We are determined to provide the latest solutions related to all subjects FREE of charge!

Please sign up to our reward program to support us in return and take advantage of the incredible listed offers.

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program


Advertise your business, and reach millions of students around the world.


All the data tables that you may search for.


For Arabic Users, find a teacher/tutor in your City or country in the Middle East.


Find the Source, Textbook, Solution Manual that you are looking for in 1 click.


Need Help? We got you covered.

Chapter 14

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.