MEC E 463 Fall 2019 Dr. Lexuan Zhong
Problem Set 3 - Solutions
4.3 Rinnai US has recently developed a new hydronic air handling unit (AHU Series 37AHA)
to heat air in residential buildings. It is expected that hot water from a tankless water heater
will be pumped through a finned-tube heating coil installed in the main branch (20 in. x 18
in.) of a ductwork system. A homeowner has calculated that 96,000 Btu/hr will be needed
to heat their home as desired. Air will enter the duct at approximately 70oF and should
leave at 150oF. Water will enter the coil system at 180oF. Design the hot water heating
coil (including the coil configuration, number of coils, size and performance of the heating
coil) for this application.
Possible Solution:
Objective
To determine the coil configuration, number of coils, size and performance of a heat exchanger to
heat air in a residential building.
Data Given or Known
i. 96,000 Btu/hr of energy is needed to heat the home.
ii. Air enters the duct at 70oF and leaves at 150oF; water enters the heating coil at 180oF.
iii. The size of the main duct branch is 20 in. x 18 in.
Assumptions/Limitations/Constraints
i. Let the air velocity over the finned-tubes be 500 fpm. This velocity is much lower than 1200
fpm (for low-velocity air ducts) to keep duct noise low since this heating coil will be used in
a home. In addition, lower air velocity will produce lower pressure losses across the heating
coil. This will result in a smaller blower motor and lower noise during operation.
ii. Let the flow velocity of water in the tubes be about 4 fps. This is acceptable for general
building service or potable water. In addition, this velocity does not exceed the erosion limits
of any general pipe material.
iii. The tubes are arranged in a staggered fashion. This will enhance heat transfer.
iv. Let the pipe material be Type K copper. Copper has high heat transfer properties and
availability.
1
,MEC E 463 Fall 2019 Dr. Lexuan Zhong
v. The 180o return bends (regular) will be soldered to make the tube connections. Soldering or
brazing is typically done for heating/cooling coils.
vi. Let the tube outer diameter be 0.402 in. This is on the order of a ⅜ in. nominal diameter,
which is a reasonable tube size for heating coils used for this application.
vii. The thickness of the tube is about 0.049 in., which is the thickness of ⅜-in. nominal diameter
Type K copper (Table A.6). Then the tube inner diameter is 0.304 in. (Or assume tube
inner diameter of 0.305 in using Type K ¼ in NPS as a reference)
viii. Negligible elevation head. Assume that all components are on the same level.
ix. This will be a counter-flow arrangement. The counter flow normally has a higher energy
transfer efficiency.
x. Entrance/exit losses of the air over the coils will be neglected. Other losses will be much
larger.
xi. Let the fin material be copper. This should enhance the heat transfer performance of the
system.
xii. A small pump in the tankless water heater package will be needed to circulate the hot water
through the finned-tube heating coil. So, use an inlet and outlet header attached directly to
each tube circuit to reduce the total losses and pump head.
Sketch
A sketch will be provided to show the tube flow circuitry after the design analysis.
Analysis
Determine the overall heat transfer coefficient (UFT)
The ε-NTU method will be used in this design. To determine the effectiveness (ε) and NTU, the
overall heat transfer coefficient (UFT) must be determined first. The overall heat transfer coefficient
for finned tubes is:
1 1 Rfi 1
= + + Rfo + .
U FT hi ( Ai / Ao ) ( Ai / Ao ) ho so
To find the heat transfer coefficient for flow in the tube, an appropriate equation must be chosen
for the Nusselt number. The Reynolds and Prandtl numbers will be needed.
2
, MEC E 463 Fall 2019 Dr. Lexuan Zhong
In general,
𝑐p,w 1𝐵𝑡𝑢/𝑙𝑏·℉ (𝑎𝑡 32−220℉)
= 0.24𝐵𝑡𝑢/𝑙𝑏·℉(𝑎𝑡 0−200℉) ≈ 4.
𝑐p,a
Therefore, if ∆Tair = (150 – 70)oF = 80oF, then ∆Twater ≈ 20oF. Assume Tw,o = (180 - 20)oF = 160oF.
The average water temperature is:
Tw =
(180 + 160)o F = 170o F .
2
For water at 170oF, the Reynolds number is:
Re Di = =
( )
V Di 60.79 lb/ft 3 (4 ft/s)(0.304 in ) 1 ft
x = 24,809 .
2.483 x 10-4 lb/ft-s 12 in
The Prandtl number is:
Pr = 2.90.
The flow in the tubes is fully turbulent and Pr is between 0.7 and 160. So, the Dittus-Boelter
equation is valid for the determination of the heat transfer coefficient in the tube. From the Dittus-
Boelter correlation equation,
hi Di
Nu = = 0.023Re 0D.8 Pr 0.3 .
k i
0.023k 0.8 0.3
hi = Re D Pr
Di i
0.023(0.386 Btu/hr-ft-R ) 12 in
hi = x x (24,809)0.8 (2.90)0.3
0.304 in 1 ft
hi = 1581Btu/hr-ft 2 - o F
The fouling resistance can be found in Table C.3.
Assume distilled water above 122oF: Rfi = 0.00114 hr-ft2-oF/Btu.
3
Problem Set 3 - Solutions
4.3 Rinnai US has recently developed a new hydronic air handling unit (AHU Series 37AHA)
to heat air in residential buildings. It is expected that hot water from a tankless water heater
will be pumped through a finned-tube heating coil installed in the main branch (20 in. x 18
in.) of a ductwork system. A homeowner has calculated that 96,000 Btu/hr will be needed
to heat their home as desired. Air will enter the duct at approximately 70oF and should
leave at 150oF. Water will enter the coil system at 180oF. Design the hot water heating
coil (including the coil configuration, number of coils, size and performance of the heating
coil) for this application.
Possible Solution:
Objective
To determine the coil configuration, number of coils, size and performance of a heat exchanger to
heat air in a residential building.
Data Given or Known
i. 96,000 Btu/hr of energy is needed to heat the home.
ii. Air enters the duct at 70oF and leaves at 150oF; water enters the heating coil at 180oF.
iii. The size of the main duct branch is 20 in. x 18 in.
Assumptions/Limitations/Constraints
i. Let the air velocity over the finned-tubes be 500 fpm. This velocity is much lower than 1200
fpm (for low-velocity air ducts) to keep duct noise low since this heating coil will be used in
a home. In addition, lower air velocity will produce lower pressure losses across the heating
coil. This will result in a smaller blower motor and lower noise during operation.
ii. Let the flow velocity of water in the tubes be about 4 fps. This is acceptable for general
building service or potable water. In addition, this velocity does not exceed the erosion limits
of any general pipe material.
iii. The tubes are arranged in a staggered fashion. This will enhance heat transfer.
iv. Let the pipe material be Type K copper. Copper has high heat transfer properties and
availability.
1
,MEC E 463 Fall 2019 Dr. Lexuan Zhong
v. The 180o return bends (regular) will be soldered to make the tube connections. Soldering or
brazing is typically done for heating/cooling coils.
vi. Let the tube outer diameter be 0.402 in. This is on the order of a ⅜ in. nominal diameter,
which is a reasonable tube size for heating coils used for this application.
vii. The thickness of the tube is about 0.049 in., which is the thickness of ⅜-in. nominal diameter
Type K copper (Table A.6). Then the tube inner diameter is 0.304 in. (Or assume tube
inner diameter of 0.305 in using Type K ¼ in NPS as a reference)
viii. Negligible elevation head. Assume that all components are on the same level.
ix. This will be a counter-flow arrangement. The counter flow normally has a higher energy
transfer efficiency.
x. Entrance/exit losses of the air over the coils will be neglected. Other losses will be much
larger.
xi. Let the fin material be copper. This should enhance the heat transfer performance of the
system.
xii. A small pump in the tankless water heater package will be needed to circulate the hot water
through the finned-tube heating coil. So, use an inlet and outlet header attached directly to
each tube circuit to reduce the total losses and pump head.
Sketch
A sketch will be provided to show the tube flow circuitry after the design analysis.
Analysis
Determine the overall heat transfer coefficient (UFT)
The ε-NTU method will be used in this design. To determine the effectiveness (ε) and NTU, the
overall heat transfer coefficient (UFT) must be determined first. The overall heat transfer coefficient
for finned tubes is:
1 1 Rfi 1
= + + Rfo + .
U FT hi ( Ai / Ao ) ( Ai / Ao ) ho so
To find the heat transfer coefficient for flow in the tube, an appropriate equation must be chosen
for the Nusselt number. The Reynolds and Prandtl numbers will be needed.
2
, MEC E 463 Fall 2019 Dr. Lexuan Zhong
In general,
𝑐p,w 1𝐵𝑡𝑢/𝑙𝑏·℉ (𝑎𝑡 32−220℉)
= 0.24𝐵𝑡𝑢/𝑙𝑏·℉(𝑎𝑡 0−200℉) ≈ 4.
𝑐p,a
Therefore, if ∆Tair = (150 – 70)oF = 80oF, then ∆Twater ≈ 20oF. Assume Tw,o = (180 - 20)oF = 160oF.
The average water temperature is:
Tw =
(180 + 160)o F = 170o F .
2
For water at 170oF, the Reynolds number is:
Re Di = =
( )
V Di 60.79 lb/ft 3 (4 ft/s)(0.304 in ) 1 ft
x = 24,809 .
2.483 x 10-4 lb/ft-s 12 in
The Prandtl number is:
Pr = 2.90.
The flow in the tubes is fully turbulent and Pr is between 0.7 and 160. So, the Dittus-Boelter
equation is valid for the determination of the heat transfer coefficient in the tube. From the Dittus-
Boelter correlation equation,
hi Di
Nu = = 0.023Re 0D.8 Pr 0.3 .
k i
0.023k 0.8 0.3
hi = Re D Pr
Di i
0.023(0.386 Btu/hr-ft-R ) 12 in
hi = x x (24,809)0.8 (2.90)0.3
0.304 in 1 ft
hi = 1581Btu/hr-ft 2 - o F
The fouling resistance can be found in Table C.3.
Assume distilled water above 122oF: Rfi = 0.00114 hr-ft2-oF/Btu.
3
