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## Q. 14.2

Aheat exchanger is considered to recover heat from the exhaust air of a 5,000-cfm laboratory make-up air system located in Denver, Colorado. The indoor temperature within the laboratory is kept at $T_{in}$ = 70°F. The exhaust air temperature is $T_{ea}$ = 120°F.
(a) Estimate the amount of outdoor air that needs to by-pass the heat recovery system under winter design conditions ($T_{oa}$ =-10°F) to ensure that the supply air temperature is equal to the indoor temperature (i.e., $T_{sa} = T_{in}$ ).
(b) Determine the simple payback period if the installation cost of the heat recovery system is $1/cfm. For Denver, the average winter season outdoor temperature is $T_{oa}$ =42.4°F and the cost of gas is$0.10/CCF. The heat recovery system is operated 24 hours/day, 271 days/year (winter season). The heat content of gas is 840 Btu/$ft^{3}$ (Denver is located at an altitude of about 5,280 ft).

## Verified Solution

The by-pass factor f can be determined using the fact that the supply temperature provided to the laboratory is the result of mixing two air streams as depicted in the diagram below:

• Outdoor air (representing a fraction f of the total air supply and kept at the temperature $T_{oa}$ = –10°F)

• Heated air [coming from the heat recovery system at the temperature $T_{ha}$ (to be determined), and representing a fraction (1 – f) of the total supply air]

First, the heated air temperature is calculated using the definition of the heat recovery system effectiveness provided by Eq. (14.3):

$ε=\frac {(T_{2}-T_{1})}{(T_{4}-T_{1})}$        (14.3)

$T_{ha} =T_{oa}+ε.( T_{ea}- T_{oa})= −10 °F+ 0.70 * (120°F+10°F ) = 81°F$

Then the fraction f can be determined by setting $T_{sa} = T_{in}$:

$T_{in} = T_{oa} . f +T_{ha} .(1− f )$

Thus the fraction f is:

$f=\frac {(T_{in}-T_{ha})}{(T_{oa}-T_{ha})}=\frac {70-81}{-10-81}=0.12$

Using the average winter conditions, the energy rate saved by the heat recovery system can be estimated as following:

$ΔE =\dot{m}_{a} .c_{p} .(T_{sa} −\overline{\bar{T}}_{oa} )$

where $T_{sa} T_{in}= 70°F$ (assuming that the waste energy is recovered only to temper the make-up air; more thermal energy can actually be recovered if it can be used for other purposes such as space heating) and $\overline{\bar{T}}_{oa}= 42.4°F$. Thus:

$ΔE = (5,000cfm)*(0.91Btu/hr .°F.cfm )*(70°F − 42.4°F) =124,200 Btu/hr$

The savings in fuel use ΔFU can then be calculated using a gas boiler efficiency of 80 percent and a number of operating hours of $N_{h}$ = 6,504 hrs/yr (=24 hrs/day * 271 days/yr):

$ΔFU =\frac {ΔE.N_{h}}{η_{b}}=\frac{(124,200 Btu/hr)*( 6,504 hrs/yr)]}{0.80}=1.007*10^{6}Btu/yr=21,021CCF/yr$

Thus, the simple payback period for installing the waste heat recovery system is:

$SP=\frac{Initial— Cost}{Annual— Savings}=\frac{\1cfm*5,000 cfm }{\0.1CCF*21,021CCF/yr}=2.1 years$

Therefore, the installation of the heat recovery system in the laboratory is cost-effective.