# HVAC - Reheat Load?



## navyasw02

Can someone explain the best way to calculate reheat load? I was trying to do one of the Lindeburg practice problems and I found the two points on the psych chart, found the two values of enthalpy and multiplied it by the mass flow rate in lbm/hr to get the answer. The solution uses q=mcp(T2-T1). My answer was way high and I wouldve selected the wrong value, so maybe I just didnt have my psych chart microscope turned up high enough again. Theoretically, the two methods should give the same solution.


----------



## jmbeck

navyasw02 said:


> Can someone explain the best way to calculate reheat load? I was trying to do one of the Lindeburg practice problems and I found the two points on the psych chart, found the two values of enthalpy and multiplied it by the mass flow rate in lbm/hr to get the answer. The solution uses q=mcp(T2-T1). My answer was way high and I wouldve selected the wrong value, so maybe I just didnt have my psych chart microscope turned up high enough again. Theoretically, the two methods should give the same solution.


The reason the solution used mdot*Cp*dT is due to the fact that reheat is typically sensible heat only. At standard conditions (e.g. 0.075 lbm/ft^3, 0.24 btu/lbm*F), this reduces to the rule of thumb 1.08 * CFM * delta T. (0.075 lbm/ft^3 * 0.24 btu/lbm*F * 60 min/hr).

What values were given in the problem? If you were given the two temperatures and volumetric flow, and sensible heating only (i.e. no humidification), it's a 10 second problem.


----------



## navyasw02

jmbeck said:


> navyasw02 said:
> 
> 
> 
> Can someone explain the best way to calculate reheat load? I was trying to do one of the Lindeburg practice problems and I found the two points on the psych chart, found the two values of enthalpy and multiplied it by the mass flow rate in lbm/hr to get the answer. The solution uses q=mcp(T2-T1). My answer was way high and I wouldve selected the wrong value, so maybe I just didnt have my psych chart microscope turned up high enough again. Theoretically, the two methods should give the same solution.
> 
> 
> 
> The reason the solution used mdot*Cp*dT is due to the fact that reheat is typically sensible heat only. At standard conditions (e.g. 0.075 lbm/ft^3, 0.24 btu/lbm*F), this reduces to the rule of thumb 1.08 * CFM * delta T. (0.075 lbm/ft^3 * 0.24 btu/lbm*F * 60 min/hr).
> 
> What values were given in the problem? If you were given the two temperatures and volumetric flow, and sensible heating only (i.e. no humidification), it's a 10 second problem.
Click to expand...

The values given were 8000ft3/min supply air delivered to a space at 60db/51wb. condition of the air leaving the cooling coil is 53db 52wb. Should've been pretty easy, but I spent an hour trying to figure out what I did wrong.

Any other useful thumbrules and/or problem solving methods and tips? I know absolutely jack about HVAC other than where to find a few equations in the MERM.


----------



## jmbeck

Wow, that is kind of a crappy question. I don't remember anything that tricky on the test.

With that process, you are heating and dehumidifying. It can be done in one process by a dessicant heat wheel, but I've not seen one used for reheat.

Typically, reheat is used to allow the coil to dehumidify the air, and sensible heat is added so as not to change the temperature of the room. Also, it can be used when multiple spaces with different cooling profiles are on one system, for better control.

In this process, you're going from 53F DB / 52F WB / 21.43 (btu/lbm) / 56.2015 grains/lbm

to

60F DB / 51F WB / 20.821 (btu/lbm) / 41.3232 grains/lbm

As you can see, you've raised temperature (53F to 61F), while removing moisture (56.2015 grains to 41.3232 grains).

So, let's use your equation. Qt = mdot * (delta h)

At standard conditions, the specific volume of air is 13.333 ft^3/lbm. So, mdot = [(60 min/hr) / 13.333 ft^3/lbm] * Vf = 4.5 (lbm*min/ ft^3 * hr) * Vf (in CFM)

Therefore, total heat load is approximated as *4.5 * CFM * (delta h). *(there's one of those to remember!)

From the above values, 4.5 (lbm*min / ft^3 * hr) * 8000 (ft^3/min) * (21.43 - 20.821) btu/lbm = 21,924 btu/hr

The sensible component is calculated by this equation. Qs = mdot * Cp * (delta T)

We know that mdot = 4.5 (lbm*min/ ft^3 * hr) * Vf (in CFM). For standard conditions, Cp for air is .240 btu/lbm * F.

Combining these, we get 1.08 (btu * min / ft^3 * F * hr) * Vf (ft^3/min) * delta T (F), or *1.08 * CFM * (delta T).*

From the above, 1.08 (btu * min / ft^3 * F * hr) * 8000 (ft^3/min) * (60F - 53F) = 60,480 btu/hr

For the latent component, Ql = mdot * specific h * (delta W).

At standard conditions, specific enthalpy for air is assumed at 1075.5 btu/lbm (or so).

So, 4.5 (lbm*min/ ft^3 * hr) * Vf (in CFM) * 1075.5 btu/lbm = 4840 (btu * min / ft^3 * hr) * Vf (ft^3 / min) * delta W (lbmwater/lbmair)

or, *4840 * CFM * (delta W)*

4840 (btu * min / ft^3 * hr) * 8000 (ft^3/min) * [( 56.2015 - 41.3232 ) grains/lbma * 1 lbmw/7000 grains] = 82,298 btu/hr

Qt = Qs + Ql

Qt = 21,924 btu/hr

Qs = 60,480 btu/hr

Ql = 82,298 btu/hr

Hmmm.....

Here is where you have to think about the process. You're raising temperature (adding heat) and lowering moisture content (removing heat, or energy, if you see it better that way).

With that, (60,480 btu/hr) + (-82,298 btu/hr) = -21,818 btu/hr, which is approximately what we found for Qt.

If the question didn't ask what was the sensible heat load, then it's just something you have to know. Typically, heating is a sensible process only.

A better question would have been 53/52 heated to 60/55. That is a sensible process only.


----------



## navyasw02

Wow thanks for the great response. This is a Lindeburg practice problem so hopefully, the real deal wont be this complicated. Only 3 more weeks and I can brain dump all this stuff again.


----------

