Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-20
Thread Content
Process Engineering Section – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-20: Discussion question: 332. There is a volume of air with pressure P1=0.14 MPa, temperature T1=173.15 K, and compression coefficient Z=1. Calculate the density of this air under operating conditions (given that the density of dry air at t20=20°C and Pn=0.1013 MPa is ρ20=1.2046 kg/m3). 5 points will be given to those who provide the correct answer; additional points will be awarded for a correct calculation process or detailed explanation
From P1*V1/P2*V2=T1/T2, we obtain ρ1=P1*T2*ρ2/P2*T1=0.14*293.15*1.2046/0.1013*173.15=2.8186 (kg/m3)
P1*V1/P2*V2=T1/T2; ρ1=P1*T2*ρ2/P2*T1=0.14*293.15*1.2046/0.1013*173.15=2.8186 (kg/m3)
There is air with a pressure of P1=0.14 MPa, a temperature of T1=173.15 K, and a compression coefficient of Z=1. Calculate the density of this air under operating conditions (given that the density of dry air at t20=20°C and Pn=0.1013 MPa is ρ20=1.2046 kg/m3). Solution: ρ1=ρ20 × (P1T20/T1PnZ) = 1.2046 × ((0.14×293.15)/(173.15×0.1013×1)) = 2.8186 kg/m3. Answer: The density of this air under operating conditions is 2.8186 Kg/m3.
2.8186Kg/m3
P1*V1/P2*V2=T1/T2; ρ1=P1*T2*ρ2/P2*T1=0.14*293.15*1.2046/0.1013*173.15=2.8186 (kg/m3)
ρ1 = P1*T2*ρ2/P2*T1 = 0.14*293.15*1.2046/0.1013*173.15 = 2.8186 (kg/m3)
There is air with a pressure of P1=0.14 MPa, a temperature of T1=173.15 K, and a compression coefficient of Z=1. Calculate the density of this air under operating conditions (given that the density of dry air at t20=20°C and Pn=0.1013 MPa is ρ20=1.2046 kg/m3). Solution: ρ1=ρ20 × (P1T20/T1PnZ) = 1.2046 × ((0.14×293.15)/(173.15×0.1013×1)) = 2.8186 kg/m3. Answer: The density of this air under operating conditions is 2.8186 Kg/m3.
There is air with a pressure of P1=0.14 MPa, a temperature of T1=173.15 K, and a compression coefficient of Z=1. Calculate the density of this air under operating conditions (given that the density of dry air at t20=20°C and Pn=0.1013 MPa is ρ20=1.2046 kg/m3). Solution: ρ1=ρ20 × (P1T20/T1PnZ) = 1.2046 × ((0.14×293.15)/(173.15×0.1013×1)) = 2.8186 kg/m3. Answer: The density of this air under operating conditions is 2.8186 Kg/m3.
P1*V1/P2*V2=T1/T2; ρ1=P1*T2*ρ2/P2*T1=0.14*293.15*1.2046/0.1013*173.15=2.8186 (kg/m3)
P1*V1/T1=P2*V2/T2, ρ=1/V, ρ1=P1*T2*ρ2/P2*T1=2.8186 (kg/m3)