MHRV Heat Exchanger Area

An Eccentric Anomaly: Ed Davies's Blog

In my recent MHRV Ventilation Requirements post I established that I need the heat exchanger in the ventilation to recover about 1200 watts of sensible and latent heat from the outgoing airflow to warm the cruel cold air coming in from the outside.

Next it's worth working out the area of interface needed between the outgoing and incoming airflows to achieve this. This might seem strange without having first decided on the design of and materials in the exchanger but actually, as I'll show, the actual materials used don't make a lot of difference.

Temperature Difference

The first thing to do is to establish the nominal temperature difference across the exchanger. In a previous post I showed that the temperature difference (in somewhat idealized circumstances) is determined by:

ΔT = (1 - e) (Ti - To)

My 1200 W figure is with a temperature difference of 20 °C. Aiming for 90% efficiency gives a temperature difference of (1 - 0.9)(20 - 0) = 2 °C so the required conductance is 1200 W / 2 °C = 600 W/K.

Hopefully in practice actual flow rates will be somewhat less than the design flow most of the time. In that case the recovered power is lower, the temperature difference will be smaller and so the overall efficiency has the opportunity to be a little higher.

However, there's a fundamental limit to the actual efficiency which can be achieved. The incoming air (denoted Ts) can't be warmed to a higher temperature than the outgoing internal air (Ti) yet it will typically be drier and so contain less energy. However good the heat exchanger is it will likely not achieve 100% energy recovery.

Thermal Resistance

The thermal resistance of the heat exchanger consists of two components: the thermal resistance of the exchanger material itself (eg, polycarbonate) and the boundary layer resistances of the two interfaces on each side.

The Professional Plastics' document Thermal Properties of Plastic Materials gives the thermal conductivity of polycarbonate as 0.19 to 0.22 W/(m·K) at 23 °C. Wikipedia repeats those numbers, likely from the same source I'd guess. The Engineering ToolBox says 0.19 W/(m·K) at 25 °C. Let's use 0.20 W/(m·K).

Assuming the polycarbonate sheet walls are 1 mm thick then the resistance will be 0.001 m / 0.2 W·m⁻¹·K⁻¹ = 0.005 m²·K/W.

The thermal resistances of the air/polycarbonate are a bit harder to assess accurately which is a pity as they're more significant.

BS EN ISO 6946:1997, apparently, gives various resistances (in m²·K/W)) depending on whether the surface is inside or outside (ie, just convection or wind affected as well) and which direction the heat is flowing (which affects convection):

Inside 0.10 0.13 0.17
Outside 0.04 0.04 0.04
Underfloor - 0.13 0.17

BRE Digest 108, also apparently, gives similar but different resistances:

Inside 0.10 0.12 0.14
Outside 0.04 0.06 0.04
Underfloor - - 0.22

Even the lowest value of 0.04 m²·K/W doubled for the two interfaces is 16 times the resistance of the polycarbonate. This is why I don't think the choice of material is very significant.

I think it's reasonable to base an exchanger design on outdoor resistances as airflows will likely be at a noticeable number of metres per second with significant turbulence. Still, it's more conservative to choose a higher resistance as that will require a larger exchanger area. I'm going to plunk for the larger of the two outside horizontal resistances: 0.06 m²·K/W giving a total resistance of 0.06 + 0.005 + 0.06 = 0.125 m²·K/W.

This calculation is done on the assumption of 1 mm of polycarbonate between the airflows. However, even if a much more conductive and potentially thinner material (eg, copper) was selected it wouldn't make much difference. In the limit a heat superconductor (with zero thermal resistance) would only reduce the resistance from 0.125 m²·K/W to 0.12 m²·K/W - the thermal resistance of the boundary layers dominates for any sensible choice of materials.

This applies even if we pick the lower British Standard boundary layer resistance of 0.04 m²·K/W where polycarbonate and super conductor exchangers would have resistances of 0.085 and 0.08 m²·K/W respectively, which is less than 7% difference.

Area Calculation

Multiplying the required conductivity (600 W/K) by the thermal resistance (0.125 m²·K/W) gives a required area of 75 m².

Using twinwall polycarbonate for this is not out of the question. As both sides could be used that'd be less than 40 m² with prices around £12/m² for 6 mm so a total of £480. Still, it's enough to be well worth some thought as to cheaper options.