Question 11.CSGP.133: A jet ejector, a device with no moving parts, functions as t......
A jet ejector, a device with no moving parts, functions as the equivalent of a coupled turbine-compressor unit (see Problems 9.157 and 9.168). Thus, the turbine-compressor in the dual-loop cycle of Fig. P11.122 could be replaced by a jet ejector. The primary stream of the jet ejector enters from the boiler, the secondary stream enters from the evaporator, and the discharge flows to the condenser. Alternatively, a jet ejector may be used with water as the working fluid. The purpose of the device is to chill water, usually for an air-conditioning system. In this application the physical setup is as shown in Fig. P11.133. Using the data given on the diagram, evaluate the performance of this cycle in terms of the ratio Q _{ L } / Q _{ H }.
a. Assume an ideal cycle.
b. Assume an ejector efficiency of 20% (see Problem 9.168).
\begin{aligned}& T _1= T _7=10^{\circ} C \\& T _2=150^{\circ} C \\& T _4=30^{\circ} C \\& T _9=20^{\circ} C\end{aligned}
Assume T _5= T _{10}
(from mixing streams 4 & 9).
a) \dot{ m }_1+\dot{ m }_2=\dot{ m }_3 \text {; ideal jet ejector }
s_1^{\prime}=s_1 \& s_2^{\prime}=s_2\left(1^{\prime} \& 2^{\prime} \text { at } P_3=P_4\right)
then, \dot{ m }_1\left( h _1^{\prime}- h _1\right)=\dot{ m }_2\left( h _2- h _2^{\prime}\right)
From s _2^{\prime}= s _2=0.4369+ x _2^{\prime} \times 8.0164 ; \quad x _2^{\prime}=0.7985
Also h _4=125.79 \,kJ / kg , \quad h _7=42.01 \,kJ / kg , \quad h _9=83.96 \,kJ / kg
Mixing of streams 4 & 9 ⇒ 5 & 10:
\left(\dot{ m }_1+\dot{ m }_2\right) h _4+\dot{ m }_7 h _9=\left(\dot{ m }_7+\dot{ m }_1+\dot{ m }_2\right) h _{5=10}
Flash chamber \left(\text { since } h _6= h _5\right): \quad\left(\dot{ m }_7+\dot{ m }_1\right) h _{5=10}=\dot{ m }_1 h _1+\dot{ m }_7 h _1
⇒ using the primary stream \dot{ m }_2=1 \,kg / s :
Solving, \dot{ m }_7=202.627 \& h _5=84.88 \,kJ / kg
LP pump: – w _{\text {LP P }}=0.0010(4.246-1.2276)=0.003\, kJ / kg
h _8= h _7- w _{\text {LP P }}=42.01+0.003=42.01 \,kJ / kg
Chiller: \dot{ Q }_{ L }=\dot{ m }_7\left( h _9- h _8\right)=202.627(83.96-42.01)=8500 \,kW \quad\left(\text { for } \dot{ m }_2=1\right)
HP pump: – w _{ HPP }=0.001002(475.8-4.246)=0.47 \,kJ / kg
h _{11}= h _{10}- w _{ HPP }=84.88+0.47=85.35 \,kJ / kg
Boiler: \dot{ Q }_{11}=\dot{ m }_{11}\left( h _2- h _{11}\right)=1(2746.5-85.35)=2661.1 \,kW
\Rightarrow \dot{ Q }_{ L } / \dot{ Q }_{ H }=8500 / 2661.1= 3 . 1 9 4
b) Jet eject. eff. =\left(\dot{ m }_1 / \dot{ m }_2\right)_{ ACT } /\left(\dot{ m }_1 / \dot{ m }_2\right)_{ IDEAL }=0.20
Solving, \dot{ m }_7=39.762 \quad \& \quad h _5= h _{10}=85.69 \,kJ / kg
Then, \dot{ Q }_{ L }=39.762(83.96-42.01)=1668 \,kW
\begin{aligned}& h _{11}=85.69+0.47=86.16 \,kJ / kg \\& \dot{ Q }_{ H }=1(2746.5-86.16)=2660.3 \,kW \\& \& \dot{ Q }_{ L } / \dot{ Q }_{ H }=1668 / 2660.3= 0 . 6 2 7\end{aligned}