. A liquid mixture containing 25.0 wt% n-octane (ρoct = 703 kg/m3 ) and the balance...

. A liquid mixture containing 25.0 wt% n-octane (ρoct = 703 kg/m3 ) and the balance n-decane (ρdec = 730 kg/m3 ) flows into a tank mounted on a balance. The mass in kg indicated by the scale is plotted against time. The data fall on a straight line that passes through the points (t = 3 min, m = 225 kg) and (t = 10 min, m = 510 kg). (a) Estimate the volumetric flow rate of the liquid mixture. (b) What does the empty tank weigh?

Solutions

Expert Solution

A liquid mixture contains 25 wt% n-octane and 75% n-decane

Average density of mixture can be calculated as = (0.25 * 703) +(0.75 * 730)

= 723.25 (kg/m³)

Now it is given that , in (10-3=7) minutes (510-225=285) kg of weight of tank is increased, means 285 kg of liquid mixture is poured into the tank in 7 minutes

Volume of 285 kg of liquid mixture = (285 / 723.25)

= 0.39405 m³

Means 0.39405 m³ of liquid mixture is poured into the tank in 7 minutes

Therefore volumetric flowrate = 0.39405 / 7

= 0.0563 (m³ / minutes)

b.)

Now as we know that

time (minutes)	Mass (kg)
3	225
10	510

So, equation of straight line satisfying these points

.....................................(1)

Equation 1 is the required equation

Where 'm' is the mass of tank in kg

't' is time in minutes

As, we know that at time = 0, tank must be empty as no liquid had been poured into the tank at time = 0 minutes.So, weight indicated at time =0 must be the weight of empty tank.

So, putting t = 0 in equation 1, we get

So,mass of empty tank at time =0 minutes is 102.857 kg.

Amanda K answered 1 year ago

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