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Len [333]
3 years ago
6

Refrigerant 134a enters a well-insulated nozzle at 200 lbf/in.2, 200°F, with a velocity of 120 ft/s and exits at 50 lbf/in.2 wit

h a velocity of 1500 ft/s. For steady-state operation, and neglecting potential energy effects, determine the temperature, in °F, and the quality of the refrigerant at the exit.
Physics
1 answer:
riadik2000 [5.3K]3 years ago
5 0

Answer:

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

Explanation:

From superheated R 134 a properties table

At 200 lb/in^2 and 200 degree F

h_1 = 138.99 Btu/lbm

steady flow energy equation is givena s

h_1 + \frac{v_1^2}{2}  = h_2 + \frac{v_2^2}{2}

138.99 + \frac{120^2}{2\times 25037} = h_2 + \frac{1500^2}{2 \times 25037}

h_2 = 94.344 Btu/lbm

At 90 lb/in2 Tsat = 72.78 degree F

h_f = 35.715 Btu/lbm

hfg  = 77.345 Btu/lbm

h = hf + x hfg

94.344 = 35.715+ x \times 77.345

solving for x we get

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

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For this case we can use the Compton shift equation given by:

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For this case we know the following values:

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So then if we replace we got:

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Part b

For this cas we can calculate the wavelength of the phton with this formula:

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Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

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