15 pts R134a enters a throttle as a saturated liquid at 36 C

(15 pts) R-134a enters a throttle as a saturated liquid at 36 °C at a mass flow rate of 0.26 kg/s and is throttled to -8 °C. It then enters a counter-flow heat exchanger and exits as a saturated vapor at the same pressure as its inlet to the heat exchanger. 3. A separate stream of H20 enters the heat exchanger at 20 °C, 1 bar and exits at 10 °C,1 bar. Make the typical assumptions for these devices. Find the volumetric flow rate of HhO at its exit from the heat exchanger, Vs (m3/s). R-134a Saturated Liquid T1-36\"C mi = 0.26 kg/s Saturated Vapor T,--8°C T-10 C Ps-1 bar H20 T-20 C P4=1 bar

Solution

we are going to do a balance of energy and mass

so we need the entlaphy for all the points

point 1

R 134 a

T=36°C, satured liquid

h1=102.33 KJ/kg

the entalphy of the point 1 is the same in the point 2, so

h2= 102.33

T2=-8°C, with this 2 values we can know the preassure :

P2= 217.08 kPa,

so in the point 3

P3=P2=217.08 kPa and satured vapor, so the enthalpy using talbs is:

h3=249.72 KJ/kg

know with the water I recomend you using the app convertpad, in this you can know the properties of water without using talbs, so the point 4 is:

T4=20°C and P4=1 bar, so h4= 83.953 KJ/kg

T5=10°C and P5= 1 bar so h5= 42.09 KJ/kg and v5= 0.000995 m^3/kg

now we do the energy mass analysis

flow water * entalphy = flow r 134a * entalphy

flow water* ( 83.95-42.09)=0.26*(249.72-102.33)

flow of water= 0.915466 kg/ s

si the flow of volume is

flow of volume=( 0.000995 m^3/kg)*(0.915466 kg/s) = 0.838078 m^3/s