Energy and the first law for process systems

With the material balances in hand, Theo turns to energy. The plant's boiler raises about 95 t/h of steam at 10 bar, and the steam heats the beer column reboilers, the evaporators, the cook process and the molecular sieve superheaters. The energy manager wants to know how much natural gas the boiler needs, how much power the feed pump draws, and how much water the desuperheater must inject. All three answers come from the energy balance, the second great conservation law in this course. This lesson sets up the energy balance for closed and open systems and shows how to use tabulated property data; Lessons 11 and 12 apply it to heating, phase change and reaction.

Forms of energy

A system can hold energy in three forms that matter for process balances. Kinetic energy is due to the motion of the system as a whole, \(E_k = \tfrac{1}{2}mv^2\). Potential energy is due to its position in a gravitational field, \(E_p = mgz\). Internal energy \(U\) is everything else: the motion and interactions of the molecules, and the energy stored in chemical bonds. Internal energy cannot be measured absolutely, only relative to a chosen reference state, but changes in it can be calculated precisely.

Energy crosses a system boundary in two ways. Heat \(Q\) is energy that flows because of a temperature difference. Work \(W\) is energy that flows because of any other driving force, such as a rotating shaft, a moving piston or an electric current. Heat and work exist only in transit; a system contains internal energy, not heat. In this course \(Q\) is positive when heat flows into the system, and \(W\) is positive when the system does work on its surroundings, the usual chemical engineering convention.

The closed-system energy balance

For a closed system, one with no mass crossing its boundary such as a batch fermenter between filling and emptying, the general balance of Lesson 1 becomes the first law of thermodynamics. Energy is neither generated nor consumed, so the change in the system's energy equals the net energy transferred in as heat minus the net energy transferred out as work.

Several simplifications are common. If the system's temperature, phase and composition do not change, \(\Delta U \approx 0\). If it does not accelerate, \(\Delta E_k = 0\). If it does not rise or fall, \(\Delta E_p = 0\). If it is perfectly insulated, the process is adiabatic and \(Q = 0\). If no moving parts or currents cross the boundary, \(W = 0\). Applying these carefully to a well-chosen system is most of the work.

Open systems: flow work and enthalpy

Most process units are open systems with streams flowing in and out. Pushing a stream into a system takes work, because the fluid ahead must be displaced; pushing one out does work on the surroundings. This flow work equals \(PV\) for each mass of fluid crossing the boundary. The remaining work, done through rotating shafts or other moving parts, is called shaft work \(W_s\). Pumps and compressors receive shaft work (\(W_s < 0\)); turbines deliver it (\(W_s > 0\)).

Because internal energy and flow work always travel together with a stream, it is convenient to combine them into a single property, the enthalpy, \(H = U + PV\). On a mass basis, specific enthalpy \(\hat{H} = \hat{U} + P\hat{V}\). With enthalpy, the energy balance for an open system at steady state takes its most useful form.

Each \(\Delta\) term is the total for all outlet streams minus the total for all inlet streams. For a unit with one inlet and one outlet and a mass flow \(\dot{m}\), the balance per unit mass is \(\Delta\hat{H} + \Delta v^2/2 + g\Delta z = \hat{Q} - \hat{W}_s\). In most chemical process units the kinetic and potential energy terms are tiny compared with the enthalpy change, and the balance reduces to \(\Delta\dot{H} = \dot{Q} - \dot{W}_s\). The program at the end of the lesson checks this for the boiler.

State properties and reference states

Internal energy and enthalpy are state properties: their values depend only on the state of the substance, fixed by its temperature, pressure, phase and composition, not on how it reached that state. That means a change in enthalpy can be calculated along any convenient path between the two states, a fact used heavily in Lessons 11 and 12. Heat and work are not state properties; they depend on the path.

Because only changes in enthalpy matter, tables list values relative to a chosen reference state where \(\hat{H}\) or \(\hat{U}\) is set to zero. The steam tables used worldwide set the internal energy of liquid water at its triple point, 0.01 °C and 0.611 kPa, to zero. Any reference works, as long as you use the same one for every stream of a given substance in one balance. Mixing values from tables with different references is a common and costly mistake.

Steam tables

Water and steam are so widely used that their properties are tabulated in great detail. Saturated steam tables give, at each temperature or pressure on the boiling curve, the specific volume, internal energy and enthalpy of saturated liquid and saturated vapor, and their difference, the latent heat of vaporization. Superheated steam tables give properties of vapor at temperatures above saturation for a range of pressures. Liquid water well below its boiling point is nearly incompressible, and its enthalpy is close to that of saturated liquid at the same temperature.

A steam quality \(x\) describes a wet mixture of saturated liquid and vapor: the fraction of the mass that is vapor. Its enthalpy is \(\hat{H} = (1 - x)\hat{H}_L + x\hat{H}_V\). Steam leaving a turbine or a long, poorly insulated line often has a quality below one, and the liquid fraction carries much less energy than the vapor.

The mechanical energy balance

For pumps, pipes and other equipment where liquids flow with little heating, a different form of the energy balance is more useful. For an incompressible fluid at nearly constant temperature, the enthalpy change can be split into the pressure term \(\Delta P/\rho\) and a friction loss \(\hat{F}\), the mechanical energy converted to internal energy by friction. The result is the mechanical energy balance, which becomes the Bernoulli equation when there is no friction and no shaft work.

# Steady-flow energy balances on the plant steam system
# Approximate steam-table enthalpies (kJ/kg), reference: liquid water at the triple point
h = {'water 25 C': 104.8, 'water 105 C (feedwater)': 440.2,
     'sat. liquid 10 bar': 762.6, 'sat. vapor 10 bar': 2777.7,
     'superheated 10 bar, 250 C': 2943.0, 'sat. vapor 3.5 bar': 2732.0}

# 1. How big are kinetic and potential energy changes compared with enthalpy changes?
v1, v2 = 2.0, 30.0              # m/s: feedwater pipe and steam header velocities
dz = 15.0                       # m rise from boiler feed pump to steam drum
dEk = (v2 ** 2 - v1 ** 2) / 2 / 1000     # kJ/kg
dEp = 9.81 * dz / 1000                   # kJ/kg
dH = h['sat. vapor 10 bar'] - h['water 105 C (feedwater)']
print(f"Per kg of steam: dH = {dH:.0f} kJ/kg, dEk = {dEk:.3f} kJ/kg, dEp = {dEp:.3f} kJ/kg")

# 2. Boiler duty and fuel use
steam = 95_000.0                # kg/h of saturated steam at 10 bar
Q = steam * (dH + dEk + dEp) / 3600      # kW
print(f"Boiler duty {Q / 1000:.1f} MW")
eff = 0.84                      # boiler efficiency (fraction of fuel HHV reaching the steam)
hhv = 39.0                      # MJ per normal cubic meter of natural gas, approximate
gas = Q / 1000 / eff / hhv * 3600        # Nm3/h
print(f"Natural gas at {eff:.0%} efficiency: {gas:.0f} Nm3/h ({gas * 24 / 1000:.0f} thousand Nm3/day)")

# 3. Boiler feed pump: mechanical energy balance for an incompressible liquid
rho = 955.0                     # kg/m3 at 105 C
P_in, P_out = 120.0, 1250.0     # kPa absolute
friction = 25.0                 # J/kg lost to friction in the piping
w = (P_out - P_in) * 1000 / rho + 9.81 * dz + friction   # J/kg
m = steam / 3600                # kg/s
eta = 0.70
print(f"Pump: shaft work {w / 1000:.2f} kJ/kg, fluid power {m * w / 1000:.1f} kW, motor input at {eta:.0%}: {m * w / 1000 / eta:.1f} kW")

# 4. Desuperheating: how much 105 C water turns 250 C steam into saturated steam at 10 bar?
hs, hw, hsat = h['superheated 10 bar, 250 C'], h['water 105 C (feedwater)'], h['sat. vapor 10 bar']
ratio = (hs - hsat) / (hsat - hw)        # kg water per kg superheated steam
print(f"Desuperheater: {ratio:.4f} kg water per kg of superheated steam")

Per kg of steam: dH = 2338 kJ/kg, dEk = 0.448 kJ/kg, dEp = 0.147 kJ/kg
Boiler duty 61.7 MW
Natural gas at 84% efficiency: 6780 Nm3/h (163 thousand Nm3/day)
Pump: shaft work 1.36 kJ/kg, fluid power 35.8 kW, motor input at 70%: 51.1 kW
Desuperheater: 0.0707 kg water per kg of superheated steam

Mixing streams: a desuperheater

Energy balances on adiabatic mixing are especially simple: with no heat, no shaft work and negligible kinetic and potential energy changes, the total enthalpy of the streams entering equals the total enthalpy leaving. Desuperheaters use this. Superheated steam from a turbine or a boiler superheater is sprayed with water, which evaporates and cools the steam to saturation, because process heaters transfer heat far better from condensing saturated steam than from superheated vapor.