Steady-flow devices

Walk through the Northgate plant and you pass, in order, almost every steady-flow device in an engineering thermodynamics text. Steam leaves the boilers and enters the turbine. A pressure-reducing valve stands beside it as a bypass. The district main is a long pipe. The deaerator is a mixing chamber; the building heat exchangers, the condenser on the turbine's gland seals and the air heater are heat exchangers; the feedwater pumps raise the water's pressure; the air compressor feeds the instrument-air system; and the blow-off nozzles on the air heater clean its surfaces. Each is analyzed with the same steady-flow energy equation from Lesson 7, simplified by recognizing which terms matter. This lesson takes them one at a time.

Nozzles and diffusers

A nozzle increases the velocity of a fluid at the expense of its pressure; a diffuser does the reverse. For subsonic flow a nozzle narrows in the flow direction and a diffuser widens. Neither involves work, and the fluid passes through so quickly that heat transfer is usually negligible. Potential energy changes are negligible. The kinetic energy change is the whole point, so it must be kept. The energy equation becomes

so any increase in kinetic energy is paid for by a decrease in enthalpy. For an ideal gas with constant specific heats, \(h_1 - h_2 = c_p(T_1 - T_2)\), and the fluid cools as it accelerates. Nozzles are at the heart of steam and gas turbines, where they turn enthalpy into the high-velocity jets that drive the blades; diffusers slow the air entering a jet engine's compressor, raising its pressure.

Turbines

A turbine extracts shaft work from a fluid expanding through it. In a steam or gas turbine, the fluid passes through rows of stationary nozzles and moving blades; each nozzle accelerates the fluid, and the blades turn its momentum into torque on the shaft. Across the whole machine, kinetic energy changes are usually small compared with the enthalpy drop, potential energy changes are negligible, and heat losses are small because the machine is insulated and the steam passes through in a fraction of a second. The energy equation reduces to

Compressors, pumps and fans

Compressors, pumps and fans increase the pressure of a fluid by doing work on it. Compressors handle gases and can produce large pressure ratios; fans handle gases with small pressure rises, mainly to move them; pumps handle liquids. The energy equation is the turbine's in reverse: \(\dot{W}_{\text{in}} = \dot{m}(h_2 - h_1) + \dot{Q}_{\text{out}}\). Compressors are often cooled, by water jackets or fins, because cooler gas needs less work to compress (as Lesson 2's polytropic example showed), so the heat term may be significant.

For a pump handling liquid water, the enthalpy rise is mostly the \(v\Delta P\) term of Lesson 5, since the temperature barely changes. The ideal pump work per unit mass is therefore \(w = v(P_2 - P_1)\), and a real pump needs more because of friction and leakage, as measured by its efficiency.

Throttling valves

A throttling valve is any restriction that causes a significant pressure drop: a partly open valve, a porous plug, a capillary tube, an orifice. There is no work, heat transfer is negligible because the device is small and the flow is fast, and kinetic and potential energy changes are usually small. The energy equation reduces to the striking result

so a throttling valve is an isenthalpic device. Enthalpy is constant, but other properties change. For an ideal gas, whose enthalpy depends only on temperature, throttling leaves the temperature unchanged. For real gases and for steam, the temperature can rise or fall. Liquids throttled below their saturation pressure partly flash into vapor and cool sharply, which is how the expansion valve in a refrigerator produces its cold refrigerant.

Mixing chambers

A mixing chamber combines two or more streams into one. It might be a T-junction in a pipe or, as at Northgate, a large vessel such as the deaerator, where steam is injected into incoming water. There is no work, the chamber is usually well insulated, and kinetic and potential energy changes are negligible. The mass and energy balances become

Heat exchangers

A heat exchanger transfers energy between two streams without mixing them. The simplest is a tube within a tube; the building exchangers are shell-and-tube units with steam condensing in the shell and hot water flowing through the tubes; the boilers' air heaters are plate or rotary exchangers. There is no work, and kinetic and potential energy changes are negligible. If the exchanger as a whole is well insulated, the energy lost by the hot stream equals the energy gained by the cold one:

The choice of control volume matters. Drawn around the whole exchanger, there is no heat transfer across the boundary and the equation above follows. Drawn around one stream only, the heat transfer from the other stream crosses the boundary and appears as \(\dot{Q}\), as in the hospital example of Lesson 7.

Pipe and duct flow

Pipes and ducts carry fluids from one device to another. They do no work unless a pump or fan is included in the control volume, and heat transfer may or may not be important: negligible in a short, insulated pipe, but significant in a long steam main, as Lesson 7 showed, or deliberately large in a boiler tube. Kinetic energy changes are usually small; potential energy changes matter for liquids lifted through large heights. Writing the energy balance for each section of pipe and adding the results is a standard way to find heat losses or pump power in a distribution system.

The program below collects the calculations from this lesson in one place, each reduced to its characteristic one-line energy balance.

# Steady-flow energy balances for the main devices in the plant
def lerp(x, x0, x1, y0, y1):
    return y0 + (x - x0) / (x1 - x0) * (y1 - y0)

mdot = 12.0                                   # kg/s, full-load steam flow

# Turbine: 4 MPa 400 C -> 1 MPa 250 C, heat loss 30 kW, negligible KE and PE changes
h1, h2, Qloss = 3214.5, 2943.1, 30.0
W_turb = mdot * (h1 - h2) - Qloss
print(f"turbine power: {W_turb:.0f} kW")

# Pressure-reducing valve bypassing the turbine: h constant, find T at 1 MPa
T_prv = lerp(3214.5, 3158.2, 3264.5, 350, 400)       # 1 MPa row: 350 C and 400 C
print(f"PRV outlet at 1 MPa: h = 3214.5 kJ/kg, T = {T_prv:.1f} C")

# Deaerator (mixing chamber): steam at 1 MPa 250 C + condensate at 80 C -> saturated liquid at 120 kPa
h_steam, h_cond, h_out = 2943.1, 335.0, 439.4
y = (h_out - h_cond) / (h_steam - h_cond)
print(f"deaerator: steam fraction y = {y:.4f}, steam flow {y * mdot:.3f} kg/s")

# Feedwater pump: 120 kPa saturated liquid to 4.2 MPa, pump efficiency 0.75
v, dP, eta = 0.001047, 4200 - 120, 0.75
w_ideal = v * dP
print(f"pump: ideal {w_ideal:.2f} kJ/kg, actual {w_ideal / eta:.2f} kJ/kg, power {mdot * w_ideal / eta:.1f} kW")

# Throttling calorimeter: wet steam from 1 MPa throttled to 100 kPa, measured 130 C
h_cal = lerp(130, 99.61, 150, 2675.0, 2776.6)       # 100 kPa: saturated steam and 150 C
x = (h_cal - 762.5) / 2014.6
print(f"calorimeter: h = {h_cal:.1f} kJ/kg, quality of main steam x = {x:.4f}")

# Building heat exchanger: 1.8 MW into water heated from 60 C to 85 C
m_water = 1800 / (4.18 * (85 - 60))
print(f"building loop water flow: {m_water:.2f} kg/s")

# Air nozzle: 300 kPa, 200 C, 30 m/s -> 380 m/s, adiabatic, cp = 1.005
T2 = 200 - (380**2 - 30**2) / 2 / 1000 / 1.005
print(f"nozzle exit temperature: {T2:.1f} C")

turbine power: 3227 kW
PRV outlet at 1 MPa: h = 3214.5 kJ/kg, T = 376.5 C
deaerator: steam fraction y = 0.0400, steam flow 0.480 kg/s
pump: ideal 4.27 kJ/kg, actual 5.70 kJ/kg, power 68.3 kW
calorimeter: h = 2736.3 kJ/kg, quality of main steam x = 0.9797
building loop water flow: 17.22 kg/s
nozzle exit temperature: 128.6 C