Chapter 3, Problem 3.165P

To determine

The rate of shaft work done for the device.

The direction of shaft work done for the device.

Answer to Problem 3.165P

The rate of shaft work done for the device is $15.3\text{kW}$.

The direction of shaft work done for the device is negative . The work is done on the shaft for the device.

Explanation of Solution

Given information:

The flow of water is steady and isothermal. The diameter at the section (1) is $9\text{cm}$, flow rate at section (1) is $220{\text{m}}^3/\text{h}$ and the pressure at section (1) is $150\text{kPa}$ . The diameter at the section (2) is $7\text{cm}$, flow rate at section (2) is $100{\text{m}}^3/\text{h}$ and the pressure at section (2) is $225\text{kPa}$ . The diameter at the section (3) is $4\text{cm}$, and the pressure at section (3) is $265\text{kPa}$.

Write the expression for the flow rate at section (3).

$Q_3=Q_1-Q_2$ …… (I)

Here, the flow rate at section (1) is $Q_1$, the flow rate at section (2) is $Q_2$ and the flow rate at section (3) is $Q_3$.

Write the expression for the steady and isothermal work.

$-\dot{W}=-{\displaystyle \int \left(h+\frac{p}{\rho }+\frac{V^2}{2}+gz\right)}\rho \left(V.n\right)dA$ …… (II)

Here, the shaft work done per unit time by the device is $\dot{W}$, the specific enthalpy is $h$, the velocity is $V$, the datum height is $z$, the density is $\rho$, the acceleration due to gravity is $g$ and the area is $A$.

Since the Equation (II) is the general form of the work transfer, so the work done at individual section is written by the above expression.

Write the expression for the work transfer at section (1).

${\dot{W}}_1=-\left(\frac{p_1}{\rho }+\frac{V_1^2}{2}\right)\rho V_1{A}_1$

Here, the work transfer at section (1) is ${\dot{W}}_1$, the pressure at section (1) is $p_1$, the velocity at section (1) is $V_1$ and the area at section (1) is $A_1$.

Write the expression for the work transfer at section (2).

${\dot{W}}_2=\left(\frac{p_2}{\rho }+\frac{V_2^2}{2}\right)\rho V_2{A}_2$

Here, the work transfer at section (2) is ${\dot{W}}_2$, the pressure at section (2) is $p_2$, the velocity at section (2) is $V_2$ and the area at section (2) is $A_2$.

Write the expression for the work transfer at section (3).

${\dot{W}}_3=\left(\frac{p_3}{\rho }+\frac{V_3^2}{2}\right)\rho V_3{A}_3$

Here, the work transfer at section (3) is ${\dot{W}}_3$, the pressure at section (3) is $p_3$, the velocity at section (3) is $V_3$ and the area at section (3) is $A_3$.

Write the expression for net work transfer by the device.

${\dot{W}}_{\text{net}}=-\left({\dot{W}}_1+{\dot{W}}_2+{\dot{W}}_3\right)$ …… (III)

Here, the net work done is ${\dot{W}}_{\text{net}}$.

Substitute $-\left(\frac{p_1}{\rho }+\frac{V_1^2}{2}\right)\rho V_1{A}_1$ for ${\dot{W}}_1$, $\left(\frac{p_2}{\rho }+\frac{V_2^2}{2}\right)\rho V_2{A}_2$ for ${\dot{W}}_2$ and $\left(\frac{p_3}{\rho }+\frac{V_3^2}{2}\right)\rho V_3{A}_3$ for ${\dot{W}}_3$ in Equation (III).

${\dot{W}}_{\text{net}}=-\left(\frac{p_1}{\rho }+\frac{V_1^2}{2}\right)\rho V_1{A}_1+\left(\frac{p_2}{\rho }+\frac{V_2^2}{2}\right)\rho V_2{A}_2+\left(\frac{p_3}{\rho }+\frac{V_3^2}{2}\right)\rho V_3{A}_3$ …… (IV)

Write the expression for the flow rate at section (1).

$Q_1=A_1{V}_1$ …… (V)

Write the expression for the area at section (1).

$A_1=\frac{\pi }{4}{D}_1^2$

Here, the diameter is $D_1$.

Substitute $\frac{\pi }{4}{D}_1^2$ for $A_1$ in Equation (V).

$Q_1=\frac{\pi }{4}{D}_1^2{V}_1$ …… (VI)

Write the expression for the flow rate at section (2).

$Q_2=A_2{V}_2$ …… (VII)

Write the expression for the area at section (2).

$A_2=\frac{\pi }{4}{D}_2^2$

Here, the diameter is $D_2$.

Substitute $\frac{\pi }{4}{D}_2^2$ for $A_2$ in Equation (VII).

$Q_2=\frac{\pi }{4}{D}_2^2{V}_2$ …… (VIII)

Write the expression for the flow rate at section (3).

$Q_3=A_3{V}_3$ …… (IX)

Write the expression for the area at section (2).

$A_3=\frac{\pi }{4}{D}_3^2$

Here, the diameter is $D_3$.

Substitute $\frac{\pi }{4}{D}_3^2$ for $A_3$ in Equation (IX).

$Q_3=\frac{\pi }{4}{D}_3^2{V}_3$ …… (X)

Substitute $A_1{V}_1$ for $Q_1$, $A_2{V}_2$ for $Q_2$ and $A_3{V}_3$ for $Q_3$ in Equation (IV).

${\dot{W}}_{\text{net}}=-\left(\frac{p_1}{\rho }+\frac{V_1^2}{2}\right)\rho Q_1+\left(\frac{p_2}{\rho }+\frac{V_2^2}{2}\right)\rho Q_2+\left(\frac{p_3}{\rho }+\frac{V_3^2}{2}\right)\rho Q_3$ …… (XI)

Calculation:

Substitute $220{\text{m}}^3/\text{h}$ for $Q_1$ and $100{\text{m}}^3/\text{h}$ for $Q_2$ in Equation (I).

$\begin{array}{c}{Q}_3=\left(220{\text{m}}^3/\text{h}\right)-\left(100{\text{m}}^3/\text{h}\right)\\ =120{\text{m}}^3/\text{h}\end{array}$

Substitute $220{\text{m}}^3/\text{h}$ for $Q_1$ and $9\text{cm}$ for $D_1$ in Equation (VI).

$\begin{array}{l}\left(220{\text{m}}^3/\text{h}\right)=\frac{\pi }{4}{\left(9\text{cm}\right)}^2{V}_1\\ V_1=\frac{\left(220{\text{m}}^3/\text{h}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)}{\frac{\pi }{4}\left(81{\text{cm}}^2\right)\left(\frac{1{\text{m}}^2}{10000{\text{cm}}^2}\right)}\\ V_1=\frac{0.061{\text{m}}^3/\text{s}}{6.36\times {10}^{-3}{\text{m}}^2}\\ V_1=9.58\text{m}/\text{s}\end{array}$

Substitute $100{\text{m}}^3/\text{h}$ for $Q_2$ and $7\text{cm}$ for $D_2$ in Equation (VI).

$\begin{array}{l}\left(100{\text{m}}^3/\text{h}\right)=\frac{\pi }{4}{\left(7\text{cm}\right)}^2{V}_2\\ V_2=\frac{\left(100{\text{m}}^3/\text{h}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)}{\frac{\pi }{4}\left(49{\text{cm}}^2\right)\left(\frac{1{\text{m}}^2}{10000{\text{cm}}^2}\right)}\\ V_2=\frac{0.028{\text{m}}^3/\text{s}}{3.848\times {10}^{-3}{\text{m}}^2}\\ V_2=7.28\text{m}/\text{s}\end{array}$

Substitute $120{\text{m}}^3/\text{h}$ for $Q_3$ and $4\text{cm}$ for $D_3$ in Equation (VI).

$\begin{array}{l}\left(120{\text{m}}^3/\text{h}\right)=\frac{\pi }{4}{\left(4\text{cm}\right)}^2{V}_3\\ V_3=\frac{\left(120{\text{m}}^3/\text{h}\right)\left(\frac{1\text{h}}{3600\text{s}}\right)}{\frac{\pi }{4}\left(16{\text{cm}}^2\right)\left(\frac{1{\text{m}}^2}{10000{\text{cm}}^2}\right)}\\ V_3=\frac{0.033{\text{m}}^3/\text{s}}{1.26\times {10}^{-3}{\text{m}}^2}\\ V_3=26.26\text{m}/\text{s}\end{array}$

Substitute $150\text{kPa}$ for $p_1$, $225\text{kPa}$ for $p_2$, $265\text{kPa}$ for $p_3$, $9.58\text{m}/\text{s}$ for $V_1$, $7.28\text{m}/\text{s}$ for $V_2$, $26.26\text{m}/\text{s}$ for $V_3$, $998\text{kg}/{\text{m}}^3$ for $\rho$, $220{\text{m}}^3/\text{h}$ for $Q_1$, $100{\text{m}}^3/\text{h}$ for $Q_2$ and $120{\text{m}}^3/\text{h}$ for $Q_3$ in Equation (XI).

Further solve the above expression.

$\begin{array}{c}{\dot{W}}_{\text{net}}=-\left(15324.10\text{W}\right)\left(\frac{1\text{kW}}{1000\text{W}}\right)\\ =-15.3\text{kW}\end{array}$

Since the sign of the work done by the device is negative, so the work done on the shaft for the given device.

Thus, the work is done on the system.

Conclusion:

The rate of shaft work done for the device is $15.3\text{kW}$.

The direction of shaft work done for the device is negative . The work is done on the shaft for the device.