In the two-pipe system, water flows continuously from the inner pipe and air from the outer pipe (steady-state). ˙The pressure change in both pipes is negligible. In addition, heat interaction with the environment, kinetic and potential energy changes are also neglected. In this system shown in Figure 1, the water flowing from the inlet number 1 is in the state of saturated vapor (gas). In this case, calculate the outlet temperature (T2) for the given system, using the ideal gas approximation for air. (Hint: it will be solved iteratively.)
In the two-pipe system, water flows continuously from the inner pipe and air from the outer pipe (steady-state). ˙The pressure change in both pipes is negligible. In addition, heat interaction with the environment, kinetic and potential energy changes are also neglected. In this system shown in Figure 1, the water flowing from the inlet number 1 is in the state of saturated vapor (gas). In this case, calculate the outlet temperature (T2) for the given system, using the ideal gas approximation for air. (Hint: it will be solved iteratively.)
![1
saturated
water vapor
P₁=1 bar
m₁=10 kg/s
3
Air
P3=1 bar
T3-1200 K
m3=5 kg/s
4
T₁=T₂
2
T₂](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe5cd22a2-4ff1-4a93-9af3-2bc2dd53b106%2Fbfe0d7f8-bd40-41bd-8f9d-6fe17abddb70%2Fv6fmr3k_processed.jpeg&w=3840&q=75)
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Step 2 of the solution is incorrect for the answer.
calculations does not add up
The outlet temperature will be:
→T2=m˙1×T1+m˙3×T3m˙1+m˙3 =10×372.78+5×120010+5 =3727.8+600015=648.52 K ..................answer
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