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skelet666 [1.2K]
2 years ago
8

R-134a vapor enters into a turbine at 250 psia and 175°F. The temperature of R-134a is reduced to 20°F in this turbine while its

specific entropy remains constant. Determine the change in the enthalpy of R-134a as it passes through the turbine.
Engineering
1 answer:
attashe74 [19]2 years ago
5 0

Answer:

Δ enthalpy = -23 Btu/Ibm

Explanation:

Given data:

Pressure ( P1 ) = 250 psi

Initial Temperature ( T1 ) = 175°F

Final temperature ( T2 ) = 20°F

<u>Calculate the change in the enthalpy of R-134a </u>

From R-134 table

h1 = 129.85 Btu/Ibm

s1 = 0.23281 Btu/Ibm.R

note : entropy is constant  

hence ; s1 = s2

by interpolation  ; h2 = 106.95

Δ enthalpy = h2 - h1

                  =  ( 106.95 - 129.85 ) = -23 Btu/Ibm

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Answer:

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Explanation:

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Now

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As we know that

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

1 = (1,000 × 0.25 × d) ÷ 0.0001

d = (1 × 0.001) ÷ (1,000 × 0.25)

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d = (250,000 × 0.001) ÷ (1,000 × 0.25)

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An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°
Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

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for 0.225% carbon concentration following formula is used

\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})

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x = 2.39\times 10^{-3} m

x = 0.002395 mm

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