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Given: r=1.2 m, Q=2 kW, L_v=510 kJ/kg, V_v=5 m³, R=461 J/kg·K, T=160°C=433 K
Tank surface area: A = 4 r² = 4*3.1416*1.2² = 18.1 m²
Boil-off rate: = Q / L_v = 2000 / 510,000 = 0.00392 kg/s
Pressure rise rate (ideal gas): = ( *R*T)/V_v = 0.00392*461*433 /5 = 156 J/(m³·s) Pa/s
= 156 Pa/s
Added 15 days ago|2/18/2026 11:49:31 AM
This answer has been confirmed as correct and helpful.
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dP/dt = ( * R * T) / V
dP/dt = (0.24448 mol/s * 8.314 J/(mol·K) * 433.15 K) / 5 m³
dP/dt = (0.24448 * 8.314 * 433.15) / 5
dP/dt = 882.06 / 5
dP/dt 176.41 Pa/s
The pressure rise rate in the sealed 5 m³ vapor space is approximately 176.41 Pa/s (or 0.1764 kPa/s).
Added 15 days ago|2/18/2026 11:49:40 AM
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[Deleted]
Added 15 days ago|2/18/2026 12:16:10 PM