Question 11.8: 100,000 lb/h of saturated vapor consisting of 30 mole percen...
100,000 lb/h of saturated vapor consisting of 30 mole percent n-butane and 70 mole percent n-pentane are to be condensed at a pressure of 75 psia. A single-pass horizontal shell-and-tube condenser will be used with the condensing vapor in the shell. Cooling water with a range of 85-120ºF will flow in the tubes. The flow area across the tube bundle is a_{s} = 0.572 ft², the equivalent diameter is D_{e} = 0.06083 ft and the baffle spacing is B/d_{s}ds = 0.45. The bundle contains 107 ft² of external surface area per foot of length. The overall heat-transfer coefficient between the vapor–liquid interface and coolant is U_{D} = 120 Btu/h . ft².º F. Use the Bell–Ghaly method to calculate the required surface area and length of the condenser.
For convenience, the condensing range (183.5–168 ºF) is divided into three intervals: 168ºF to 173ºF, 173ºF to 178ºF and 178ºF to 183.5ºF. Equilibrium flash calculations are then performed at each of the above four temperatures using a flowsheet simulator (PRO/II by SimSci-Esscor). The results are summarized in the following table. (Note that physical properties of the condensate are not required because U_{D} is given in this problem.)
| T_{V}\left({ }^{\circ} F \right. | m_{V}( lbm / h ) | Duty (Btu/h) | C_{P, V}\left( Btu / lbm \cdot{ }^{\circ} F \right) | k_{V}\left( Btu / h \cdot ft \cdot{ }^{\circ} F \right) | \mu_{V}( cp ) |
| 183.5 | 100,000 | 0 | 0.486 | 0.01194 | 0.00853 |
| 178 | 54,590 | 6,373,860 | 0.483 | 0.01185 | 0.00851 |
| 173 | 24,850 | 10,661,170 | 0.48 | 0.01177 | 0.0085 |
| 168 | 0 | 14,314,980 | 0.477 | 0.01168 | 0.00848 |
The flow rate of the cooling water is obtained from the overall energy balance:
\dot{m}_{c}=\frac{q}{C_{P,c} \Delta T_{c}}=\frac{14,314,980}{1.0(120-85)}=408,999 lbm / hNext, the temperature profile for the cooling water is determined using energy balances over the individual temperature intervals. Counter-current flow and constant heat capacity of the cooling water are assumed. For the interval from 168ºF to 173ºF, we have:
\Delta q=14,314,980-10,661,170=3,653,810 Btu / h\Delta T_{c}=\frac{\Delta q}{m_{c} C_{P f}}=\frac{3,653,810}{408,999 \times 1.0}=8.93 ^{\circ} F
T_{c, \text { out }}=85+8.93=93.93 ^{\circ}F
Similarly, for the interval from 173ºF to 178ºF we obtain:
\Delta q=10,661,170-6,373,860=4,287,310 Btu / h\Delta T_{c}=\frac{4,287,310}{408,999 \times 1.0}=10.48 ^{\circ}F
The temperature profiles of the vapor and cooling water are summarized in the following table.
| T_{V} (ºF) | T_{c} (ºF) |
| 183.5 | 120 |
| 178 | 104.41 |
| 173 | 93.93 |
| 168 | 85 |
The calculations for the temperature interval from 168ºF to 173ºF are performed next. The rate of sensible heat transfer to cool the vapor in this interval is calculated using the average vapor flow rate and heat capacity for the interval. The averages are:
\bar{C}_{P, V}=0.5(0.477+0.480)=0.4785 Btu / lbm \cdot{ }^{\circ} F\left(\dot{m}_{V}\right)_{a v e}=0.5(0+24,850)=12,425 lbm / h
Hence, the rate of sensible heat transfer is:
\Delta q_{V}=\left(\dot{m}_{V}\right)_{a v e} \bar{C}_{P, V} \Delta T_{V}=12,425 \times 0.4785 \times 5=29,727 Btu / hThe value of L for the interval is:
\Lambda=\Delta q_{V} / \Delta q=29,727 / 3,653,810=0.0081The Reynolds number for the vapor is computed next using the average vapor flow rate and average viscosity for the interval.
\bar{\mu}_{V}=0.5(0.00848+0.00850)=0.00849 cpR e_{V}=\frac{D_{e}\left\{\left(\dot{m}_{V}\right)_{ave} / a_{s}\right\}}{\mu_{\nu}}=\frac{0.06083\{12,425 / 0.572\}}{0.00849 \times 2.419}=64,339
Equations (3.20) and (3.21) are used to calculate the heat-transfer coefficient for the vapor phase:
h_0=j_H\left(\frac{k}{D_e}\right)Pr^{1/3}\left(\frac{\mu }{\mu _w}\right)^{0.14} (3.20)
j_{H}=0.5\left(1+B / d_{5}\right)\left(0.08 R e^{0.6821}+0.7 R e^{0.1772}\right) (3.21)
=0.5(1+0.45)\left\{0.08(64,339)^{0.6821}+0.7(64,339)^{0.1772}\right\}
j_{H}=114.09
\bar{k}_{V}=0.5(0.01168+0.01177)=0.011725 Btu / h \cdot ft \cdot{ }^{\circ} F
P_{r_{V}}=\frac{\bar{C}_{ P , V} \bar{\mu}_{V}}{\bar{k}_{V}}=\frac{0.4785 \times 0.00849 \times 2.419}{0.011725}=0.8381
h_{V}=\frac{j_{H} \bar{k}_{V} P_{V}^{1 / 3} \phi}{D_{e}}=\frac{114.09 \times 0.011725(0.8381)^{1 / 3} \times 1.0}{0.06083}
h_{V}=20.7 Btu / h \cdot ft ^{2} \cdot{ }^{\circ} F
The heat-transfer area required for the temperature interval is obtained from the finite difference form of Equation (11.79) using the average vapor and coolant temperatures
\Delta A=\left\{\frac{1+U_{D} \Lambda / h_{V}}{U_{D}\left(T_{V, a v e}-T_{c, a v e}\right)}\right\} \Delta q\Delta A=\left\{\frac{1+120 \times 0.0081 / 20.7}{120(170.5-89.465)}\right\} \times 3,653,810
\Delta A=393.4 ft ^{2}
The above calculations are repeated for the other two temperature intervals to obtain the results shown in the following table.
| Temperature Interval | |||
| Paramete | 168–173ºF | 173–178ºF | 178–183.5ºF |
| \left(\dot{m}_{V}\right)_{a v e}(lbm/h) | 12,425 | 39,720 | 77,295 |
| \Delta q_{V} (lbm/h) | 29,727 | 95,626 | 205,972 |
| Δq(lbm/h) | 3,653,810 | 4,287,310 | 6,373,860 |
| ∧ | 0.0081 | 0.0223 | 0.0323 |
| R e_{V} | 64,339 | 205,315 | 398,839 |
| j_{H} | 114.09 | 248.23 | 388.46 |
| h_{V}(Btu/h . ft².º F) | 20.7 | 45.5 | 71.7 |
| T_{\text {Vrave }}(ºF) | 170.5 | 175.5 | 180.75 |
| T_{\text {ouse }}(ºF) | 89.465 | 99.17 | 112.205 |
| ΔA (ft²) | 393.4 | 495.6 | 816.8 |
The total heat-transfer area required in the exchanger is the sum of the incremental areas:
A=393.4+495.6+816.8 \cong 1706 ft ^{2}The required length of the exchanger is:
L=\frac{A}{(A / L)}=\frac{1706}{107}=15.9 \cong 16 ft