orifice(d, mid, T0, patm)
        if m / (patm / (R * T0)) > Qair: hi = mid
        else: lo = mid
    print(f"Filling at {Vfill} m/s ({Qair*1000:.0f} L/s of air): {d*1000:.0f} mm orifice needs {(mid-patm)/1000:6.1f} kPa above atmosphere{' (choked)' if ch else ''}")

Critical pressure ratio p*/p0 = 0.5283; T*/T0 = 0.8333
 M     A/A*    p/p0    T/T0
 0.2   2.964  0.9725  0.9921
 0.5   1.340  0.8430  0.9524
 0.8   1.038  0.6560  0.8865
 1.0   1.000  0.5283  0.8333
 1.5   1.176  0.2724  0.6897
 2.0   1.688  0.1278  0.5556
 2.5   2.637  0.0585  0.4444
 3.0   4.235  0.0272  0.3571
Exit Mach subsonic branch: 0.372, p_e/p0 = 0.9088
Exit Mach supersonic branch: 2.000, p_e/p0 = 0.1278
5 mm orifice at 50 kPa gauge: 4.14 g/s (3.5 L/s of free air), not choked
5 mm orifice at 300 kPa gauge: 11.62 g/s (9.7 L/s of free air), choked
5 mm orifice at 600 kPa gauge: 20.39 g/s (17.0 L/s of free air), choked
Filling at 0.3 m/s (85 L/s of air): 25 mm orifice needs   48.2 kPa above atmosphere
Filling at 0.3 m/s (85 L/s of air): 50 mm orifice needs    2.9 kPa above atmosphere
Filling at 0.3 m/s (85 L/s of air): 75 mm orifice needs    0.6 kPa above atmosphere

The table that the program prints is a short version of the isentropic flow tables printed in every gas dynamics text. They make hand calculations quick: given a Mach number, read off A/A*, p/p₀ and T/T₀; given an area ratio or a pressure ratio, find the Mach number. For computer work, the relations are simply solved numerically, as the program does by bisection.

Choking and the maximum mass flow

The mass flow through a passage is \(\dot{m} = \rho A V\). Writing ρ and V in terms of the stagnation state and the Mach number, and evaluating at a throat where M = 1, gives the maximum possible mass flow through that throat for a given stagnation pressure and temperature:

Once the throat is sonic, the flow is said to be choked. Lowering the downstream pressure further cannot increase the mass flow, because the sonic throat is the place where information travels upstream at exactly the flow speed: news of the lower downstream pressure cannot reach the reservoir. The mass flow then depends only on the upstream stagnation pressure and temperature and on the throat area. This is a very useful property. Choked orifices, called critical flow nozzles or sonic nozzles, are used as flow standards because they deliver a mass flow that is independent of conditions downstream.

For air discharging into the atmosphere through a simple orifice, the flow chokes when the absolute upstream pressure exceeds about 1.89 times the absolute downstream pressure, the reciprocal of 0.528. At Omaha's atmospheric pressure of about 97.4 kPa, that is an internal gauge pressure above about 87 kPa. The transmission main runs at several hundred kilopascals, so air released from it under pressure always leaves at sonic speed in the orifice's vena contracta.

The converging nozzle and back pressure

Consider a converging nozzle fed from a large reservoir at stagnation pressure p₀ and discharging into a space at back pressure p_b. With p_b equal to p₀, there is no flow. As p_b is lowered, flow starts and increases, and the exit pressure equals the back pressure. When p_b reaches the critical value 0.528 p₀, the exit becomes sonic and the mass flow reaches its maximum. Lowering p_b further changes nothing inside the nozzle: the exit stays sonic at 0.528 p₀, and the gas expands outside the nozzle in a pattern of expansion waves and shocks, the diamonds visible in the exhaust of a jet engine or a compressed-air blowgun. A converging nozzle can never deliver a supersonic jet at its exit.

The converging-diverging nozzle

To produce supersonic flow, the passage must continue to widen after the sonic throat. Carl Gustaf de Laval developed this shape for steam turbines in the 1880s, and it is used today in rocket engines, steam turbines, supersonic wind tunnels and some ejectors. A converging-diverging nozzle has a design back pressure at which the flow accelerates smoothly through Mach 1 at the throat and leaves the exit at the design Mach number, with the exit pressure equal to the back pressure. The exit Mach number depends only on the exit-to-throat area ratio.

The program examines the blowdown nozzle in the university gas dynamics laboratory where Jun took his compressible flow course. It has an exit area 1.6875 times its throat area. Two isentropic solutions fit that area ratio: a subsonic exit at Mach 0.37 with p_e/p₀ = 0.909, where the flow speeds up to just below sonic in the throat and slows again, and a supersonic exit at Mach 2.0 with p_e/p₀ = 0.128. At back pressures between these values, neither isentropic solution can match the back pressure, and a shock wave forms in the diverging section or outside the exit. Lesson 10 locates it.

Flow through air valves

Air valves for water mains come in three kinds. An air-release valve has a small orifice, typically a few millimeters across, controlled by a float; it releases the small pockets of air that collect in a pressurized main. An air and vacuum valve has a large orifice, comparable to a fraction of the pipe diameter, that lets large volumes of air out during filling and in during draining, but closes when water arrives and cannot open again while the main is pressurized. A combination valve has both. AWWA C512 covers these valves, and AWWA Manual M51 describes how to locate and size them.

The program examines both duties for one high point. With a 5 mm release orifice and a discharge coefficient of 0.62, the flow at 50 kPa gauge is not choked, because the pressure ratio is less than 1.89, and the valve releases about 4 g/s of air. At 300 and 600 kPa it is choked and releases about 12 and 20 g/s, or roughly 10 and 17 liters of free air per second. Because the choked mass flow is proportional to the absolute upstream pressure, the valve releases air faster at higher line pressure, which is helpful.

For filling, the main is filled at a controlled velocity of 0.30 m/s, a common practice limit for large mains, which displaces 85 L/s of air. The program finds the internal pressure needed to push that much air out through orifices of three sizes. A 25 mm orifice would need about 48 kPa above atmospheric pressure, and the air pocket would cushion and then slam as water reached the valve. A 50 mm orifice needs only about 3 kPa, and a 75 mm orifice about 0.6 kPa. Many design guides limit the pressure differential across an air and vacuum orifice during filling to a few kilopascals, roughly 2 psi or about 14 kPa at most, to avoid slam and premature closure; the 50 mm orifice meets that comfortably.

Other applications of choked flow

Choked flow appears wherever gas leaves a high-pressure system. Safety relief valves on compressed-air receivers and steam boilers are sized with choked-flow equations, because at their set pressure the flow through them is sonic; the receiver's relief valve must pass the compressor's full output without the pressure rising more than a set percentage above the set point. Leaks in compressed-air systems are almost always choked, so a leak's flow is proportional to the absolute pressure and the leak area, and plant energy audits estimate leak losses that way: a single 3 mm leak at 700 kPa gauge wastes several liters of free air per second, which adds up to thousands of kilowatt-hours a year. Gas-turbine and jet-engine nozzles, rocket nozzles, steam-turbine blade passages and the throats of supersonic wind tunnels are all designed with the relations in this lesson.

In the United States, safety relief valves on air receivers are certified under the ASME Boiler and Pressure Vessel Code, with their capacity stamped on the valve, and state inspectors check them. Air valves on Platte Bluffs Water's mains are specified to AWWA C512 and located following AWWA Manual M51.