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Sunny_sXe [5.5K]
2 years ago
9

A 0.2-m^3 rigid tank equipped with a pressure regulator contains steam at 2MPa and 320C. The steam in the tank is now heated. Th

e regulator keeps the steam pressure constant by letting out some steam, but the temperature inside rises. Determine the amount of heat transferred when the steam temperatures reaches 540C.
Engineering
1 answer:
prohojiy [21]2 years ago
7 0

Answer:

Q=486.49 KJ/kg

Explanation:

Given that

V= 0.2 m³

At initial condition

P= 2 MPa

T=320 °C

Final condition

P= 2 MPa

T=540°C

From steam table

At P= 2 MPa and T=320 °C

h₁=3070.15 KJ/kg

At P= 2 MPa and T=540°C

h₂=3556.64  KJ/kg

So the heat transfer ,Q=h₂ - h₁

Q= 3556.64 - 3070.15  KJ/kg

Q=486.49 KJ/kg

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=\frac{40}{9.8*10^3*0.2} = 20.4 m

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fro Re = 7.53*10^5 and f = 0.0560

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