Question

A piece of hot metal should be cooled rapidly (quenched) to 25°C, which requires removal of 1000 kJ from the metal. The cold space that absorbs the energy could be one of three possibilities:

(1) Submerge the metal into a bath of liquid water and ice, thus melting the ice.

(2) Let saturated liquid R-22 at −20°C absorb the energy so that it becomes saturated vapor.

(3) Absorb the energy by vaporizing liquid nitrogen at 101.3 kPa pressure.

a.  Calculate the change of entropy of the cooling media for each of the three cases.

b.  Discuss the significance of the results.

Answer 1

Continuity Eq.: m₂ − m_A – m_B = 0

Energy Eq. m₂u₂ − m_Au_A – m_Bu_B = –₁W₂

Energy Eq.  m₂u₂ − m_Au_A – m_Bu_B = _–1W₂

Entropy Eq  m₂s₂ − m_As_A – m_Bs_B = ∫ dQ/T + ₁S_{2 gen} 

Process: P = Constant => ₁W₂ = ∫ PdV = P(V₂ - V₁) Q = 0

Substitute the work term into the energy equation and rearrange to get

m₂u₂ + P₂V₂ = m₂h₂ = m_Au_A + m_Bu_B+ PV₁ = m_Ah_A + m_Bh_B 

where the last rewrite used PV₁ = PV_A + PV_B.

State A1:  h_A= 3487.03 kJ/kg, s_A= 8.5132 kJ/kg K

State B1:  h_B = 2706.63 kJ/kg, s_B= 7.1271 kJ/kg K

Energy equation gives:

State 2: P₂, h₂ = 3096.83 kJ/kg => s₂ = 7.9328 kJ/kg K; T₂ = 312.2°C

With the zero heat transfer we have

₁S_{2 gen} = m₂s₂ − m_As_A – m_Bs_B

= 2 x 7.9328 – 1 x 8.5132 – 1 × 7.1271 = 0.225 kJ/K