Lecture 3: Compressor Refrigeration Cycle Steam Cycle (4-8) CHE441 Lecture 3: Compressor Refrigeration Cycle Steam Cycle (4-8)

Introduction of Compressor A gas compressor is a mechanical device that increases the pressure of a gas by reducing its volume. Compressors are similar to pumps: both increase the pressure on a fluid and both can transport the fluid through a pipe. As gases are compressible, the compressor also reduces the volume of a gas. Liquids are relatively incompressible.

Compressor Curves Surge is defined as the operating point at which centrifugal compressor peak head capability and minimum flow limits are reached. Backflow occurs. https://www.youtube.com/watch?v=W8DMun06NXg

Gas Compression Relationship Gas compression can be expressed in terms of pressure and temperature variation as PVk = constant A compression process typically following this pressure volume relation is known as polytropic process. If k=1, the process is isothermal (constant temperature). If k=γ=CP/CV, the compression is adiabatic compression (no heat exchange with the surrounding). Most gas compressions generally follow the adiabatic curve. Hence compressor equations are also based on adiabatic curve with k=γ, PVγ = constant Let subscripts 1 and 2 stand for inlet and outlet process conditions of the compressor. Then the pressure ratio of the compressor is P2/P1. P2/P1 = (V1/V2)k ... Polytropic compression P2/P1 = (V1/V2)γ ... Adiabatic compression

Theoretical Compression Power Closed System: 1 2 −𝑃𝑑𝑉 Flow system: 1 2 −𝑃𝑑𝑉 + ( 𝑃 2 𝑉 2 − 𝑃 1 𝑉 1 )= 1 2 𝑉𝑑𝑃 Isothermal Compressor (requires cooling): W = 1 2 𝑉𝑑𝑃 = 1 2 𝑛 𝑅𝑇/𝑃𝑑𝑃= 𝑛 𝑅𝑇𝑙𝑛( 𝑃2 𝑃1 ) Adiabatic Compressor: W = 1 2 𝑉𝑑𝑃 = 1 2 𝑃1 𝑃 1/𝛾 𝑉1𝑑𝑃= 𝛾 𝛾−1 𝑃1 1 𝛾 𝑉1 𝑃2 𝑃1 − 1 𝛾 +1 = 𝑛 𝑅𝑇1 𝛾 𝛾−1 𝑃2 𝑃1 𝛾−1 𝛾 −1

Real Power of an adiabatic compressor Power is proportional to inlet temperature T1 Power is related to pressure ratio instead of pressure increase. Questions: will multiple stage compressor with inter-cooling better than a single compressor? W = 𝑛 𝑅𝑇1 η 𝛾 𝛾−1 𝑃2 𝑃1 𝛾−1 𝛾 −1

Multiple compressor with intercooler Gamma = 1.5 Assuming that the second compressor has the same inlet temperature and efficiency as those in the first compressor 𝜕𝑊 𝜕𝑇2 =0, or 𝑃3 𝑃2 = 𝑃2 𝑃1 To approach an isothermal compressor, you need infinite number of compressors with inter-coolers!

Example Compress air from 1 bar to 10 bar. Maximum compression ratio is 3 in a compressor. At least how many compressors are required? Using the least number of compressors, calculate the compression ratio and outlet pressures at each stage. 3 stages. 1-3, 3-9, 9-10, or 1-2.15, 2.15-4.64, 4.64-10

Mollier Chart - Methane

Mollier Chart - Methane Red: two stages with intercooler Green: one stage Isentropic operation

Mollier Chart - Ethylene

Compressor Efficiency η 𝑐𝑜𝑚𝑝 = ∆ℎ 𝑖𝑑𝑒𝑎𝑙 ∆ℎ 𝑟𝑒𝑎𝑙 Enthalpy Pressure Isentropic line ∆ℎ 𝑟𝑒𝑎𝑙 ∆ℎ 𝑖𝑑𝑒𝑎𝑙 P1 P2 h1 h2,ideal h2,real If we know the suction pressure and temperature we can easily plot it on the Mollier chart and read off the enthalpy during our compressor training courses.  Assuming isentropic compression we follow the constant entropy line (S) until it intersects the discharge pressure line and then read off the temperature (and enthalpy).  If you want to be more accurate you could divide the increase in enthalpy by the isentropic efficiency to find the real increase in enthalpy.  Add the increase in enthalpy to the isentropic increase in enthalpy and draw a vertical line up from our new enthalpy value to the discharge pressure and now we read off the actual discharge temperature. Isentropic efficiency = adiabatic efficiency

Example: Calculate the horsepower for compressing 5,000 lbs/hr of ethylene from 100 psia,  40oF to 200 psia. The adiabatic efficiency of the compressor is 75%. Include your calculations and Mollier chart. (

Refrigeration System (Compressor) Steam System (Turbine)

Two Interesting Questions Put a refrigerator in an insulated, closed room. If the refrigerator door is left open, would the room temperature increase or decrease after a long time? Is the following system feasible? 1. Only work in, Q increase?

Clausius Statement 2nd Law of Thermodynamics It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.

Work and Heat Heat to work: the efficiency cannot be 100%, second law of thermodynamics. Work to heat: can be higher than 100%. Use “Coefficient of Performance (COP)” instead of “efficiency” Work to heat: can be higher than 100%. Use “Coefficient of Performance (COP)” instead of “efficiency”? HPS to turbine and produce work and LPS.

Heat Pump and Refrigerator For refrigeration: COP = QL/Win For heat pump: COP = QH/Win

Refrigeration Cycle For refrigeration: COP = QL/Win For heat pump: COP = QH/Win

Air Conditioning System https://www.archtoolbox.com/materials-systems/hvac/how-air-conditioners-work.html

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Multi-Stage Refrigeration System Large commercial or industrial refrigeration systems may have multiple expansion valves and multiple evaporators in order to refrigerate multiple users at different temperature levels. (e.g. ethane, propane) Cold box http://bv.com/Home/news/solutions/energy/producing-lng-from-coal https://www.google.ch/patents/US6637237

Example: Design a refrigeration system, which supplies two levels of ammonia refrigerant at 10oC and at  10oC to two exchangers requiring duties of 850 and 2,500 kW, respectively. Draw a process flow diagram (PFD) showing major equipment, flow rates in kg/s of refrigerant, duty of each heat exchanger, duty, and horsepower of each compressor. Also, show temperatures and pressures of all streams. There are no inter-coolers between compressors. The compressed gas from the last stage compressor can be cooled to saturated liquid at 30oC. Use 75% adiabatic efficiency for all compressors.

PFD

CATT3: Computer Aided Thermodynamic Tables 3 Available via the following link. You may download and install in your own computer. Gibbs Phase Rule: F = C-P+2 F = 3 – P for single component. For superheated vapor or subcooled liquid, F = 2, i.e., specifying T and P uniquely determine the state. For saturated liquid/vapor, F = 1, i.e., specifying T (or P) uniquely determine the state. T and P are related by equations, e.g., Antoine Equation.

Mollier Chart

𝑚 1 ℎ 𝐿, 30𝐶 + 𝑄 1 = 𝑚 2 ℎ 𝐿, 10𝐶 +( 𝑚 1 − 𝑚 2 ) ℎ 𝑉, 10𝐶 Solving Strategy Calculate backwards (from the last level to the first level) 𝑚 2 = 𝑄 2 /(ℎ 𝑉, −10𝐶 − ℎ 𝐿, 10𝐶 ) 𝑚 1 ℎ 𝐿, 30𝐶 + 𝑄 1 = 𝑚 2 ℎ 𝐿, 10𝐶 +( 𝑚 1 − 𝑚 2 ) ℎ 𝑉, 10𝐶

Properties at all States from CATT3   Specific Type Temp Pressure Volume Enthalpy Entropy Quality Phase C MPa m3/kg kJ/kg kJ/kg/K 1 Ammonia 30 1.167 0.00168 322.4 1.2 Saturated Liquid 2 0.1105 1466 4.974 Saturated Vapor 3 10 0.6152 0.0016 227 0.8779 4 0.2054 1452 5.204 5 -10 0.2909 0.001534 134.4 0.5408 6 0.4181 1431 5.467 7 39.43 0.2343 1530 Superheated Vapor 8 52.64 0.2465 1563 5.57 9 38.85 0.2338 1529 5.463 87.57 0.1417 1628 11 100.8 0.1482 1661 5.554

Vapor Fraction in a Mixture - Lever Rule fV hV + (1-fV) hL = h = fV h + (1-fV) h, or fV (hV – h) = fL (h-hL) = (1-fV)(h-hL) Example: Calculate the vapor fraction of the mixture after the first valve. hL hV h

Compressor Step 1: Solve the ideal compressor outlet enthalpy ℎ 𝑜, 𝑖𝑑𝑒𝑎𝑙 = h(P, S). P is based on the pressure at the upper level. S is based on the compressor inlet Step 2: Solve the real compressor outlet enthalpy ℎ 𝑜, 𝑟𝑒𝑎𝑙 using the definition of compressor efficiency η = ∆ℎ 𝑖𝑑𝑒𝑎𝑙 ∆ℎ 𝑟𝑒𝑎𝑙 = ℎ 𝑜, 𝑖𝑑𝑒𝑎𝑙 − ℎ 𝑖 ℎ 𝑜,𝑟𝑒𝑎𝑙 − ℎ 𝑖 Step 3: Solve real compressor outlet temperature T = T(H,P) The compressor power = 𝑚 ∆ℎ 𝑟𝑒𝑎𝑙

Mixer Heat Exchanger Using mixing rule: ℎ= 𝑚 1 ℎ 1 + 𝑚 2 ℎ 2 𝑚 1 + 𝑚 2 ℎ= 𝑚 1 ℎ 1 + 𝑚 2 ℎ 2 𝑚 1 + 𝑚 2 Based on h and P, T can be calculated. Heat Exchanger Example Stop here? Entropy could use mixing rule as well? Q = 𝑚 (ℎ 2 − ℎ 1 )

Introduction of Turbine A turbine is a rotary mechanical device that extracts energy from a fluid flow and converts it into useful work.

Steam Turbine and Gas Turbine

Rankine Cycle Start a new session for turbine?

Note the Difference in Compressor and Turbine Efficiencies Enthalpy Pressure Isentropic line ∆ℎ 𝑟𝑒𝑎𝑙 ∆ℎ 𝑖𝑑𝑒𝑎𝑙 P1 P2 h1 h2,ideal h2,real η 𝑐𝑜𝑚𝑝 = ∆ℎ 𝑖𝑑𝑒𝑎𝑙 ∆ℎ 𝑟𝑒𝑎𝑙 η 𝑡𝑢𝑟𝑏 = ∆ℎ 𝑟𝑒𝑎𝑙 ∆ℎ 𝑖𝑑𝑒𝑎𝑙

Note the Difference in Compressor and Turbine Efficiencies η 𝑐𝑜𝑚𝑝 = ∆ℎ 𝑖𝑑𝑒𝑎𝑙 ∆ℎ 𝑟𝑒𝑎𝑙 Enthalpy Pressure Isentropic line ∆ℎ 𝑟𝑒𝑎𝑙 ∆ℎ 𝑖𝑑𝑒𝑎𝑙 P2 h1 h2,ideal h2,real P1 η 𝑡𝑢𝑟𝑏 = ∆ℎ 𝑟𝑒𝑎𝑙 ∆ℎ 𝑖𝑑𝑒𝑎𝑙

Turbine Power Calcs on Mollier Chart

Multi-stage turbine schematic HP MP LP G VHP stm From Blr LP stm To Proc KW

Multistage ST with intermediate reheat Reheat option HP MP LP G VHP stm From Blr LP stm To Proc KW H Reheating effectively converts 100% of supplied heat into power – a fantastic thermodynamic deal, especially if it can done with waste heat.

Superheating, reheating and regenerative Rankine cycles Matlab-based Steam Table (Xsteam) can be download here.

Regenerative Rankine Cycle

Turbine Example If the gases leave the tower at 6 atm, 25 °C, 10,000 kmol/hr, and are expanded to 1.5 atm, calculate the turbine power. If the gases are preheated to 400 °C with the reactor off-gas. Also, estimate the power recovered from the preheated gases. W=10000*8.314*(25+273)*(1.27/(1.27-1))*((1.5/6)^((1.27-1)/1.27)-1)/3600*75%=6.2kW W=10000*8.314*(400+273)*(1.27/(1.27-1))*((1.5/6)^((1.27-1)/1.27)-1)/3600*75%=14.0kW See turbine hysys simulaiton W = 𝑛 𝑅𝑇1 𝛾 𝛾−1 𝑃2 𝑃1 𝛾−1 𝛾 −1 𝛾 = 1.27 R = 8.314 T (K) W (kJ/kmol) https://www.google.com/patents/US20130216461

Steam Turbine and Steam System Combined Heat and Power (CHP) Cogeneration There seems to be a lot of confusion about these two terms. CHP applies to cogeneration as well as non-cogeneration Cogeneration is specifically defined as the “A system that makes effective use of both the heat and shaft work output (or other form of energy) from burning fuel in a Heat Engine”

This is both CHP and “Co-Generation” LATENT HEAT OF STEAM IS USED IN THE PROCESS BOILERS PROCESS FUEL KW STEAM TURBINE Condensing turbines are invariably across the pinch (BAD) So are Extraction turbines So we will consider back pressure steam turbines only LP STEAM OVERALL EFF ~ 75%

This is CHP, but not Cogeneration LATENT HEAT OF STEAM IS WASTED BOILERS PROCESS FUEL STEAM STEAM TURBINE KW CONDENSER BOILER EFF ~ 80% POWER GEN EFF ~ 30%

Typical Industrial Steam System Project example

Example – Steam Mass Balance What if the MPS demand is 20t/h less than the current supply, how to adjust the steam system?

Homework: Steam System Design a steam system for an ammonia manufacturing process. The system generates three level of steam at 1500 psia, 600 psia, and 50 psia for heating and power generation. The 1500 psia steam is superheated to 1000oF. A compressor system totaling 40,000 hp is driven by steam turbines with adiabatic efficiencies of 60% operating between 1500 and 600 psia steam. Two pumps totaling 1,400 hp are driven by steam turbines with adiabatic efficiency of 60% operating between 600 psia and 50 psia. Two exchangers requiring 600 psia steam for heating are rated at 1.0107 Btu/hr each. One exchanger (E-4) requiring 50 psia steam for heating is rated at 2.0107 Btu/hr. Excess 600 psia steam can be sold to neighboring plant. All the condensate from heat exchangers and turbine condenser is pumped to a boiler maintained at 1500 psia from which steam is generated. All condensate is saturated at 50 psia. Make-up water is available at 50 psia saturated. A process flow diagram (PFD) for the team system is show in the figure. Calculate flow rates, duties, and horsepower on all equipment listed in the table below the PFD. Include all calculations using a Steam Table and an engineering solver.

Solution using Hysys or UniSim

Detailed Result Show results slides to students?

Matlab code based on Xsteam (steam table) clear all;clc; h1 = XSteam('h_pt',1500/14.503773800721815,(1000-32)/1.8); %% enthalpy of 1500 psia, 1000F steam, [kJ/kg] s1 = XSteam('s_pt',1500/14.503773800721815,(1000-32)/1.8); %% entropy of 1500 psia, 1000F steam, [kJ/kg/C] h2i = XSteam('h_ps',600/14.503773800721815,s1); eta1 = 0.6; dh_T1 = (h1-h2i)*eta1; %% deltah h2 = h1-dh_T1; %% enthalpy of 600 psia steam eta2 = 0.6; s2 = XSteam('s_ph',600/14.5,h2); %% entropy of the 600 psia steam h3i = XSteam('h_ps',50/14.5,s2); dh_T2 = (h2-h3i)*eta2; h3 = h2-dh_T2; %% enthalpy of 50 psia steam W_T1 = 40000*0.7456998715823; %% T1 duty, [kW] m_T1 = W_T1/dh_T1 %% flow through E1 or T1, [kg/s] W_T2 = 1400*0.7456998715823; %% T2 duty, [kW] m_T2 = W_T2/dh_T2 %% flow through T2, [kg/s] Q_E2 = 10e6*0.0002930711111111; %% E2 duty, [kW] Q_E3 = 10e6*0.0002930711111111; %% E3 duty, [kW] Q_E4 = 20e6*0.0002930711111111; %% E4 duty, [kW] hl_50 = XSteam('h_px',50/14.5,0); %% enthalpy of 50 psia saturated steam, [kJ/kg] m_E2 = Q_E2/(h2-hl_50) %% flow through E2, [kg/s] m_E3 = Q_E3/(h2-hl_50) %% flow through E3, [kg/s] m_600_extra = m_T1-m_E2-m_E3-m_T2 %% flow of 600 psia free stream, [kg/s] m_E4 = Q_E4/(h3-hl_50) %% flow through E4, [kg/s] m_E5 = m_T2 - m_E4 %% flow through E5, [kg/s] Q_E5 = m_E5*(h3-hl_50) %% E5 duty, [kW] m_makeup = m_600_extra %% makeup water flow [kg/s] rho = XSteam('rhoL_p',50/14.5,0) %% density of water [kg/m3] eta_pump = 0.75; W_pump = m_T1/rho*(1500-50)/14.503773800721815*1e5/1e3/eta_pump; %% pump bhp [kW] Q_E1 = m_T1*(h1-hl_50)-W_pump %% E1 duty, [kW]

Another Steam System Problem: Design a steam system based on the following specifications: Three steam headers at 1400, 600, and 55 psia. Three compressor systems requiring 30,000, 2,100 and 12,000 hp respectively, are driven by steam turbines. One pump requiring 1461 hp is also driven by a steam turbine. Three heat exchangers require duties of 15, 25 and 30 million Btu/hr respectively to be supplied by the 600 psia steam. Two exchangers require 30 and 25 million Btu/hr respectively to be supplied by 55 psia steam. All condensates are saturated at 55 psia. High pressure steam at 1400 psia and 900O F are generated by a boiler. All turbines and pumps have adiabatic efficiencies of 75%. Draw a PFD and complete mass and energy balances. Use the CATT3, or Matlab (XSteam), or an engineering solver to calculate the total horsepower and flow rates. Hand-draw a PFD with stream numbers, known pressures, temperatures, and flowrates.

Solution: Trick: adjust Q1 so that Q1+Q2 = 30,000 hp.