Mass and energy analysis of control volumes

Almost nothing at Northgate is a closed system. Steam flows through the turbine, water through the pumps, flue gas through the stack, air through the compressor. To analyze these devices we draw a control volume around each and keep track of the mass and energy crossing its surface. The laws are the same as for closed systems: mass is conserved and energy is conserved. What changes is the bookkeeping, because mass entering and leaving carries energy with it, and pushing that mass across the boundary takes work. This lesson develops the mass and energy balances for control volumes and specializes them to steady flow, the condition in which the plant spends almost all its time.

Conservation of mass

For a closed system, mass conservation is automatic: the mass is fixed by definition. For a control volume, the principle reads: the net mass transferred into the control volume during a time interval equals the change in the mass inside it.

The mass flow rate \(\dot{m}\) through a cross-section of area \(A\), where the fluid has density \(\rho\) and mean velocity \(V\) normal to the section, is \(\dot{m} = \rho V A = VA/v\). The volume flow rate is \(\dot{V} = VA = \dot{m}v\). In a pipe the velocity varies across the section, from zero at the wall to a maximum on the centerline; \(V\) here is the average velocity, defined so that \(\rho VA\) gives the correct mass flow.

In steady flow, nothing inside the control volume changes with time, so its mass is constant and \(dm_{CV}/dt = 0\). The total mass flow in then equals the total mass flow out. For a single stream, \(\dot{m}_1 = \dot{m}_2\), or \(\rho_1V_1A_1 = \rho_2V_2A_2\). Note that it is mass, not volume, that is conserved: when steam expands through a turbine its density falls, so its volume flow rate rises even though its mass flow rate is unchanged. Only for incompressible liquids are volume flows conserved too.

Flow work

Unlike a closed system, a control volume has mass crossing its boundary, and that takes work. Imagine a small slug of fluid of volume \(V\) about to enter the control volume. The fluid behind it, at pressure \(P\), pushes it in, acting with force \(PA\) through the distance \(L = V/A\). The work done is \(PA \times V/A = PV\), or per unit mass, \(Pv\). This flow work, or flow energy, is energy transferred into the control volume with each kilogram entering; similarly, the fluid in the control volume does flow work \(Pv\) per kilogram to push fluid out at each exit.

Flow work is sometimes listed as a form of energy of the flowing fluid and sometimes as a work interaction. The second view is more physical, but the first is more convenient, and the bookkeeping comes out the same either way. With flow work counted as part of the energy of the stream, each kilogram of flowing fluid carries

This is where enthalpy earns its keep. The combination \(u + Pv\) appears automatically for every stream crossing the boundary, and having a single tabulated property for it simplifies every control-volume analysis. The energy carried by a stream per unit time is \(\dot{m}\theta = \dot{m}(h + V^2/2 + gz)\).

The energy balance for a control volume

For a control volume, energy crosses the boundary as heat, as work (shaft, electrical, boundary if the control volume deforms), and with mass. Excluding flow work from the work term, since it is now inside \(\theta\), the general energy balance in rate form is

This one equation covers turbines, compressors, boilers, tanks being filled, and everything else. It is easier to apply in two special forms. The first, steady flow, is the subject of the rest of this lesson and the next. The second, uniform flow for devices whose contents change over time, such as a tank being filled, is the subject of Lesson 9.

Steady-flow processes

In steady flow, no property at any point in the control volume changes with time, so its mass and energy content are constant: \(dm_{CV}/dt = 0\) and \(dE_{CV}/dt = 0\). The heat transfer and power are also steady. The balances become

For a device with one inlet (state 1) and one outlet (state 2), the mass flow is the same at both, and dividing by \(\dot{m}\) gives the steady-flow energy equation per unit mass:

Here \(q = \dot{Q}/\dot{m}\) and \(w = \dot{W}/\dot{m}\) are heat and work per kilogram of fluid flowing through. Mind the units: \(h\) is in kJ/kg, while \(V^2/2\) and \(gz\) come out in J/kg and must be divided by 1000.

Which terms matter?

In most devices some terms in the steady-flow energy equation are negligible, and recognizing which ones is half the skill. Kinetic energy changes are small unless velocities are large: a change from 2 m/s to 30 m/s is only 0.45 kJ/kg. They matter in nozzles and diffusers, whose whole purpose is to change velocity, and occasionally at turbine exhausts. Potential energy changes are almost always negligible in thermal devices, at about 0.1 kJ/kg per 10 m of elevation; they matter for water pumped up tall buildings or through hydroelectric plants. Heat transfer is negligible in well-insulated devices and in those where the fluid passes through too quickly to exchange much heat, such as turbines and compressors. Work is zero in devices with no shafts, wires or moving boundaries, such as heat exchangers, valves, nozzles and pipes.

The program below makes this concrete for the steam that heats a building 40 m above the plant. It compares the enthalpy change between steam in the main and condensate leaving the building with the changes in kinetic and potential energy, and finds the steam flow a building needs for a given heat load.

# Mass flow, velocity and the size of each energy term for flows in the plant
import math

def velocity(mdot, v, D):
    """Mean velocity (m/s) for mass flow mdot (kg/s), specific volume v (m3/kg), pipe bore D (m)."""
    A = math.pi * D**2 / 4
    return mdot * v / A

flows = [   # name, mdot kg/s, v m3/kg, bore m
    ("steam main, 1 MPa 250 C", 12.0, 0.23275, 0.400),
    ("condensate return, 85 C",  12.0, 0.001032, 0.150),
    ("boiler outlet, 4 MPa 400 C", 12.0, 0.07343, 0.250),
]
print(f"{'flow':28s} {'V m/s':>7s} {'Vol m3/s':>9s} {'KE kJ/kg':>9s}")
for name, mdot, v, D in flows:
    V = velocity(mdot, v, D)
    print(f"{name:28s} {V:7.2f} {mdot * v:9.4f} {V**2 / 2 / 1000:9.3f}")

# Compare the energy terms for steam delivered to a building 40 m higher than the plant
dh = 2943.1 - 356.0             # kJ/kg released when steam at 1 MPa 250 C returns as 85 C condensate
dke = (30.0**2 - 2.0**2) / 2 / 1000
dpe = 9.81 * 40 / 1000
print(f"\nenthalpy change {dh:8.1f} kJ/kg")
print(f"kinetic energy  {dke:8.3f} kJ/kg ({100 * dke / dh:.3f} % of dh)")
print(f"potential energy{dpe:8.3f} kJ/kg ({100 * dpe / dh:.3f} % of dh)")

# Steam needed by a building that requires 1.8 MW of heat
Q = 1800.0
print(f"\nsteam for 1.8 MW: {Q / dh:.3f} kg/s = {Q / dh * 3600:.0f} kg/h")

flow                           V m/s  Vol m3/s  KE kJ/kg
steam main, 1 MPa 250 C        22.23    2.7930     0.247
condensate return, 85 C         0.70    0.0124     0.000
boiler outlet, 4 MPa 400 C     17.95    0.8812     0.161

enthalpy change   2587.1 kJ/kg
kinetic energy     0.448 kJ/kg (0.017 % of dh)
potential energy   0.392 kJ/kg (0.015 % of dh)

steam for 1.8 MW: 0.696 kg/s = 2505 kg/h

The kinetic and potential energy terms are each less than 0.02 percent of the enthalpy change. For steam heating, enthalpy is everything. The same conclusion holds for most heat exchangers, boilers, condensers and pipes: the energy balance reduces to \(\dot{Q} = \dot{m}(h_2 - h_1)\) for each stream.

Energy balances on the whole plant

Control volumes can be drawn around a single component or around a whole system. Drawing one around the entire boiler house, with natural gas and air entering, flue gas leaving, feedwater entering and steam leaving, gives the boiler efficiency directly. Drawing one around the whole district loop, including the plant and every building, shows that in steady operation the fuel energy entering must equal the heat delivered to buildings plus all the losses: up the stacks, through pipe insulation, in the cooling tower, and as electricity exported from the turbine generator. Northgate's annual energy report is exactly such a balance, and the losses it reveals guide where the plant spends its maintenance budget.