Question

The gauge pressure inside a vessel is ‐40kPa, at an elevation of 5000m. a) What is the absolute pressure? b) At this elevation, what temperature does water boil?

Answer 1

Answer:

a) Pabs = 48960 KPa

b) T = 433.332 °C

Explanation:

• Pabs = Pgauge + d*g* h

∴ d = 1000 Kg/m³

∴ g = 9.8 m/s²

∴ h = 5000 m

∴ P gauge = - 40 KPa * ( 1000 Pa / KPa ) = - 40000 Pa; Pa≡Kg/m*s²

⇒ Pabs = - 40000 Kg/ms² + ( 1000 Kg/m³ * 9.8 m/s² * 5000 m )

⇒ Pabs = 48960000 Pa = 48960 KPa

a) at that height and pressure, we find the temperature at which the water boils by means of an almost-exponential graph which has the following equation:

P(T) = 0.61094 exp ( 17.625*T / ( T + 243.04 ))......P (KPa) ∧ T (°C)....from literature

∴ P = 48960 KPa

⇒ ( 48960 KPa / 0.61094 ) = exp ( 17.625T / (T+ 243.04))

⇒ 80138.803 = exp ( 17.625T / ( T + 243.04))

⇒ Ln ( 80138.803) = 17.625T / ( T + 243.04))

⇒ 11.292 * ( T + 243.04 ) = 17.625T

⇒ 11.292T + 2744.289 = 17.625T

⇒ 2744.289 = 17.625T - 11.292T

⇒ 2744.289 = 6.333T

⇒ T = 433.332 °C