Question 12.12: Use of the Effectiveness-NTU Method The same example as the ......
Use of the Effectiveness-NTU Method
The same example as the previous one (Example 12.11) will be analyzed under a slightly modified specifications. Here, the heat exchanger area is known, but the exit temperature of the hot fluid is not. Also, we will simply assume a singlepass counterow heat transfer rate and the exit temperatures of both streams.
Given: T_{c,i} = 20°C, T_{h,i} = 110°C, U_{o} = 320 W/(m^{2} ·K), A_{o} = 20 m^{2}, \dot{m}_{c} = 70 kg/min, \dot{m}_{h} = 90 kg/min
Figure: See Figure 12.11.
Assumptions: Jacket losses are negligible.
Find: \varepsilon, T_{c,o}, T_{h,o}
Lookup values: c_{p} = 4180 J/(kg ·K)
As previously, cold-side capacitance rate is \dot{C}_{c} = 4877 W/K and the hot-side capacitance rate is \dot{C}_{h} = 6270 W/K .
The minimum capacitance is for the cold stream. The capacitance ratio is
The NTU value is found from the following definition:
NTU = \frac{U_{o} A_{o}}{\dot{C}_{min}} = \frac{320 \times 20}{4877} = 1.31Reading from Figure 12.17 (alternatively the second equation in Table 12.6 can be used), we find that the effectiveness is
\varepsilon = 0.60
It is now a simple matter to find first the heat rate and then the outlet temperatures of each stream.
From Equation 12.35,
\dot{Q} = ε \dot{C}_{min}(T_{h,i} – T_{c,i})
= 0.60 \times 4877 W/K \times (110^° C – 20^° C) = 263.4 kW
The heat rate for either stream is also given by the product of the mass flow, specific heat, and stream temperature rise:
\dot{Q} = \dot{m} c_{p} (T_{i} – T_{o})One now solves for the outlet temperatures. For the hot stream,
T_{h,o} = T_{h,i} – \frac{\dot{Q}}{\dot{C}_{max}}= 110 – \frac{263,400}{6,270} = 68.0°C
and for the cold stream,
T_{c,o} = 20 + \frac{263,400}{4,877} = 74.0°CComments
For improved accuracy, one could use the equation for effectiveness listed in Table 12.6. However, for design purposes, the curves are satisfactory.
Alternatively, manufacturers’ data can be consulted for effectiveness values.
| TABLE 12.6 | |
| Heat Exchanger Effectiveness Relations, N = NTU = \frac{U_{o} A_{o}}{\dot{C}_{min}}, C = \frac{\dot{C}_{min}}{\dot{C}_{max}} | |
| Flow Geometry | Relation |
| Double pipe | |
| Parallel flow | \varepsilon = \frac{1 – exp[-N(1+c)]}{1 + C} |
| Counter flow | \varepsilon = \frac{1 – exp[-N(1+c)]}{1 – C exp[-N(1+c)]} for C < 1 |
| \varepsilon = \frac{N}{1 + N} for C = 1 | |
| Cross-flow | |
| Both fluids unmixed | \varepsilon = 1 – exp \{ \frac{1}{Cn} [exp (-NCn) – 1] \} where n = N^{-0.22} |
| Both fluids unmixed | \varepsilon = N \left[ \frac{N}{1 – exp(-N)} + \frac{NC}{1 – exp(-NC)} \right]^{-1} |
| \dot{C}_{max} mixed, \dot{C}_{min} unmixed | \varepsilon = \frac{1}{C} [1 – exp[- C + C exp(-N)] \} |
| \dot{C}_{max} unmixed, \dot{C}_{min} mixed | \varepsilon = 1 – exp \{ – \frac{1}{C} [ 1 – exp(-NC)] \} |
| Shell and tube | |
| One shell pass; two, four, and six tube passes | \varepsilon 2 \left[ 1 + C + \sqrt{1 + C^{2}} \frac{1 + exp (-N \sqrt{1 + C^{2}})}{1 – exp (-N \sqrt{1 + C^{2}})} \right]^{-1} |
