Chapter 8, Problem 93P

To determine

(a)

The electric power consumption of the system for pumping of water.

Answer to Problem 93P

The electric power consumption of the system for pumping of water is $4160.62\text{kW}$.

Explanation of Solution

Given information:

The temperature of the water is $110°\text{C}$, density of the water is $950.6\text{kg}/{\text{m}}^{\text{3}}$, dynamic viscosity of the water is $0.255\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$, diameter of the pipe is $60\text{cm}$, roughness of the stainless steel surfaces is $2\times {10}^{-6}\text{m}$, specific heat capacity of water is $4.229\text{kJ}/\text{kg}\cdot °\text{C}$, pump-motor efficiency is $80\%$, length of the pipe is $12\text{km}$ and volume flow rate of the water is $1.5{\text{m}}^{\text{3}}/\text{s}$.

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{\rho vD}{\mu }$........... (I)

Here, Reynolds number is $\mathrm{Re}$, density of the water is $\rho$, velocity of the water is $v$, diameter of the pipe is $D$ and dynamic viscosity of the water is $\mu$.

Write the expression for the volume flow rate of the water.

  $\dot{Q}=\frac{\pi }{4}{D}^{2}v$........... (II)

Here, the volume flow rate of the water is $\dot{Q}$.

Write the expression for the friction factor for turbulent flow.

  $\frac{1}{\sqrt{f}}=-2\log \left(\frac{\frac{\epsilon }{D}}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$........... (III)

Here, friction factor for turbulent flow is $f$ and roughness of the stainless-steel surfaces is $\epsilon$.

Write the expression for the head loss through the pipe.

  $h_f=\frac{fL{v}^2}{2gD}$........... (IV)

Here, length of the pipe is $L$, gravitational acceleration is $g$ and the head loss through the pipe is $h_f$.

Write the expression for the pressure drop through the pipe.

  $\Delta P=\rho g{h}_f$........... (V)

Here, the pressure drop through the pipe is $\Delta P$.

Write the expression for the electric power consumption of the system for pumping of water.

  ${\dot{W}}_{\text{pump}}=\dot{Q}\Delta P$........... (VI)

Here, electric power consumption of the system for pumping of water is ${\dot{W}}_{\text{pump}}$.

Write the expression for the Bernoulli Equation.

  $\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+Z_1+h_{\text{pump}}=\frac{P_2}{\rho g}+\frac{v_2^2}{2g}+Z_2+h_f$........... (VII)

Here, pressure at point $1$ is $P_1$, velocity at point $1$ in $v_1$, density of the fluid is $\rho$, gravitational acceleration is $g$, elevation of point $1$ is $Z_1$, pressure at point $2$ is $P_2$, elevation of point $2$ is $Z_2$, velocity at point $2$ in $v_2$ and head loss in pump is $h_{\text{pump}}$.

Write the expression for the motor-pump efficiency.

  ${\eta }_{\text{pump}}=\frac{{\dot{W}}_{\text{pump}}}{{\dot{W}}_{\text{electric}}}$........... (VIII)

Here, the motor-pump efficiency is ${\eta }_{\text{pump}}$.

Calculation:

Substitute $1.5{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ and $60\text{cm}$ for $D$ in Equation (II).

  $\begin{array}{l}\left(1.5{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\pi }{4}{\left(60\text{cm}\right)}^{2}v\\ \left(1.5{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\pi }{4}{\left(\left(60\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]\right)}^{2}v\\ v=5.305\text{m}/\text{s}\end{array}$

Substitute $950.6\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $5.305\text{m}/\text{s}$ for $v$, $0.255\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ and $60\text{cm}$ for $D$ in Equation (I).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(950.6\text{kg}/{\text{m}}^{\text{3}}\right)\left(5.305\text{m}/\text{s}\right)\left(60\text{cm}\right)}{\left(0.255\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{\left(950.6\text{kg}/{\text{m}}^{\text{3}}\right)\left(5.305\text{m}/\text{s}\right)\left(60\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}{\left(0.255\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =1.187\times {10}^7\end{array}$

Since, value of the Reynolds number is greater than $4000$, therefore flow is turbulent.

Substitute $1.187\times {10}^7$ for $\mathrm{Re}$, $2\times {10}^{-6}\text{m}$ for $\epsilon$ and $60\text{cm}$ for $D$ in Equation (III).

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-2\log \left(\frac{\frac{\left(2\times {10}^{-6}\text{m}\right)}{\left(60\text{cm}\right)}}{3.7}+\frac{2.51}{\left(1.187\times {10}^7\right)\sqrt{f}}\right)\\ \frac{-1}{2\sqrt{f}}=\log \left(\frac{\frac{\left(2\times {10}^{-6}\text{m}\right)}{\left(60\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}}{3.7}+\frac{2.51}{\left(1.187\times {10}^7\right)\sqrt{f}}\right)\\ f=0.00829\end{array}$

Substitute $0.00829$ for $f$, $12\text{km}$ for $L$, $5.305\text{m}/\text{s}$ for $v$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $60\text{cm}$ for $D$ in Equation (IV).

  $\begin{array}{c}{h}_f=\frac{\left(0.00829\right)\left(12\text{km}\right){\left(5.305\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(60\text{cm}\right)}\\ =\frac{\left(0.00829\right)\left(12\text{km}\right)\left[\frac{1000\text{m}}{1\text{km}}\right]{\left(5.305\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(60\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}\\ =238\text{m}\end{array}$

Substitute $238\text{m}$ for $h_f$, $950.6\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (V).

  $\begin{array}{c}\Delta P=\left(950.6\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(238\text{m}\right)\\ =\left(2218\times {10}^3\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\right)\left(1{\text{m}}^{-2}\right)\left[\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right]\left[\frac{1\text{Pa}}{1\text{N}/{\text{m}}^{\text{2}}}\right]\left[\frac{1\text{kPa}}{{10}^3\text{Pa}}\right]\\ =2219\text{kPa}\end{array}$

Substitute $1.5{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ and $2219\text{kPa}$ for $\Delta P$ in Equation (VII).

  $\begin{array}{c}{\dot{W}}_{\text{pump}}=\left(1.5{\text{m}}^{\text{3}}/\text{s}\right)\left(2219\text{kPa}\right)\\ =\left(1.5{\text{m}}^{\text{3}}/\text{s}\right)\left(2219\text{kPa}\right)\left[\frac{{10}^3\text{Pa}}{1\text{kPa}}\right]\left[\frac{1\text{N}/{\text{m}}^{\text{2}}}{\text{1}\text{Pa}}\right]\left[\frac{1\text{J}}{1\text{N}\cdot \text{m}}\right]\left[\frac{1\text{W}}{1\text{J}/\text{s}}\right]\\ =\left(3327\times {10}^3\text{W}\right)\left[\frac{1\text{kW}}{1000\text{W}}\right]\\ =3328.5\text{kW}\end{array}$

Substitute $3328.5\text{kW}$ for ${\dot{W}}_{\text{pump}}$ and $80\%$ for ${\eta }_{\text{pump}}$ in Equation (VIII).

  $\begin{array}{l}80\%=\frac{3328.5\text{kW}}{{\dot{W}}_{\text{electric}}}\\ \left(80\%\right)\left[\frac{{10}^{-2}}{1\%}\right]=\frac{3328.5\text{kW}}{{\dot{W}}_{\text{electric}}}\\ {\dot{W}}_{\text{electric}}=4160.62\text{kW}\end{array}$

Conclusion:

The electric power consumption of the system for pumping is $4160.62\text{kW}$.

To determine

(b)

The daily cost of the power consumption of the system.

Answer to Problem 93P

The daily cost of the power consumption of the system is $\$5991.2928{\text{day}}^{-1}$.

Explanation of Solution

Calculation:

The unit cost of the electricity is $\$0.06{\text{kWh}}^{-1}$.

Write the expression for the amount of daily cost of the power consumption.

  $A=ct{\dot{W}}_{\text{electric}}$........... (IX)

Here, amount of daily cost of the power consumption is $A$, unit cost is $c$ and time is $t$.

Substitute $\$0.06{\text{kWh}}^{-1}$ for $c$, $4160.62\text{kW}$ for ${\dot{W}}_{\text{electric}}$ and $24\text{hr}/\text{day}$ for $t$ in Equation (IX).

  $\begin{array}{c}A=\left(\$0.06{\text{kWh}}^{-1}\right)\left(24\text{hr}/\text{day}\right)\left(4160.62\text{kW}\right)\\ =\$5991.2928{\text{day}}^{-1}\end{array}$

Conclusion:

The daily cost of the power consumption of the system is $\$5991.2928{\text{day}}^{-1}$.

To determine

(c)

The frictional heating during flow will make up drop in temperature or not.

Answer to Problem 93P

The frictional heating during flow will make up drop in temperature.

Explanation of Solution

Given information:

The drop-in temperature of geothermal water during flow is $0.5°\text{C}$.

Write the expression for the rise in temperature of geothermal water.

  ${\dot{W}}_{\text{pump}}=\rho \dot{Q}{C}_P\Delta T$........... (X)

Here, rise in temperature of geothermal water during flow is $\Delta T$.

Calculation:

Substitute $950.6\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $3328.5\text{kW}$ for ${\dot{W}}_{\text{pump}}$, $1.5{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ and $4.229\text{kJ}/\text{kg}\cdot °\text{C}$ for $C_P$ in Equation (X).

  $\begin{array}{l}3328.5\text{kW}=\left(950.6\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.5{\text{m}}^{\text{3}}/\text{s}\right)\left(4.229\text{kJ}/\text{kg}\cdot °\text{C}\right)\Delta T\\ \left(3328.5\text{kW}\right)\left[\frac{1\text{kJ}/\text{s}}{1\text{kW}}\right]=\left(950.6\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.5{\text{m}}^{\text{3}}/\text{s}\right)\left(4.229\text{kJ}/\text{kg}\cdot °\text{C}\right)\Delta T\\ \Delta T=0.5519°\text{C}\end{array}$

Since, rise in temperature of geothermal water due to frictional heating is greater than the drop in temperature of geothermal water during flow, therefore the frictional heating during flow will make up drop in temperature.

Conclusion:

The frictional heating during flow will make up drop in temperature.