def Ft_aircooler(Thi, Tho, Tci, Tco, Ntp=1, rows=1):
r'''Calculates log-mean temperature difference correction factor for
a crossflow heat exchanger, as in an Air Cooler. Method presented in [1]_,
fit to other's nonexplicit work. Error is < 0.1%. Requires number of rows
and tube passes as well as stream temperatures.
.. math::
F_T = 1 - \sum_{i=1}^m \sum_{k=1}^n a_{i,k}(1-r_{1,m})^k\sin(2i\arctan R)
R = \frac{T_{hi} - T_{ho}}{T_{co}-T_{ci}}
r_{1,m} = \frac{\Delta T_{lm}}{T_{hi} - T_{ci}}
Parameters
----------
Thi : float
Temperature of hot fluid in [K]
Tho : float
Temperature of hot fluid out [K]
Tci : float
Temperature of cold fluid in [K]
Tco : float
Temperature of cold fluid out [K]
Ntp : int
Number of passes the tubeside fluid will flow through [-]
rows : int
Number of rows of tubes [-]
Returns
-------
Ft : float
Log-mean temperature difference correction factor [-]
Notes
-----
This equation assumes that the hot fluid is tubeside, as in the case of air
coolers. The model is not symmetric, so ensure to switch around the inputs
if using this function for other purposes.
This equation appears in [1]_. It has been verified.
For some cases, approximations are made to match coefficients with the
number of tube passes and rows provided.
16 coefficients are used for each case; 8 cases are considered:
* 1 row 1 pass
* 2 rows 1 pass
* 2 rows 2 passes
* 3 rows 1 pass
* 3 rows 3 passes
* 4 rows 1 pass
* 4 rows 2 passes
* 4 rows 4 passes
Examples
--------
>>> Ft_aircooler(Thi=125., Tho=45., Tci=25., Tco=95., Ntp=1, rows=4)
0.5505093604092708
References
----------
.. [1] Roetzel, W., and F. J. L. Nicole. "Mean Temperature Difference for
Heat Exchanger Design-A General Approximate Explicit Equation." Journal
of Heat Transfer 97, no. 1 (February 1, 1975): 5-8.
doi:10.1115/1.3450288
'''
dTlm = LMTD(Thi=Thi, Tho=Tho, Tci=Tci, Tco=Tco)
rlm = dTlm / (Thi - Tci)
R = (Thi - Tho) / (Tco - Tci)
# P = (Tco-Tci)/(Thi-Tci)
if Ntp == 1 and rows == 1:
coefs = _crossflow_1_row_1_pass
elif Ntp == 1 and rows == 2:
coefs = _crossflow_2_rows_1_pass
elif Ntp == 1 and rows == 3:
coefs = _crossflow_3_rows_1_pass
elif Ntp == 1 and rows == 4:
coefs = _crossflow_4_rows_1_pass
elif Ntp == 1 and rows > 4:
# A reasonable assumption
coefs = _crossflow_4_rows_1_pass
elif Ntp == 2 and rows == 2:
coefs = _crossflow_2_rows_2_pass
elif Ntp == 3 and rows == 3:
coefs = _crossflow_3_rows_3_pass
elif Ntp == 4 and rows == 4:
coefs = _crossflow_4_rows_4_pass
elif Ntp > 4 and rows > 4 and Ntp == rows:
# A reasonable assumption
coefs = _crossflow_4_rows_4_pass
elif Ntp == 2 and rows == 4:
coefs = _crossflow_4_rows_2_pass
else:
# A bad assumption, but hey, gotta pick something.
coefs = _crossflow_4_rows_2_pass
tot = 0
atanR = atan(R)
cmps = range(len(coefs))
for k in cmps:
x0 = (1. - rlm)**(k + 1.)
for i in cmps:
tot += coefs[k][i] * x0 * sin(2. * (i + 1.) * atanR)
return 1. - tot
def Ft_aircooler(Thi=None, Tho=None, Tci=None, Tco=None, Ntp=1, rows=1):
r'''Calculates log-mean temperature difference correction factor for
a crossflow heat exchanger, as in an Air Cooler. Method presented in [1]_,
fit to other's nonexplicit work. Error is < 0.1%. Requires number of rows
and tube passes as well as stream temperatures.
.. math::
F_T = 1 - \sum_{i=1}^m \sum_{k=1}^n a_{i,k}(1-r_{1,m})^k\sin(2i\arctan R)
R = \frac{T_{hi} - T_{ho}}{T_{co}-T_{ci}}
r_{1,m} = \frac{\Delta T_{lm}}{T_{hi} - T_{ci}}
Parameters
----------
Thi : float
Temperature of hot fluid in [K]
Tho : float
Temperature of hot fluid out [K]
Tci : float
Temperature of cold fluid in [K]
Tco : float
Temperature of cold fluid out [K]
Returns
-------
Ft : float
Log-mean temperature difference correction factor []
Notes
-----
This equation assumes that the hot fluid is tubeside, as in the case of air
coolers. The model is not symmetric, so ensure to switch around the inputs
if using this function for other purposes.
This equation appears in [1]_. It has been verified.
For some cases, approximations are made to match coefficients with the
number of tube passes and rows provided.
16 coefficients are used for each case; 8 cases are considered:
* 1 row 1 pass
* 2 rows 1 pass
* 2 rows 2 passes
* 3 rows 1 pass
* 3 rows 3 passes
* 4 rows 1 pass
* 4 rows 2 passes
* 4 rows 4 passes
Examples
--------
Example 1 as in HTFS manual.
Example 2 as in [1]_; author rounds to obtain a slightly different result.
>>> Ft_aircooler(Thi=93., Tho=52., Tci=35, Tco=54.59, Ntp=2, rows=4)
0.9570456123827129
>>> Ft_aircooler(Thi=125., Tho=45., Tci=25., Tco=95., Ntp=1, rows=4)
0.5505093604092708
References
----------
.. [1] Roetzel, W., and F. J. L. Nicole. "Mean Temperature Difference for
Heat Exchanger Design-A General Approximate Explicit Equation." Journal
of Heat Transfer 97, no. 1 (February 1, 1975): 5-8.
doi:10.1115/1.3450288
'''
dTlm = LMTD(Thi=Thi, Tho=Tho, Tci=Tci, Tco=Tco)
rlm = dTlm / (Thi - Tci)
R = (Thi - Tho) / (Tco - Tci)
# P = (Tco-Tci)/(Thi-Tci)
if Ntp == 1 and rows == 1:
coefs = _crossflow_1_row_1_pass
elif Ntp == 1 and rows == 2:
coefs = _crossflow_2_rows_1_pass
elif Ntp == 1 and rows == 3:
coefs = _crossflow_3_rows_1_pass
elif Ntp == 1 and rows == 4:
coefs = _crossflow_4_rows_1_pass
elif Ntp == 1 and rows > 4:
# A reasonable assumption
coefs = _crossflow_4_rows_1_pass
elif Ntp == 2 and rows == 2:
coefs = _crossflow_2_rows_2_pass
elif Ntp == 3 and rows == 3:
coefs = _crossflow_3_rows_3_pass
elif Ntp == 4 and rows == 4:
coefs = _crossflow_4_rows_4_pass
elif Ntp > 4 and rows > 4 and Ntp == rows:
# A reasonable assumption
coefs = _crossflow_4_rows_4_pass
elif Ntp == 2 and rows == 4:
coefs = _crossflow_4_rows_2_pass
else:
# A bad assumption, but hey, gotta pick something.
coefs = _crossflow_4_rows_2_pass
tot = 0
atanR = atan(R)
for k in range(len(coefs)):
for i in range(len(coefs)):
tot += coefs[k][i] * (1 - rlm)**(k + 1) * sin(2 * (i + 1) * atanR)
return 1 - tot
#print Ft_aircooler(Thi=66.82794, Tho=47.52647, Tci=15.4667, Tco=51.5889) # some example?
#print [Ft_aircooler(Thi=93., Tho=52., Tci=35, Tco=54.59, Ntp=2, rows=4)] # , 0.957045612383, works, _crossflow_4_rows_2_pass
#print [Ft_aircooler(Thi=125., Tho=45., Tci=25., Tco=95., Ntp=1, rows=4)] # , _crossflow_4_rows_1_pass, 0.550509360409, close enough
#print [[Ft_aircooler(Thi=125., Tho=80., Tci=25., Tco=95., Ntp=i, rows=j) for i in range(1,6)] for j in range(1, 6)] # , _crossflow_4_rows_1_pass, 0.550509360409, close enough