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A nuclear power plant generates waste thermal energy at a rate of 1000 MW = 1000 × 106 W. Assume the heat of vaporization of a water is 2.256 × 106 J/kg.A.)If this energy is transferred by hot water passing through tubes in the water in an evaporative cooling tower, how much water must evaporate to cool the plant per second?B.)How much water must evaporate to cool the plant per day?

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davidlcastiblanco

Answer:

a . m = 443.26 kg

b . m = 38. 3 x 10 ⁶ kg

Explanation:

1 Watts = 1 J / s

Thermal energy 1000 MW

a.

Water transferred by hot water per second

m = H * Δ t / lₙ

m = ( 1000 x 10 ⁶ J /s * 1 s ) / ( 2.256 x 10 ⁶ J / kg )

m = 443.26 kg

b.

Water transferred by hot water per day

Δt = 1 day * 24 hr / 1 day * 60 minute / 1 hr * 60 s / 1 minute

Δt = 86400 s

m = H * Δ t / lₙ

m = ( 1000 x 10 ⁶ J /s * 86400 s ) / ( 2.256 x 10 ⁶ J / kg )

m = 38297872.34

m = 38. 3 x 10 ⁶ kg

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