Question 4 of 8: Two-stage ammonia plant with a flash chamber

Question 4: Two-stage ammonia plant with a flash chamber (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A two-stage R-717 (ammonia) plant with a flash chamber acting both as an intercooler and as the intermediate expansion vessel.

Find. The plant sketch and its cycle on the appended pressure–enthalpy diagram; the mass fraction of vapour leaving the flash chamber; the coefficient of performance; and the refrigerant mass flow through the condenser at 450 kW of refrigeration.

Approach. Take one kilogram through the condenser as the basis. The first throttle fixes the flash fraction directly from a lever rule on enthalpy; an energy balance at the mixing tee fixes the HP suction state; the two compressions follow from the isentropic enthalpy rises divided by the stage efficiency; and the refrigerating effect is carried only by the liquid fraction.

1. Read the state points. Ammonia properties at the three pressures, on the standard reference datum used by the appended chart: $h_{1}=1591.6$ kJ/kg and $s_{1}=6.396\ \text{kJ/(kg}\cdot\text{K)}$ at 2 bar and $-16^{\circ}\text{C}$; $h_{3}=1611.8$ kJ/kg for saturated vapour at 5 bar; $h_{8}=364.9$ kJ/kg for saturated liquid at 5 bar; $h_{6}=491.9$ kJ/kg for saturated liquid at 12 bar.
2. Part (c), first answer — the flash fraction. The first throttle is isenthalpic, $h_{7}=h_{6}=491.9$ kJ/kg, and the flash chamber separates that wet mixture at 5 bar into its saturated phases. The vapour fraction is the quality of state 7: $$x=\frac{h_{7}-h_{8}}{h_{3}-h_{8}}=\frac{491.9-364.9}{1611.8-364.9} =\boxed{x=0.1018\quad(10.2\ \%\ \text{by mass})}$$ So of every kilogram circulating through the condenser, 0.102 kg flashes to vapour at the intermediate pressure and by-passes the evaporator entirely, while 0.898 kg goes on as saturated liquid.
3. Low-pressure compression. Isentropic to 5 bar from $s_{1}=6.396$ gives $h_{2s}=1714.5$ kJ/kg, so the real discharge is $$h_{2}=h_{1}+\frac{h_{2s}-h_{1}}{\eta_{is}} =1591.6+\frac{1714.5-1591.6}{0.90}=1728.1\ \text{kJ/kg}$$ at $50.3^{\circ}\text{C}$ — comfortably superheated, as it must be if the flash chamber is to desuperheat it.
4. Mixing at the intermediate pressure. The 0.898 kg of hot LP discharge meets 0.102 kg of saturated vapour off the chamber; an adiabatic mixing balance on one kilogram of total flow gives $$h_{4}=(1-x)\,h_{2}+x\,h_{3}=0.8982(1728.1)+0.1018(1611.8)=1716.3\ \text{kJ/kg}$$ which is $45.3^{\circ}\text{C}$ at 5 bar — the flash vapour has removed 5 K of superheat before the second stage sees the gas. That desuperheating is the entire point of the arrangement.
5. High-pressure compression. From $s_{4}=6.401$ isentropic to 12 bar gives $h_{5s}=1860.2$ kJ/kg, so $$h_{5}=1716.3+\frac{1860.2-1716.3}{0.90}=1876.2\ \text{kJ/kg}$$ at $121.3^{\circ}\text{C}$. For comparison, a single-stage machine working between the same evaporator and condenser pressures would discharge at $125.9^{\circ}\text{C}$ — ammonia's very high discharge temperatures are the practical reason two-stage plant is used at all below about $-15^{\circ}\text{C}$.
6. Refrigerating effect and work. Only the liquid fraction reaches the evaporator, and it enters after the second throttle at $h_{9}=h_{8}=364.9$ kJ/kg: $$q_{e}=h_{1}-h_{9}=1591.6-364.9=1226.7\ \text{kJ/kg of evaporator flow}$$ Per kilogram through the condenser the refrigeration is $(1-x)\,q_{e}=0.8982(1226.7)=1101.7$ kJ/kg, while the work is $$w=(1-x)(h_{2}-h_{1})+(h_{5}-h_{4})=0.8982(136.6)+159.9=122.7+159.9=282.6\ \text{kJ/kg}$$
7. Part (c), second answer — coefficient of performance. $$\mathrm{COP}=\frac{(1-x)\,q_{e}}{w}=\frac{1101.7}{282.6}=\boxed{3.90}$$ A single-stage machine on the same duty would return 3.71, so the flash-chamber arrangement buys about 5 % on efficiency as well as the reduction in discharge temperature. Against the Carnot value between $-18.8^{\circ}\text{C}$ and $31.0^{\circ}\text{C}$, 5.11, the plant achieves 76 % — reasonable for a real machine with 90 % isentropic stages.
8. Part (c), third answer — condenser mass flow. The 450 kW load is carried by the 1101.7 kJ per kilogram of condenser flow: $$\dot m_{c}=\frac{\dot Q_{e}}{(1-x)\,q_{e}}=\frac{450}{1101.7} =\boxed{0.408\ \text{kg/s through the condenser}}$$ of which $0.8982(0.408)=0.367$ kg/s passes through the evaporator and the LP compressor, and 0.042 kg/s is the flash vapour.
9. Check the plant balance. Total compressor power is $450/3.90=115.4$ kW, so the condenser must reject $450+115.4=565.4$ kW. Directly, $\dot m_{c}(h_{5}-h_{6})=0.408(1876.2-491.9)=565.4$ kW — the two agree exactly, which is the surest sign the mixing balance and the flash fraction are both right.