Question #268178

A reservoir contains 2.83 cubic meter of CO at 6895 kPa and 23.6 degrees C. A cylindrical tank with h=1.5D is filled from the reservoir to a pressure of 3697 kPa and a pressure of 3697 kPa and a temperature of 15.4 degrees C, while the pressure and temperature in the reservoir decreases to 6205 kPa and 18.3 degrees C respectively.a) the total mass of CO transferred to the tank in kg, b) the total volume of the tank in cubic meter, c) the diameter of the tank in mm


Expert's answer

Solution;

Given;

Vco=2.83m3V_{co}=2.83m^3

Pr1=6895kPaP_{r1}=6895kPa

Pr2=6205kPaP_{r2}=6205kPa

Pt=3697kPaP_t=3697kPa

Tr2=18.3°c=291.3KT_{r2}=18.3°c=291.3K

Tt=15.4°c=288.4KT_t=15.4°c=288.4K

Tr1=23.6°c=296.6KT_{r 1}=23.6°c=296.6K

(a)

Calculate mass in reservoir before transferring;

mr1=Pr1VcoRTr1=m_{r1}=\frac{P_{r1}V_{co}}{RT_{r1}}= 6895×2.830.2968×296.6=221.64kg\frac{6895×2.83}{0.2968×296.6}=221.64kg

Mass in reservoir at end of transfer;

mr2=6205×2.830.2968×291.3=203.09kgm_{r2}=\frac{6205×2.83}{0.2968×291.3}=203.09kg

Mass of CO transfer to the tank is;

mt=mr1mr2=221.64203.09=18.56kgm_t=m_{r1}-m_{r2}=221.64-203.09=18.56kg

(b)

Volume of the tank;

PtVt=mtRTtP_tV_t=m_tRT_t

Vt=mtRTtPt=18.56×0.2968×288.43697V_t=\frac{m_tRT_t}{P_t}=\frac{18.56×0.2968×288.4}{3697}

Vt=0.43m3V_t=0.43m^3

(c)

Diameter of the tank;

V=πr2hV=πr^2h

0.43=π4×D2×1.5D0.43=\frac{π}{4}×D^2×1.5D

D3=0.365m3D^3=0.365m^3

D=0.7147mD=0.7147m

D=714.7mmD=714.7mm




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