Unsteady-flow processes: filling and emptying

Steady flow describes the plant at full load, but some of the most important events at Northgate are not steady at all. Every few minutes the air compressor starts, charges the instrument-air receiver from its low setpoint to its high setpoint, and stops. Each autumn, the cold district main is warmed and filled with steam. The boiler drum is filled with water before firing, and blown down to remove sludge. A relief valve opens and a vessel empties. In each case, the mass and energy inside the control volume change with time. This lesson extends the control-volume balances to such transient, or unsteady-flow, processes.

The general balances

Lesson 7 gave the mass and energy balances for a control volume in rate form. Integrating them over the duration of a process, from an initial state 1 to a final state 2, gives

where \(\theta = h + V^2/2 + gz\) is the energy carried per unit mass by each stream. In general the properties of the entering and leaving streams can change during the process, and so can the state inside the control volume, which makes the integrals hard to evaluate. Two idealizations make most practical problems tractable.

The uniform-flow model

In a uniform-flow process, the properties of each stream crossing the boundary are constant in time and uniform over the inlet or outlet area, although the flow rates may vary; and the state inside the control volume, while it changes with time, is uniform throughout at any instant. A tank filled from a large supply line meets the first condition well, because the line pressure and temperature hardly change; the second condition holds approximately if the gas inside mixes well. With these assumptions the integrals become simple products, and with kinetic and potential energies neglected, the energy balance becomes

Note carefully: the streams carry enthalpy \(h\), because they do flow work as they cross the boundary, while the contents of the control volume store internal energy \(u\). This asymmetry is the source of most of the interesting results in transient problems, and of most of the mistakes.

The result does not depend on the line pressure or the size of the tank. That is a hallmark of uniform-flow problems: often the answer comes out in terms of properties alone, once the mass and energy balances are combined.

Charging the instrument-air receiver

The Northgate receiver is never evacuated; it cycles between about 600 and 900 kPa absolute. Each time the compressor starts, warm air from the aftercooler flows in and compresses the air already in the tank. The temperature rises during charging and then falls slowly as the tank loses heat to the compressor room, so the pressure sags a little after the compressor stops. Graham had noticed this sag and wondered whether the check valve was leaking.

The program below repeats this calculation and then simulates the charging step by step, allowing heat to flow from the warming air to the tank wall and the room. In each small time step it adds the energy brought in by the entering air, subtracts the heat lost, and recomputes the temperature from the new internal energy and mass. It finishes with the evacuated-tank result.

# Charging the 2.0 m3 instrument-air receiver from 600 kPa to 900 kPa (absolute)
R, cp = 0.287, 1.005
cv = cp - R
V, P1, T1 = 2.0, 600.0, 293.15          # m3, kPa, K (tank starts at room temperature)
P2, T_in = 900.0, 313.15                # target pressure; air arrives at 40 C from the aftercooler
m1 = P1 * V / (R * T1)

# 1. Adiabatic, uniform flow: (m2 - m1) cp T_in = m2 cv T2 - m1 cv T1, with m2 = P2 V / (R T2)
#    -> cv P2 V / R - m1 cv T1 = cp T_in (P2 V / (R T2) - m1), solve for T2
T2 = cp * T_in * P2 * V / R / (cv * P2 * V / R - m1 * cv * T1 + cp * T_in * m1)
m2 = P2 * V / (R * T2)
print(f"adiabatic: m1 = {m1:.2f} kg, m2 = {m2:.2f} kg, added {m2 - m1:.2f} kg, T2 = {T2 - 273.15:.1f} C")
print(f"after cooling to 20 C the pressure settles at {m2 * R * T1 / V:.0f} kPa")

# 2. Time-stepping with heat loss to the tank wall and room: Q_out = UA (T - T_room)
mdot, UA, T_room = 0.12, 0.060, 293.15  # kg/s, kW/K, K
m, T, t, dt = m1, T1, 0.0, 0.1
while m * R * T / V < P2:
    Q_out = UA * (T - T_room)
    U = m * cv * T + (mdot * cp * T_in - Q_out) * dt   # energy balance over dt
    m += mdot * dt
    T = U / (m * cv)
    t += dt
print(f"with heat loss: reaches {P2:.0f} kPa after {t:.0f} s, T = {T - 273.15:.1f} C, m = {m:.2f} kg")

# 3. Filling an evacuated rigid tank from a supply line: T2 = k T_line (adiabatic)
k = cp / cv
print(f"evacuated tank filled from a 20 C line: T2 = {k * 293.15 - 273.15:.1f} C")

adiabatic: m1 = 14.26 kg, m2 = 19.03 kg, added 4.77 kg, T2 = 56.4 C
after cooling to 20 C the pressure settles at 801 kPa
with heat loss: reaches 900 kPa after 41 s, T = 53.8 C, m = 19.18 kg
evacuated tank filled from a 20 C line: T2 = 137.2 C

With a modest wall heat loss the final temperature is 53.8 °C rather than 56.4 °C, and slightly more air goes in before the pressure switch trips. The adiabatic calculation is a good first estimate, and the simulation shows how to refine it once heat transfer can be estimated. Note also that the tank air ends up hotter than the 40 °C air entering it: flow work again.

Filling with steam

The same balances apply to steam, with properties from the tables instead of the ideal-gas relations. Each autumn, before the heating season, a section of the district main is isolated, drained, and then refilled with steam. The first steam into the cold, empty pipe condenses on the walls, which is why warming is done slowly through a small bypass valve with the drains open. But consider an idealized version: an insulated, evacuated vessel filled from the main.

Emptying and blowdown

When mass leaves a control volume during a transient, the exit stream's enthalpy usually changes as the contents change, and the uniform-flow assumption for the exit must be applied with care. For a tank discharging gas, the gas remaining inside expands as the tank empties and cools, so the exit enthalpy falls. If the process is slow and the remaining gas expands reversibly and adiabatically, its temperature follows the isentropic relation of Lesson 11; otherwise, an average exit enthalpy is used, or the process is integrated step by step as in the program.

Boiler blowdown is a liquid example. A small fraction of the boiler water, carrying concentrated dissolved solids, is discharged continuously or in bursts to keep the water chemistry within limits. The blowdown water leaves as saturated liquid at boiler pressure, and when it is throttled to atmospheric pressure, part of it flashes into steam. Many plants, including Northgate, pass blowdown through a flash tank to recover that steam for the deaerator, and then through a heat exchanger to recover energy from the remaining hot water.

Choosing between steady and unsteady analysis

The same device can call for different models at different times. A boiler at constant load is a steady-flow device; the same boiler during a cold start, when its contents are heating up and its water level is changing, is not. The test is whether the mass and energy inside the control volume change significantly during the period of interest. If they do, use the transient balances; if they change only slightly compared with the flows through the device, steady flow is a good approximation even when conditions are drifting slowly. For the receiver, which doubles its mass in under a minute, the transient analysis is essential; for the boiler at full load, steady flow is fine.