ENGINEERING FUNDAMENTALS
ENGINEERING FUNDAMENTALS
6th Edition
ISBN: 9781337705011
Author: MOAVENI
Publisher: CENGAGE L
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Chapter 4, Problem 27P
To determine

Divide and show the steps of Problem Example 9.3 in chapter 9 from the textbook into the format of “Given”, “Find” and “Solution” sections.

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Answer to Problem 27P

The mass of water stored after 5 minutes in tank (a) is 600 kg and in tank (b) is 300 kg, and the time taken to fill tank (a) and tank (b) of volume 12m3 is 100 min, 200 min respectively and it is divided into “Given”, “Find” and “Solution” sections.

Explanation of Solution

Given:

Volume of each tank (V) is 12m3.

Density of water (ρ) is 1000kgm3.

Rate at which fluid enters tank is 2kgs.

Find:

The mass of water stored after 5 minutes in each tanks and also find the time taken to fill the tanks of volume 12m3.

Solution:

Refer figure 1. Water enters the tank (a) at rate of 2kgs

Formula for the relationship of conservation of mass is,

(Therateofaccumulationordepletionofthemassoffluidwithinthegivencontrolvolume)=(Therateatwhichthefluidentersacontrolvolume)(Therateatwhichthefluidleavesthecontrolvolume) (1)

Consider the expression for rate of accumulation or depletion of fluid in the control volume is,

(Therateofaccumulationordepletionofthemassoffluidwithinthegivencontrolvolume)=(changesofmassinsidethecontrolvolumechangeintime)                             (2)

Substitute equation (2) in equation (1).

(changesofmassinsidethecontrolvolumechangeintime)=(Therateatwhichthefluidentersacontrolvolume)(Therateatwhichthefluidleavesthecontrolvolume) (3)

In figure 1, there is no disposal of water from the tank through any opening. Therefore, the rate at which the fluid leaves the control volume is zero. Therefore, equation (3) becomes,

(changesofmassinsidethecontrolvolumechangeintime)=(2kgs)(0) (4)

As the water is filled inside the control volume at rate of 2kgs, then there will be change in mass inside the control volume after 5 minutes. Therefore,

(changesofmassinsidethecontrolvolumeafter5minutes)=(2kgs)(5min)(60s1min)            [1min=60s]=600kg

Formula to determine how much mass each tank can hold is,

mass=densityofwater×volume=1000kgm3×12m3=12,000kg

Formula to determine the time taken to fill the tank (a) is,

Timetaken=massoftankflowofwater=12,000kg2kgs=6000s

Convert the unit of time taken from seconds to minutes.

Timetaken=(6000s)(1min60s)[1min=60s]=100min

Refer to Figure 2. Water enters the tank (b) at rate of 2kgs and leaves the tank (b) at rate of 1kgs.

From equation (3),

(changesofmassinsidethecontrolvolumechangeintime)=(2kgs)(1kgs) (5)

As the water is filled inside the tank at rate of 2kgs and leaves the tank at rate of 1kgs , then there will be change in mass inside the control volume after 5 minutes. Therefore, equation (5) becomes,

(changesofmassinsidethecontrolvolumechangeintime)=[(2kgs)(1kgs)](5min)(60s1min)[1min=60s]=(1kgs)(5min)(60s1min)=300kg

Formula to determine the time taken to fill the tank (b) is,

Timetaken=massoftankflowofwater=12,000kg[(2kgs)(1kgs)]=12,000kg1kgs=12,000s

Convert the unit of time taken from seconds to minutes.

Timetaken=(12,000s)(1min60s)[1min=60s]=200min

Conclusion:

Hence, the mass of water stored after 5 minutes in tank (a) is 600 kg and in tank (b) is 300 kg, and the time taken to fill tank (a) and tank (b) of volume 12m3 is 100 min, 200 min respectively and it is divided into “Given”, “Find” and “Solution” sections.

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