| 5B-2 : | Heat Transfer Required to Keep the Energy in a Flow System Constant | 4 pts |
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| Water vapor enters a tank at a rate of 32.4 kg/min at 250oC and 140 kPa and leaves the tank at the same rate at 180oC and 110 kPa. The diameter of the inlet and outlet pipes are 6 cm and 15 cm, respectively. | ||||||||||||||||||
| No form of work enters
or leaves the tank other than flow work. Calculate... a.) The total rate at which energy is entering the tank in the feed stream |
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| b.) The total rate at which energy is entering the tank
in the effluent stream c.) The heat transfer rate required to keep the total energy of the water inside the tank constant. |
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| Read : | The key to this problem is the enthalpy form of the 1st Law for open systems. Once you assume that gravitational potential energy is negligible in this problem, the solution is straightforward. Kinetic energy changes are not negligible. The relationships amoung velocity, density, specific volume, volumetric flow rate and mass flow rate are also important. | |||||||||||||||||
| Diagram: | 
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| Given: | m | 32.4 | kg/min | Wsys | 0 | kW | ||||||||||||
| 0.54 | kg/s | Tout | 180 | °C | ||||||||||||||
| Tin | 250 | °C | Pout | 110 | kPa | |||||||||||||
| Pin | 140 | kPa | Dout | 15 | cm | |||||||||||||
| Din | 6 | cm | 0.15 | m | ||||||||||||||
| 0.06 | m | c.) | dEsys/dt | 0 | kW | |||||||||||||
| Find: | a.) | Ein | ??? | kW | c.) | Q | ??? | kW | ||||||||||
| b.) | Eout | ??? | kW | |||||||||||||||
| Assumptions: | 1 - | Gravitational potential energy is negligible in computing the energy entering and leaving the system. Without this assumption, we would add the same arbitrary amount of energy to both the feed and effluent streams and then assume that changes in potential energy are negligible. Either way, potential energy is negligible in this problem. | ||||||||||||||||
| Equations / Data / Solve: | ||||||||||||||||||
| The key equation for this problem in the enthalpy form of the 1st Law for open systems. | ||||||||||||||||||
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Eqn 1 | |||||||||||||||||
| Eqn 1 can be simplified for this problem because Wsys = 0 and we have assumed that changes in potential energy are negligible. | ||||||||||||||||||
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Eqn 2 | |||||||||||||||||
| Part a) & Part b) | In order to evaluate Ein and Eout, we first need to lookup the enthalpies of the inlet and outlet streams. The Steam Tables or the NIST Webbook provide the information we need. |
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| Hin | 2973.2 | kJ/kg | Hout | 2835.4 | kJ/kg | |||||||||||||
| Next, we need to evaluate the specific kinetic energies at the inlet and outlet. | ||||||||||||||||||
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Eqn 3 | |||||||||||||||||
| We can determine the velocity from the mass flow rate as follows : | ||||||||||||||||||
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Eqn 4 | Where : | 
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Eqn 5 | ||||||||||||||
| We still need the specific volumes of the water at the inlet and outlet conditions to make use of Eqn 4. The Steam Tables or the NIST Webbook provide the information we need. |
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| Vin | 1.7163 | m3/kg | Vin | 1.8883 | m3/kg | |||||||||||||
| Now, we can plug values into Eqns 5, 4 & 3, in that order. | ||||||||||||||||||
| Ain | 0.002827 | m2 | Aout | 0.017671 | m2 | |||||||||||||
| vin | 327.8 | m/s | vout | 57.7 | m/s | |||||||||||||
| gc | 1 | kg-m/N-s2 | ||||||||||||||||
| Ekin,in | 53.72 | kJ/kg | Ekin,out | 1.66 | kJ/kg | |||||||||||||
| We can now use the right-hand portion of Eqn 2 to complete parts (a) and (b) of this problem. | ||||||||||||||||||
| Ein | 1634.5 | kW | Eout | 1532.0 | kW | |||||||||||||
| Part c.) | Eqn 2 can be simplified because dEsys/dt = 0. The result can be solved for Q to obtain the following equation. | |||||||||||||||||
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Eqn 6 | |||||||||||||||||
| Plugging values into Eqn 6 yields : | Q | -102.5 | kW | |||||||||||||||
| Verify: | The assumption made in this solution cannot be verified with the given information. | |||||||||||||||||
| Answers : | a.) | Ein | 1630 | kW | c.) | Q | -103 | kW | ||||||||||
| b.) | Eout | 1530 | kW | |||||||||||||||
