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 − mA – mB = 0

Energy Eq. m2u2 − mAuA – mBuB = –1W2

Energy Eq. m2u2 − mAuA – mBuB = –1W2

Entropy Eq m2s2 − mAsA – mBsB = ∫ dQ/T + 1S2 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 = mAuA + mBuB+ PV1 = mAhA + mBhB

where the last rewrite used PV1 = PVA + PVB.

State A1: hA= 3487.03 kJ/kg, sA= 8.5132 kJ/kg K

State B1: hB = 2706.63 kJ/kg, sB= 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

1S2 gen = m2s2 − mAsA – mBsB

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