Chapter 2.8, Problem 56P

The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 × 10⁶ Btu/h. The combustion efficiency of the boiler is measured to be 0.7 by a handheld flue gas analyzer. After tuning up the boiler, the combustion efficiency rises to 0.8. The boiler operates 4200 h a year intermittently. Taking the unit cost of energy to be $13/10⁶ Btu, determine the annual energy and cost savings as a result of tuning up the boiler.

To determine

The annual energy and cost savings.

Answer to Problem 56P

The annual energy and cost savings are $\underset{\_}{2.89\times {10}^9\text{Btu}/\text{yr}}$ and $\underset{\_}{\$37,570\text{/yr}}$ respectively.

Explanation of Solution

Calculate the heat output of boiler.

${\dot{Q}}_{\text{out}}={\dot{Q}}_{\text{in}}\times {\eta }_{\text{furnace}}$ (I)

Here, heat input of boiler is ${\dot{Q}}_{\text{in}}$ and combustion efficiency is ${\eta }_{\text{furnace}}$.

Re-write Equation (I) if the current rate of heat input to the boiler is ${\dot{Q}}_{\text{in},\text{current}}$.

Write the equation for the rate of useful heat output of the boiler.

${\dot{Q}}_{\text{out}}={\left({\dot{Q}}_{\text{in}}{\eta }_{\text{furnace}}\right)}_{\text{current}}$ (II)

Write the rate of heat input to the boiler after the tune up.

${\dot{Q}}_{\text{in},\text{new}}=\frac{{\dot{Q}}_{\text{out}}}{{\eta }_{\text{furnace},\text{new}}}$ (III)

Write the rate of energy savings.

${\dot{Q}}_{\text{in},\text{saved}}={\dot{Q}}_{\text{in},\text{current}}-{\dot{Q}}_{\text{in},\text{new}}$ (IV)

Calculate the energy savings.

$\text{Energy}\text{}\text{savings}={\dot{Q}}_{\text{in,saved}}\left(t\right)$ (V)

Here, the operational hours is t.

Calculate the cost savings.

$\text{Cost}\text{}\text{savings}=\left(\text{Energy}\text{savings}\right)\left(\text{Unit}\text{cost}\text{of energy}\right)$ (VI)

Conclusion:

Substitute $5.5\times {10}^6\text{Btu/h}$ for ${\dot{Q}}_{\text{in},\text{current}}$ and 0.7 for ${\left({\eta }_{\text{furnace}}\right)}_{\text{current}}$ in Equation (II).

$\begin{array}{c}{\dot{Q}}_{\text{out}}=\left(5.5\times {10}^6\text{Btu/h}\right)\left(0.7\right)\\ =3.85\times {10}^6\text{}\text{Btu/h}\end{array}$

Substitute $3.85\times {10}^6\text{}\text{Btu/h}$ for ${\dot{Q}}_{\text{out}}$ and 0.8 for ${\eta }_{\text{furnace},\text{new}}$ in Equation (III).

$\begin{array}{c}{\dot{Q}}_{\text{in},\text{new}}=\frac{3.85\times {10}^6\text{}\text{Btu/h}}{0.8}\\ =4.81\times {10}^6\text{Btu/h}\end{array}$

Substitute $5.5\times {10}^6\text{Btu/h}$ for ${\dot{Q}}_{\text{in},\text{current}}$ and $4.81\times {10}^6\text{Btu/h}$ for ${\dot{Q}}_{\text{in},\text{new}}$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_{\text{in},\text{saved}}=5.5\times {10}^6\text{Btu/h}-4.81\times {10}^6\text{Btu/h}\\ =0.69\times {10}^6\text{Btu/h}\end{array}$

Substitute 4200 h/yr for t and $0.69\times {10}^6\text{Btu/h}$ for ${\dot{Q}}_{\text{in},\text{saved}}$ in Equation (V).

$\begin{array}{c}\text{Energy}\text{}\text{savings}=0.69\times {10}^6\text{Btu/h}\left(4200\text{h/yr}\right)\\ =2.89\times {10}^9\text{Btu/yr}\end{array}$

Substitute $2.89\times {10}^9\text{Btu/yr}$ for energy savings and $\$13/{10}^6\text{Btu}$ in Equation (VI).

$\begin{array}{c}\text{Cost}\text{}\text{savings}=\left(2.89\times {10}^9\text{Btu/yr}\right)\left(\$13/{10}^6\text{Btu}\right)\\ =\$37,570\text{/yr}\end{array}$

Thus, the annual energy and cost savings are $\underset{\_}{2.89\times {10}^9\text{Btu}/\text{yr}}$ and $\underset{\_}{\$37,570\text{/yr}}$ respectively.