Heat pipeline


 * This design project is under discussion on its talk page

The term heat pipeline may refer to a pipeline carrying steam or hot water to provide heating to customers.

But a heat pipeline also refers to a proposed kilometers long insulated pipeline, meters in diameter, running from the cold side of a heat engine (which must be kept cool to operate the engine) up a mountainside to a heat sink and radiator on the cold mountaintop. Refrigerant vapor flows up through the large pipe, and condensed liquid refrigerant flows back down through a separate cold refrigerant drain pipeline, also insulated, to the reservoir at the bottom.

As heat enters the pipeline from the heat engine, liquid refrigerant evaporates from metal fins near the surface of the liquid and bubbles up. The vapor will flow up the pipeline because vapor condenses on the cold heat sink at the top, forming a partial vacuum there, and more vapor moves up to fill the vacuum. The cold heat sink is arranged so the condensed liquid flows away from the vapor pipeline to the cold refrigerant drain pipeline, which carries a stream of rapidly flowing cold liquid down to the heat engine.

The vapor pipeline will stay no colder than the cold side of the heat engine, and the liquid drain pipeline will stay no warmer than the cold heat sink at the mountain top. For, the insulated metal walls of the vapor pipeline cannot long remain colder than the vapor, because whenever they are colder, the vapor will condense on the walls, heating them. Nor can the insulated metal walls of the liquid drain pipeline stay warmer than the descending liquid, because additional cold liquid is always flowing by, carrying heat away. If the walls are hot enough, bubbles form in the liquid and carry even more heat down, swiftly cooling the walls, which then have no way to heat up again.

The walls of the vapor pipeline can stay hotter than the cold side of the heat engine, because there is no refrigerant liquid there to evaporate and carry much heat away, and the walls of the liquid drain pipeline can stay colder than the cold heat sink, because there is no vapor to condense and warm those walls.

Insulation is needed on the vapor pipeline only to ensure that the vapor gets to the top before condensing, and hence, that the liquid flows down the separate drain pipeline instead of back down the vapor pipeline. Insulation on the liquid drain pipeline is more important, because it prevents heat flow into the cold liquid refrigerant from places other than the cold side of the heat engine.

Gravitational effects
The minimum temperature difference between the cold heat sink at the top of the mountain and the evaporating liquid at the bottom is influenced by the height of the mountain. The pressure of the vapor is higher at the bottom than the top because of gravitational effects, so the temperature at which the liquid will evaporate is a little higher at the bottom than the temperature at which it condenses at the top.

The barometric equation gives the pressure $$P_h$$ at height $$h$$ above the level where the pressure is $$P_0$$:
 * $$P_h = P_0 \exp(-Mgh/RT)$$, where

$$M$$ is the molecular weight of the gas, $$g$$ is the acceleration of gravity, $$R$$ is the gas constant, and $$T$$ is the absolute temperature.

Letting $$P_0 = 1$$ atm, $$M = \mathrm{m.w.}(C_3H_8)$$, $$h = 2$$ km, $$T = 270$$ K, we get a pressure of 0.68 atm.

1 atm exp(-((3 12 + 8)g/mol) gravity 2km/R 270K) = 0.68084605 atm

The vapor pressure of propane is 4.75 bar at 0°C. Suppose the mountain top is at the freezing point of water. We want to know the pressure at the bottom.

4.75 bar exp(-((3 12 + 8)g/mol) gravity (-2km)/R 273.15K) = 6.9457538 bar

The vapor pressure of propane at 10°C is 6.37 bar and at 15°C is 7.32 bar. Linear interpolation gives 13.3°C as the temperature with vapor pressure 6.9457 bar

tempC(10 + (15 + -10)(6.9457538 + -6.37)/(7.32 + -6.37)) = 286.18028K (10 degC + (15 degC + -10 degC)(6.9457538 bar + -6.37 bar)/(7.32 bar + -6.37 bar)) = 13.030283 degC

Consider the equilibrium case, where no more heat is being added at the bottom and there is no gas flow going on. The pressure at the top should equal the vapor pressure of propane at the cold sink temperature, i.e., 4.75 bar. Then the barometric equation gives the pressure 2km below that as 6.95 bar, if the temperature were 273K.

That means that liquid propane at the bottom will not start boiling until it reaches 13°C, and the gas which boils off will be at 13°C. That violates the assumption of constant temperature in the barometric equation.

The barometric equation for non-constant temperature uses concentrations instead of pressures. Concentration is $$C = n/V$$, and from $$PV = nRT$$ we get $$C = n/V = P/RT$$.

(4.75 bar / R tempC(0)) exp(-((3 12 + 8)g/mol) gravity (-2km)/R tempC(0)) = 305.83241 mol/m^3

Now we need to find T such that the concentration of propane vapor above boiling liquid is 305.8 mol/m^3.

We simply convert the vapor pressures into concentrations by dividing by RT, and interpolate.

(15 degC + (20 + -15)degC(305.83241 + -305.53276)/(343.40047 + -305.53276)) = 15.039565 degC

In this case, interpolation wasn't necessary. Using concentrations instead of assuming constant temperature changes the answer by 2°C. If the top of the 2km mountain is at 0°C, and the liquid at the bottom is just boiling, then the liquid and gas above it are at 15°C.

Mass of liquid flowing down the mountain
If large amounts of heat are flowing, there will also be a considerable amount of gravitational potential energy available from the flow of liquid propane in the return pipeline. From the table below, we see that propane boiling at 15°C absorbs 359.9 kJ/kg.

The work available in allowing a given mass of liquid to descend a given distance is $$mgh$$, where $$m$$ is mass, $$g$$ is the acceleration of gravity, and $$h$$ is the height, the vertical component of the distance. One kg descending 2 km produces 19.6 kJ:

1 kg gravity 2km = 19.6133 kJ

The work per unit mass is just $$gh$$, shown for various heights:

gravity 1km = 9.80665 kJ/kg gravity 2km = 19.6133 kJ/kg gravity 3km = 29.41995 kJ/kg gravity 4km = 39.2266 kJ/kg

Note that the taller (vertical component of length) the heat pipeline, the more work available. Of course, a taller pipeline also has a bigger temperature drop due to the barometric equation.

Starting with 0 degC as the temperature at the top of the pipeline, 44 g/mol as the molecular weight of propane, 209.1 mol/m^3 as the concentration of vapor above liquid propane at 0 degC, and substituting into the barometric equation, we calculate the gas concentration at given distance below the top of the pipeline. From this we find the liquid temperature which would give this vapor concentration by interpolating using the tables below.

Top temp 0 degC height 1km requires 7.3 degC at the bottom to start condensing at top: (209.14993 mol/m^3) exp(-(44 g/mol) gravity (-1 km) / R tempC(0)) = 252.91268 mol/m^3 5degC + (10degC + -5degC)(252.91268 + -238.25272)/(270.57533 + -238.25272) = 7.2677562 degC

Top temp 0 degC height 2km requires 15.0 degC at the bottom to start condensing at top: (209.14993 mol/m^3) exp(-(44 g/mol) gravity (-2 km) / R tempC(0)) = 305.8324 mol/m^3 15degC + (20degC + -15degC)(305.8324 + -305.53276)/(343.40047 + -305.53276) = 15.039564 degC

Top temp 0 degC height 3km requires 23.3 degC at the bottom to start condensing at top: (209.14993 mol/m^3) exp(-(44 g/mol) gravity (-3 km) / R tempC(0)) = 369.8251 mol/m^3 20degC + (25degC + -20degC)(369.8251 + -343.40047)/(384.03204 + -343.40047) = 23.251736 degC

Top temp 0 degC height 4km requires 33.0 degC at the bottom to start condensing at top: (209.14993 mol/m^3) exp(-(44 g/mol) gravity (-4 km) / R tempC(0)) = 447.2077 mol/m^3 30degC + (35degC + -30degC)(447.2077 + -475.39044)/(428.0842 + -475.39044) = 32.978755 degC

A heat pipeline 2 km tall carrying 1 MW heat from 15.0 degC at the bottom to 0 degC at the top will deliver 54.5 kW of gravitational power to a turbine in the liquid return pipe. (19.61 kJ/kg) / (359.9 kJ/kg) = 54.487358 kW/MW

How big is such a pipeline? Propane density is 528 kg/m^3 at 15°C 54.5 kW / (19.61 kJ/kg) = 2.7791943 kg/s 1 MW / (359.9 kJ/kg) = 2.7785496 kg/s 2.779 kg/s / (528 kg/m^3) = 0.0052632576 m^3/s 2.779 kg/s / (528 kg/m^3) = 5.2632576 l/s

That is, a heat pipeline 2 km tall delivers 10.36 kW gravitational power to the turbine for every liter per second liquid flow. 54.5 kW / (5.263 l/s) = 10.355311 kW/(l/s)

Calculations regarding power vs. size and insulation efficiency
To be able to guess at costs, we need to know how much materials are needed to build a system of given power.

We know the flow rate. What pipe diameter is required for

Some properties of propane
Gas concentration at given temperature of vapor above excess liquid: (0.70 bar / R tempC(-50)) = 37.728234 mol/m^3 (0.89 bar / R tempC(-45)) = 46.9175 mol/m^3 (1.11 bar / R tempC(-40)) = 57.260203 mol/m^3 (1.37 bar / R tempC(-35)) = 69.188722 mol/m^3 (1.68 bar / R tempC(-30)) = 83.099868 mol/m^3 (2.03 bar / R tempC(-25)) = 98.389122 mol/m^3 (2.45 bar / R tempC(-20)) = 116.40013 mol/m^3 (2.92 bar / R tempC(-15)) = 136.04295 mol/m^3 (3.45 bar / R tempC(-10)) = 157.68161 mol/m^3 (4.06 bar / R tempC(-5)) = 182.10151 mol/m^3 (4.75 bar / R tempC(0)) = 209.14993 mol/m^3 (5.51 bar / R tempC(5)) = 238.25272 mol/m^3 (6.37 bar / R tempC(10)) = 270.57533 mol/m^3 (7.32 bar / R tempC(15)) = 305.53276 mol/m^3 (8.37 bar / R tempC(20)) = 343.40047 mol/m^3 (9.52 bar / R tempC(25)) = 384.03204 mol/m^3 (10.79 bar / R tempC(30)) = 428.0842 mol/m^3 (12.18 bar / R tempC(35)) = 475.39044 mol/m^3 (13.70 bar / R tempC(40)) = 526.17896 mol/m^3 (15.34 bar / R tempC(45)) = 579.90754 mol/m^3 (17.13 bar / R tempC(50)) = 637.55628 mol/m^3

Vapor pressures etc.: http://www.care-refrigerants.co.uk/images/Therm_CARE40.gif