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Steady-State E Balances

Steady-State E Balances. Compressors, pumps, turbines and heat exchangers. Steady State Work. There are two types of steady state devices that use energy input or output as work. Pumps and compressors use energy input as work to increase the pressure of fluids

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Steady-State E Balances

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  1. Steady-State E Balances Compressors, pumps, turbines and heat exchangers

  2. Steady State Work • There are two types of steady state devices that use energy input or output as work. • Pumps and compressors use energy input as work to increase the pressure of fluids • Turbines produce energy output as work by depressurizing fluids Both are used widely in the CPI.

  3. Compressors and Pumps Compressors and pumps are steady-state flow through devices that use shaft work to increase the pressure of a process fluid. Though they work in similar ways, compressors are for gases and pumps are for liquids. The shaft work is usually supplied by an electrical motor, but sometimes comes directly from the shaft work supplied by a turbine. Compressors and pumps are used widely in the CPI for a variety of tasks.

  4. How they work Piston-Cylinder pumps and compressors. Intake stroke – intake valve opens, piston moves back and low P inlet fluid enters cylinder. Compression – valves close, piston moves forward raising fluid pressure Outlet stroke – outlet valve opens, piston moves forward expelling fluid at high pressure. Process repeats.

  5. More “work” goes in during compression than one gets back during expansion.  Gas exits at high temp and pressure Gas enters at low temp and pressure

  6. Compressor Compressors are generally assumed to be adiabatic. They are always assumed SS. KE and PE effects are ignored. Work in Tin, Pin Tout, Pout

  7. Compressor The question with a compressor is usually how much work (as electical power – Watts) will it require to compress a gas at a certain flow rate (moles/sec). The answer depends on the exit pressure and exit temperature.

  8. EB for compressor

  9. EB for compressor Molar is better for gases:

  10. Pumps • Pumps are devices for raising the pressure of liquids. Work input is used to increase the pressure at the outlet of the pump to the inlet of the pump. The E balance is the same as for compressors.

  11. A pump uses 5 kW of electricity to raise 2 kg/s of water from 1 atm, 25 C to 5 atm pressure. Estimate the temperature of the water as it exits the pump.

  12. 5 kW input 5 atm 2 kg/s 1 atm, 25 C

  13. Turbine Problem Problem 3A 1000 kg/hr of steam at 600 C and 5,000 kPa is sent through a turbine where the pressure drops to 200 kPa. The lower pressure steam exits at 300 C. Compute the work produced by the turbine.

  14. Turbines

  15. Turbines

  16. Problem 3 How much work would be obtained from turbine taking in 1000 kg/s of steam at 600 C and 5000 kPa and ejecting it at 500 kPa and 350 C? Steam 600 C 5000 kPa Steam T 350 C 500 kPa

  17. Steps: • First calculate the work or enthalpy difference across the device for the isentropic case • Next multiply this work or enthalpy difference by the efficiency (this is a turbine, actual work is 0.75 of isentropic (most work) case). • Compute Hout for the “actual” case. • Use table to estimate Tout • You can check second law just to make sure DSuniv should be positive.

  18. Step 1Calculate Hout for Isentropic Case 100% efficient Steam 600 C 5000 kPa Steam T ??? 500 kPa Sin= 7.26 kJ/kg-K Hin = 3776 kJ/kg Sout= 7.26 kJ/kg-K Hout = 2955 kJ/kg Hin-Hout = 821 kJ/kg Hypothetical Isentropic case

  19. Find Hout for Actual Case Calculate Work Steam 600 C 5000 kPa Steam T 350 C 500 kPa Hin = 3776 kJ/kg Hout = ???? kJ/kg Wout= m(Hin-Hout)

  20. Ideal gas Calculation Problem 4 How much work would be obtained processing the same number of moles/s of nitrogen gas at 600 C and 5000 kPa to 500 kPa and 350 C in a gas turbine? 1000 kg/s x 1000 moles/18 kg= 55,555 moles/s

  21. Summary • Pumps and Compressors use work to increase the pressure of liquids and gases respectively. • Turbines produce work during a decrease in pressure of a gas during a pressure drop.

  22. Condensation HTX Problem 5 A heat exchanger takes in 5 GPM of water at 38 C and releases it at 50 C. At what rate would steam at 120 C need to be condensed to accomplish this? 120 C steam in 38 C water in @ 5GPM 50 C water Out @ 5GPM 120 C liquid water out.

  23. Condensation Problem 5 What is the system? (If I don’t care about how fast the energy is transferred from the steam side to the tube, then the whole thing is the system.) 120 C steam in 38 C water in @ 5GPM 50 C water Out @ 5GPM 120 C liquid water out.

  24. Condensation 120 C steam in 38 C water in @ 5GPM 50 C water Out @ 5GPM 120 C liquid water out. System – Whole Device Get from U + PV= Cv(T-Tref) + PV 120 C get from steam tables

  25. Condensation System – Whole Device 2706 kJ/kg 505 kJ/kg

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