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A 0.05-m3 rigid tank initially contains refrigerant-134a at 0.8 MPa and 100 percent quality. The tank is connected by a valve to a supply line that carries refrigerant-134a at 1.2 MPa and 40°C. Now the valve is opened, and the refrigerant is allowed to enter the tank. The valve is closed when it is observed that the tank contains saturated liquid at 1.2 MPa. The properties of refrigerant are v1 = 0.02565 m3/kg, u1 = 246.82 kJ/kg, v2 = 0.0008935 m3/kg, u2 = 116.72 kJ/kg, and hi = 108.28 kJ/kg.A) Determine the heat transfer (Q). B) Determine the mass of the refrigerant that has entered the tank.

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Answer:

A= 203 KJ

B= 54 Kg

Explanation:

The initial specific volumes and internal energies are obtained from A-12 for a given pressure and state. The enthalpy of the refrigerant in the supply line is determined using the saturated liquid approximation for the given temperature with data from A-11. The mass that has entered the tank is:

Δm = m₂ – m₁

= V(1/α₂ – 1/α₁)

= 0.05 (1/0.0008935 – 1/ 0.025645)Kg

= 54Kg

The heat transfer is obtained from the energy balance:

ΔU= [tex]m_i_n[/tex] [tex]h_i_n[/tex]+  [tex]Q_n_e_t[/tex]

m₂u₂ – m₁u₂ = [tex]m_i_n[/tex][tex]h_i_n[/tex] + [tex]Q_n_e_t[/tex]

[tex]Q_n_e_t[/tex]= m₂u₂ – m₁u₁ – [tex]m_i_n[/tex]

= V/α₂u₂ - V/α₁u₁ – [tex]m_i_n[/tex]

=(0.05/0.0008935 . 116.72 – 0.05/0.025645 . 246.82 – 54.108.28) Kj

= 203 KJ

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