Air at 1 atmosphere and a temperature of 60 F enters 1/2-in. id tube with a velocity of 80 fps. The wall is maintained at a constant temperature of 210 F by condensing steam. Fin the convective heat-transfer coefficient for this situation if the tube is 5ft long. Find also the q

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Air at 1 atmosphere and a temperature of 60 F enters 1/2-in. id tube with a velocity of 80 fps. The wall is maintained at a constant temperature of 210 F by condensing steam. Fin the convective heat-transfer coefficient for this situation if the tube is 5ft long. Find also the q

EXAMPLE 3
Air at 1 atmosphere and a teraperature of 60°F enters al-in.-ID tube with a velocity of
80 fos. The wall is maintained at a constant temperature of 210°F by condensing steam.
ne convective heat-transfer coefficient for this situation if the tube jab it long.
As in example 2 it will be necessary to calculate the exit fluid temperature by
F
I
T-T,Lh
To-T,
+4
(19-61)
D pec
Transcribed Image Text:EXAMPLE 3 Air at 1 atmosphere and a teraperature of 60°F enters al-in.-ID tube with a velocity of 80 fos. The wall is maintained at a constant temperature of 210°F by condensing steam. ne convective heat-transfer coefficient for this situation if the tube jab it long. As in example 2 it will be necessary to calculate the exit fluid temperature by F I T-T,Lh To-T, +4 (19-61) D pec
Another expression to be satisfied is
q- hA, (T,-T)-pAuc, (T. - T)
Evaluating the Reynolds number at the tube entrance, we have
Pe Du_G/12] ft(80 fps)
0.159 x 10 fr/s
The flow is turbulent and Re is sufficiently high that equation (20-30), (20-31), or (20-32)
= 21 000
may be used.
Employing equation (20-31) and assuming an exit temperature of 190°F, giving a
mean bulk temperature of 125'F, and a film temperature of 167°F, we obtain
St-
-=0.023 Re0 Pr
pec,
[Git0(80 fps)(0.0764 lb/ft')] 02
1.45x 10 Ib./fts
=0.023(0.1416)(1.276) = 0.00416
=0.023
(0.694)-
Substitution into equation (19-61) yields
T-L- exp -4)(0.00416)
T-T,
-ехpt-1.99)-0.136
and
T-210-0.136(0) = 189.6°F
This agreement is excellent. The heat transfer coetficient is
(361 K)
h- pec, (St)
= (0.0764 Ib/t(80 fps)(0.240 Btu/lb °FX0.00416)(3600 s/hr)
= 22.0 Btu/hr t "F (125 W/m K)
For flow in short passages the correlations presented thus far must be
modified to account for variable velocity and temperature profiles along the axis
of flow. Deissler* has analyzed this region extensively for the case of turbulent
flow. The following equations may be used to modify the heat-transfer coefficients
in passages for which L/D<60:
for 2<L/D<20,
h-1+(D/L)
(20-33)
and for 20<L/D<60,
-1+6D/L
he
(20-34)
*R. G. Deissler, Trans. A.S.M.E, 77, 1221 (1955).
Transcribed Image Text:Another expression to be satisfied is q- hA, (T,-T)-pAuc, (T. - T) Evaluating the Reynolds number at the tube entrance, we have Pe Du_G/12] ft(80 fps) 0.159 x 10 fr/s The flow is turbulent and Re is sufficiently high that equation (20-30), (20-31), or (20-32) = 21 000 may be used. Employing equation (20-31) and assuming an exit temperature of 190°F, giving a mean bulk temperature of 125'F, and a film temperature of 167°F, we obtain St- -=0.023 Re0 Pr pec, [Git0(80 fps)(0.0764 lb/ft')] 02 1.45x 10 Ib./fts =0.023(0.1416)(1.276) = 0.00416 =0.023 (0.694)- Substitution into equation (19-61) yields T-L- exp -4)(0.00416) T-T, -ехpt-1.99)-0.136 and T-210-0.136(0) = 189.6°F This agreement is excellent. The heat transfer coetficient is (361 K) h- pec, (St) = (0.0764 Ib/t(80 fps)(0.240 Btu/lb °FX0.00416)(3600 s/hr) = 22.0 Btu/hr t "F (125 W/m K) For flow in short passages the correlations presented thus far must be modified to account for variable velocity and temperature profiles along the axis of flow. Deissler* has analyzed this region extensively for the case of turbulent flow. The following equations may be used to modify the heat-transfer coefficients in passages for which L/D<60: for 2<L/D<20, h-1+(D/L) (20-33) and for 20<L/D<60, -1+6D/L he (20-34) *R. G. Deissler, Trans. A.S.M.E, 77, 1221 (1955).
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