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.

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 $12{\text{m}}^{\text{3}}$ is 100 min, 200 min respectively and it is divided into “Given”, “Find” and “Solution” sections.

Explanation of Solution

Given:

Volume of each tank $\left(V\right)$ is $12{\text{m}}^{\text{3}}$.

Density of water $\left(\rho \right)$ is $1000\frac{\text{kg}}{{\text{m}}^{\text{3}}}$.

Rate at which fluid enters tank is $2\frac{\text{kg}}{\text{s}}$.

Find:

The mass of water stored after 5 minutes in each tanks and also find the time taken to fill the tanks of volume $12{\text{m}}^{\text{3}}$.

Solution:

Refer figure 1. Water enters the tank (a) at rate of $2\frac{\text{kg}}{\text{s}}$

Formula for the relationship of conservation of mass is,

$\left(\begin{array}{l}\text{The}\text{rate}\text{of}\text{accumulation}\text{or}\\ \text{depletion}\text{of}\text{the}\text{mass}\text{of}\text{fluid}\\ \text{within}\text{the}\text{given}\text{control}\text{volume}\end{array}\right)\text{=}\left(\begin{array}{l}\text{The}\text{rate}\text{at}\text{which}\\ \text{the}\text{fluid}\text{enters}\text{a}\\ \text{control}\text{volume}\end{array}\right)-\left(\begin{array}{l}\text{The}\text{rate}\text{at}\text{which}\\ \text{the}\text{fluid}\text{leaves}\text{the}\\ \text{control}\text{volume}\end{array}\right)$ (1)

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

$\left(\begin{array}{l}\text{The}\text{rate}\text{of}\text{accumulation}\text{or}\\ \text{depletion}\text{of}\text{the}\text{mass}\text{of}\text{fluid}\\ \text{within}\text{the}\text{given}\text{control}\text{volume}\end{array}\right)=\left(\frac{\begin{array}{l}\text{changes}\text{of}\text{mass}\text{inside}\\ \text{the}\text{control}\text{volume}\end{array}}{\text{change}\text{in}\text{time}}\right)$                             (2)

Substitute equation (2) in equation (1).

$\left(\frac{\begin{array}{l}\text{changes}\text{of}\text{mass}\text{inside}\\ \text{the}\text{control}\text{volume}\end{array}}{\text{change}\text{in}\text{time}}\right)\text{=}\left(\begin{array}{l}\text{The}\text{rate}\text{at}\text{which}\\ \text{the}\text{fluid}\text{enters}\text{a}\\ \text{control}\text{volume}\end{array}\right)-\left(\begin{array}{l}\text{The}\text{rate}\text{at}\text{which}\\ \text{the}\text{fluid}\text{leaves}\text{the}\\ \text{control}\text{volume}\end{array}\right)$ (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,

$\left(\frac{\begin{array}{l}\text{changes}\text{of}\text{mass}\text{inside}\\ \text{the}\text{control}\text{volume}\end{array}}{\text{change}\text{in}\text{time}}\right)\text{=}\left(2\frac{\text{kg}}{\text{s}}\right)-\left(0\right)$ (4)

As the water is filled inside the control volume at rate of $2\frac{\text{kg}}{\text{s}}$, then there will be change in mass inside the control volume after 5 minutes. Therefore,

$\begin{array}{c}\left(\begin{array}{l}\text{changes}\text{of}\text{mass}\text{inside}\text{the}\\ \text{control}\text{volume}\text{after}\text{5}\text{minutes}\end{array}\right)=\left(2\frac{\text{kg}}{\text{s}}\right)\left(5\min \right)\left(\frac{60\text{s}}{1\min }\right)\left[\therefore 1\min =60s\right]\\ =600\text{kg}\end{array}$

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

$\begin{array}{c}\text{mass}\text{=}\text{density}\text{of}\text{water×}\text{volume}\\ \text{=}\text{1000}\frac{\text{kg}}{{\text{m}}^{\text{3}}}\times 12{\text{m}}^{\text{3}}\\ =12,000\text{kg}\end{array}$

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

$\begin{array}{c}\text{Time}\text{taken}\text{=}\frac{\text{mass}\text{of}\text{tank}}{\text{flow}\text{of}\text{water}}\\ =\frac{12,000\text{kg}}{2\frac{\text{kg}}{\text{s}}}\\ =6000\text{s}\end{array}$

Convert the unit of time taken from seconds to minutes.

$\begin{array}{c}\text{Time}\text{taken}=\left(6000\text{s}\right)\left(\frac{1\min }{60\text{s}}\right)\left[\therefore 1\min =60\text{s}\right]\\ =100\min \end{array}$

Refer to Figure 2. Water enters the tank (b) at rate of $2\frac{\text{kg}}{\text{s}}$ and leaves the tank (b) at rate of $1\frac{\text{kg}}{\text{s}}$.

From equation (3),

$\left(\frac{\begin{array}{l}\text{changes}\text{of}\text{mass}\text{inside}\\ \text{the}\text{control}\text{volume}\end{array}}{\text{change}\text{in}\text{time}}\right)\text{=}\left(2\frac{\text{kg}}{\text{s}}\right)-\left(1\frac{\text{kg}}{\text{s}}\right)$ (5)

As the water is filled inside the tank at rate of $2\frac{\text{kg}}{\text{s}}$ and leaves the tank at rate of $1\frac{\text{kg}}{\text{s}}$ , then there will be change in mass inside the control volume after 5 minutes. Therefore, equation (5) becomes,

$\begin{array}{c}\left(\frac{\begin{array}{l}\text{changes}\text{of}\text{mass}\text{inside}\\ \text{the}\text{control}\text{volume}\end{array}}{\text{change}\text{in}\text{time}}\right)=\left[\left(2\frac{\text{kg}}{\text{s}}\right)-\left(1\frac{\text{kg}}{\text{s}}\right)\right]\left(5\min \right)\left(\frac{60\text{s}}{1\min }\right)\left[\therefore 1\min =60\text{s}\right]\\ =\left(1\frac{\text{kg}}{\text{s}}\right)\left(5\min \right)\left(\frac{60\text{s}}{1\min }\right)\\ =300\text{kg}\end{array}$

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

$\begin{array}{c}\text{Time}\text{taken}\text{=}\frac{\text{mass}\text{of}\text{tank}}{\text{flow}\text{of}\text{water}}\\ =\frac{12,000\text{kg}}{\left[\left(2\frac{\text{kg}}{\text{s}}\right)-\left(1\frac{\text{kg}}{\text{s}}\right)\right]}\\ =\frac{12,000\text{kg}}{1\frac{\text{kg}}{\text{s}}}\\ =12,000\text{s}\end{array}$

Convert the unit of time taken from seconds to minutes.

$\begin{array}{c}\text{Time}\text{taken}=\left(12,000\text{s}\right)\left(\frac{1\min }{60\text{s}}\right)\left[\therefore 1\min =60\text{s}\right]\\ =200\min \end{array}$

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 $12{\text{m}}^{\text{3}}$ is 100 min, 200 min respectively and it is divided into “Given”, “Find” and “Solution” sections.