*From Gene Wengert, forum technical advisor:*

Heat transfer in lumber drying: There are three equations that express what is happening when wood dries in a conventional drying operation.

Equation 1: Heat Transfer

Heat energy is transferred from the air to the wood surface in the boundary layer. The differential equation for this is

dQ = h (Ta - Ts) dA

where

dQ = amount of energy transferred at location dA

h = convective heat transfer coefficient

Ta = air temperature

Ts = wood surface temperature (= wet-bulb temperature if wood is soaking wet)

dA = small differential area being studied

Equation 2: Evaporation

The energy that is transferred to the wood is used for heating the wood and for evaporation. Except during initial heating, very little energy is used for heating. It is all used for evaporation. Therefore, if we can determine the amount of water that is lost from a certain time period, we can figure out how much energy was used to evaporate this amount of water.

dQ = w (dV) H [d(MC) / dt]

where

w = density of wood

dV = small differential volume of wood being studied

H = latent heat of vaporization

MC = moisture content %

t = time

[d(MC) / dt] = drying rate

Equation 3: Air Cooling

As the air gives up its heat energy, it will cool. The amount of cooling is determined by using the specific heat, which is BTUs required to change the temperature of a given volume by 1 F. Therefore,

dQ = a Cp dR dT

where

a = density of air

Cp = specific heat of air

dR = differential volume rate of flow of air

dR = (vel = velocity) x (th = sticker thickness / 2) x (z = the perpendicular air flow spacing, such as sticker spacing)

Combinations

The dQ in all equations are equal (First Law of Thermodynamics). You can combine equations two and three and then integrate the equation across the width of a load (L) of lumber and get the drop in temperature from one side of the load to the other (TDAL = temperature drop across the load) being proportional to the average drying rate d(MC) / dt. Th = lumber thickness

TDAL = {[(w) (th) (L) (H)] / [(a) (Cp) (vel) (st)]} [average d(MC) / dt]

= (coefficient) (average drying rate)

Equations one and three can be combined in a similar manner to give

TDAL = {[(h) (L)] / (a) (Cp) (vel) (st/2)} (average Ta - Ts)

= (coefficient) (average Ta - Ts)

These last two equations can be combined to give the average drying rate.

Forum Responses

(Commercial Kiln Drying Forum)
*From contributor P:*

Does the temperature of the water vapor go up at constant pressure? Why can't the water vapor transfer heat energy to the wood surface in the boundary layer?

The term "air" means moist air, so the small amount of water vapor in the air is not treated separately. As you reach 212 F, then you must begin to use steam tables. Water vapor will obey the gas laws; that is, PV = nRT.

The boundary layer contains both water (vapor) and air molecules. If the boundary layer on the surface of the lumber is at the wet bulb temperature then both the dry air and water vapor molecules are at the wet bulb temperature. The air and water (vapor) molecules away from the boundary layer are at the dry bulb temperature. There is a temperature gradient (DB-WB) for both the dry air and the water (vapor) molecules.

What about influence of air speed when heating lumber?

Air speed simply strips away some of the boundary layers.

If air speed around wood is very low then you have a "simple" conductive heat transfer form air to wood. If air speeds up then you can heat wood faster because new hot air is always around. So, at what air speed (usually) improving air speed you do not gain important advantages?

What I mean is there are boundary layers of air and water (vapor) around the wood. Heat must conduct through this layer to the wood. With higher velocity the layers are stripped away. The faster the air the more layers get stripped away. With fewer boundary layers the heat takes less time to conduct to the wood surface. Also, the heat from the air loses less heat before it gets to the wood - therefore heating it quicker.

I am using the word air to mean the nitrogen, oxygen and all the other molecules in dry air.

The term boundary layer is defined as that region where the velocity is essentially zero at the surface to where the velocity is 99% of the free stream velocity. Consequently, at the top of the boundary layer, the temperature is essentially the free stream temperature and at the bottom of the boundary layer the temperature is (for wet wood) the wet-bulb temperature.

Contributor P - you have divided the boundary layer into sub layers. We do not do that. It is treated as one entity and that allows us to use the first equation. When these equations use the term air that includes nitrogen, oxygen, water vapor and etc.

Contributor W - the effect of air flow is included with the term "vel". Actually, we are concerned about both the volume of air and also its velocity. The sticker thickness is included to accommodate the volume. As the air speed increases, experiments will show that convective heat transfer coefficient also increases; velocity is included in that term as well.

Now, if you understand the initial discussion, then what happens to these equations when the process is limited or controlled by the rate with which moisture can move to the surface? (In hardwoods this is likely after about 10% MC loss from the lumber.) This means that heat flow (dQ) is not controlling. So, the piece of lumber will not need as much heat or energy, because there is not so much drying going on. The lumber (T抯) will therefore be at a temperature above the WB temperature. In fact, near the end of drying T抯 will be very close to the DB temperature.

The boundary layer contains air molecules. If the bottom of the boundary layer is at the wet bulb temperature and the top of the boundary layer is at the dry bulb temperature then there is a temperature gradient (DB-WB). Since the term air contains nitrogen, oxygen, water vapor, etc. this proves that some of the energy in the water (vapor) molecules heats the lumber correct?

As the TDAL indicates, the air is cooler when it exits a pile of lumber. The energy loss in this air includes any and all molecules in the air. Because there are so few water vapor molecules in air at 130 F or so, the contribution of the water vapor molecules is not overwhelming, compared to 212 F.

I think I know what you mean. The humidity ratio is small at lower temperatures. For example at 200F dry bulb and a 180F wet bulb the humidity ratio is 641g/1000g. Since there is only 641g of water vapor to every 1000g dry air then the dry air you think is transferring the most heat to the lumber?

Does the flow rate of water out of the lumber increase with hotter lumber?

The flow rate does increase with hotter lumber. Do you think that 1000 g of water has 641 g of water? If so, that is not correct. The relative humidity is a ratio of the amount of water in air compared to the maximum amount of water possible at that temperature.

The humidity ratio is a lot different than the relative humidity ratio. The humidity ratio is the weight of water vapor/weight of dry air - 641g of water vapor/1000g of dry air.

The humidity ratio is sometimes called the moisture content of the air. Do you not agree air at 200(F) and a wet bulb of 180(F) has a moisture content of 641g/1000g?

At 200 F, the specific humidity is 2.33 kg/kg. At 200 F and 180 F DB and WB, the RH is 64%, so the specific humidity is therefore about 1.5. Note that your value of humidity of 641/1000 = 64%, so I believe you have somehow calculated RH and not specific humidity (weight of water vapor / weight of air). Note that my three equations are correct within the limits stated. Equation 1 is basic heat transfer; equation 2 is evaporation of water, although it ignores the heat of wetting unless you incorporate it into "H"; and equation 3 is again basic heat transfer. Check with your teacher. The key is to use the correct coefficients, as stated.

I see how you thought I calculated the RH and not the specific humidity. At 200(F) and 180(F) DB and WB, the RH is 64%, but the software I was using calculated the specific humidity of 0.641. The software I have must be wrong. What I don't understand is if the specific humidity is 1.5 then thinking about heat transfer and boundary layers the energy from the water vapor must be transferring more heat than the nitrogen, oxygen, etc. It weighs more and the differential temperature exists between the air and the wood surface. I know you said (not in the exact words) the air contains water vapor and the nitrogen, oxygen, etc. Water vapor is not treated separately but if not then wouldn't the heat capacity of the air go up with more water vapor in the air since water vapor has a higher heat capacity?

Let抯 make this easier. Do you know of any websites or easy to find books on heat transfer to wood?

You are correct that the specific heat of water vapor is great than that of dry air. As you approach 212 F, with the increasing amount of vapor, the specific heat of the water vapor begins to dominate. You will note that I frequently mentioned that I was using 130 F, and other low temperatures, where the contribution of water vapor is small (at 90 F, specific humidity is 0.027 kg of vapor per kg of air). With respect to the equations I gave, you would use the specific heat of moist air at whatever temperature you are at and also use the appropriate density. This approach accommodates the effect of the heat carried by the moisture, even up to 212 F.

Heat transfer is heat transfer whether it is in metal, glass, plastic, wood, bamboo, or whatever. For mass transfer, a porous capillary solid is close to wood. All these are discussed in basic heat and mass transfer texts. One such text is Introduction To Heat Transfer by Frank P. Incropera, David P. DeWitt.

To make note again I know you referenced to 130(F) a few times. I also did not want you to think that I though your equations were wrong. If great amount of heat is carried by the moisture at temperatures close to 200(F) then higher humidity (assuming no saturation or condensation) would evaporate more water at a time.

The rate that water evaporates (or paint dries or any liquid evaporates) is a function of the difference between the vapor pressure or the air and the vapor pressure of the liquid. Instead of focusing on the energy level of the air, look up the vapor pressure of the air and then look up the vapor pressure of water at the surface temperature of the wood. Compare that at various conditions and you will start to get a better picture. Boundary layer effects and air flow all effect the rate of heat transfer and thus the surface temperature but the factor that controls the rate of evaporation of anything,(mercury, water, alcohol, paint, sweat and etc.) is vapor pressure deficit. I think you are looking at the wrong physical properties in trying to figure this out.

In a gas fired dry kiln. The charge of lumber top to bottom and end to end would have more even heat if the heat from the heat duct came out the low pressure side. Why are gas fired kilns not built with a heat duct switch, switching the heat to the low pressure side?

Most direct-fired kilns I have seen have the heat risers in the middle of the kiln as they are not on the high or low pressure side.

I have seen some with the gas heat duct on one side and perhaps performance would be better with switching but the cost would be high and maintenance would be more while the improvement would probably not be measureable or significant.

I have seen two types of gas fired kilns with a smoke bomb inside to see the air flow. Gas fired kilns with heat coming out high and low pressure sides and gas fired kilns with heat coming out the low pressure side only. The smoke on the low pressure side kilns is a lot more even from both end to end and top to bottom. This smoke test showed such a big difference. The quality must make a big difference.

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