heating water

C ratio
= C min
C max
= 2499
2705 = 0.924
Figure 4-17
manufacturer's data
NTU 6000= UA (
= 150 )20
C min
2499 = 1.2
5000
⎡ ∆T
q 2
= q 2
⎤ ⎡140 − 50 ⎤
1 ⎢ ⎥ = 3,261
⎣ ∆T
⎢
1 ⎦ ⎣140 − 65
⎥
⎦
= 3,913
Btu
hr
Although principle #1 is helpful for quick performance
estimates, it should not replace the use of thermal rating
data supplied by the manufacturer when it is available.
Heat output (Btu/hr)
1− e ( − NTU )(1−C ratio )
Principle #2: Increasing the fluid flow rate through
4000
the coil will marginally increase the heating capacity
1− (C ratio
)e ( = 1− e ( −1.2)(1−0.924)
0.0872
− NTU )(1−C ratio ) (
=
1− (0.924)e −1.2 )(1−0.924)
0.1565 = 0.557
of a fan-coil or air handler.
3000
Some heating professionals “instinctively” disagree with
this statement. They reason that because the water moves
2000
through the coil at a faster speed, it has less time during
ε =
q = ε C min
θ =
1000
0
0 20 40 60 80 100 120 140 160
∆T Entering water temp. - air temp. (ºF)
LMTD
( )( T hotin
− T coldin ) = 0.557( 2499) ( 150 − 60) = 125,280 Btu
hr
Figure 4-18
constant air
flow rate
For example, assume a fan-coil has a known output of 5,000
Btu/hr when supplied with 180ºF water and operating with
room air entering ⎡ ∆T at 65ºF. The designer wants to estimate
the qfan-coil’s 2
= q 2
⎤
1 ⎢ output ⎥ when supplied with 140ºF water, with
all other conditions ⎣ ∆T 2 ⎦ (e.g., room air temperature, water flow
rate and air flow rate) remaining the same. Formula 4-16
yields the estimated output:
constant
inlet air
temperature
constant
inlet water
temperature
⎡ ∆T
q 2
= q 2
⎤ ⎡140 − 65⎤
1 ⎢ ⎥ = 5,000
⎣ ∆T
⎢
1 ⎦ ⎣180 − 65
⎥
⎦
= 3,261
Btu
hr
A proportional relationship between any two quantities will
result in a straight line that passes through or very close
to the origin when the data is plotted on a graph. Figure
4-17 shows this to be true when the heat output data for a
small fan-coil is plotted against the difference between the
entering water and entering air temperatures.
This relationship can also be used to estimate the heat
output from a fan-coil or air handler when the entering
air temperature is above or below normal comfort
temperatures. For example, if a fan-coil can deliver 3,261
Btu/hr with an entering water temperature of 140°F and
entering air temperature of 65°F, its output while heating
a garage maintained at 50°F could be estimated using the
same formula:
Heat output (Btu/hr)
4500
4000
10
9
3500
8
3000
heat output 7
2500
6
2000
∆T
5
4
1500
3
1000
2
500
1
0
0
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
water flow rate (gpm)
∆T (ºF)
48