Answer:
a)P(abs)=104.15 KPa
b)P(abs)=103.082 KPa
Explanation:
Given that
P(atm) = 1.013 x 10⁵ Pa
P(atm) = 101.3 KPa
Density ρ = 1188 kg/m³
We know that
Absolute pressure = Gauge pressure + Atmospheric pressure
P(abs)=P(gauge)+P(atm)
When h= 0.24 m
Gauge pressure
P(gauge) = ρ g h
P(gauge) = 1188 x 10 x 0.24 = 2851.2 Pa
P(gauge) = 2.851 KPa
P(abs)=P(gauge)+P(atm)
P(abs)=101.3 + 2.851 KPa
P(abs)=104.15 KPa
When h= 0.15 m
Gauge pressure
P(gauge) = ρ g h
P(gauge) = 1188 x 10 x 0.15 = 1782 Pa
P(gauge) = 1.782 KPa
P(abs)=P(gauge)+P(atm)
P(abs)=101.3 + 1.782 KPa
P(abs)=103.082 KPa