Chapter 17, Problem 76P

To determine

The fraction of cost due to heat loss and the time taken for the insulation to pay itself.

Explanation of Solution

Given:

Height of the heater is 2 m.

Diameter of the heater is 40 cm.

Temperature of hot water is $55°\text{C}$.

Temperature of the room is $27°\text{C}$.

The heater transfer coefficient on the inner surface of the heater is $50\text{W}/{\text{m}}^2\cdot \text{K}$.

The heater transfer coefficient on the outer surface of the heater is $12\text{W}/{\text{m}}^2\cdot \text{K}$.

Diameter of the sheet metal tank is 46 cm.

Thermal conductivity of foam is $0.03\text{W}/\text{m}\cdot \text{K}$.

Unit cost is $\$0.08/\text{kWh}$.

Annual cost is $\$280$.

Thickness of insulation is 3cm.

Thermal conductivity of fiberglass is $0.035\text{W}/\text{m}\cdot \text{K}$.

Calculation:

The total thermal resistance is,

  $\begin{array}{c}{R}_{total}=R_i+R_{foam}+R_o\\ =\frac{1}{h_i{A}_i}+\frac{\ln \left(D_2/D_1\right)}{2\pi kL}+\frac{1}{h_o{A}_o}\\ =\frac{1}{\left(50\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(\pi \times 0.40\text{ m}\times 2\text{ m}\right)}+\frac{\ln \left(\frac{23}{20}\right)}{2\pi \left(0.03\text{W}/\text{m}\cdot \text{K}\right)\left(\text{2 m}\right)}\\ +\frac{1}{\left(12\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(\pi \times 0.46\text{ m}\times 2\text{ m}\right)}\\ =0.4075°\text{C}/\text{W}\end{array}$

The rate of heat transfer is,

  $\begin{array}{c}\dot{Q}=\frac{T_w-T_{\infty }}{R_{total}}\\ =\frac{\left(55-27\right)°\text{C}}{0.4075°\text{C}/\text{W}}\\ =68.71\text{ W}\end{array}$

The heat lost per year is,

  $\begin{array}{c}Q=\dot{Q}\Delta t\\ =\left(68.71\text{ W}\right)\left(365\times 24\text{h}\right)\\ =601.9\text{kWh}\end{array}$

The cost due to energy lost is,

  $\begin{array}{c}\text{Cost}=\left(601.9\text{kWh}\right)\left(\$0.08/\text{kWh}\right)\\ =\$48.15\end{array}$

Calculate the fraction of cost due to heat loss.

  $\begin{array}{c}f=\frac{\$48.15}{\$280}\\ =17.2\%\end{array}$

Thus, the fraction of cost due to heat loss is $17.2\%$.

The total thermal resistance when insulation is added is,

  $\begin{array}{c}{R}_{total}=R_i+R_{foam}+R_{fiberglass}+R_o\\ =\frac{1}{h_i{A}_i}+\frac{\ln \left(D_2/D_1\right)}{2\pi k_{1}L}+\frac{\ln \left(D_3/D_2\right)}{2\pi k_{2}L}+\frac{1}{h_o{A}_o}\\ =\frac{1}{\left(50\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(\pi \times 0.40\text{ m}\times 2\text{ m}\right)}+\frac{\ln \left(\frac{23}{20}\right)}{2\pi \left(0.03\text{W}/\text{m}\cdot \text{K}\right)\left(\text{2 m}\right)}\\ +\frac{\ln \left(\frac{26}{23}\right)}{2\pi \left(0.035\text{W}/\text{m}\cdot \text{K}\right)\left(\text{2 m}\right)}+\frac{1}{\left(12\text{W}/{\text{m}}^2\cdot \text{K}\right)\left(\pi \times 0.52\text{ m}\times 2\text{ m}\right)}\\ =0.6830°\text{C}/\text{W}\end{array}$

The heat transfer rate when insulation is added is,

  $\begin{array}{c}\dot{Q}=\frac{T_w-T_{\infty }}{R_{total}}\\ =\frac{\left(55-27\right)°\text{C}}{0.6830°\text{C}/\text{W}}\\ =41.00\text{ W}\end{array}$

The savings in energy is,

  $\begin{array}{c}\text{Saving}=68.71\text{W}-41.00\text{W}\\ \text{=27}\text{.71W}\end{array}$

Calculate the time required for the insulation to pay itself.

  $\begin{array}{l}\text{Cost}=\left(0.02771\text{kW}\right)\left(t\right)\left(\$0.08/\text{kWh}\right)\\ t=\frac{\$30}{\left(0.02771\text{kW}\right)\left(\$0.08/\text{kWh}\right)}\\ t=13533\text{h}\\ t=19\text{ months}\end{array}$

Thus, the time required for the insulation to pay itself is $19\text{ months}$.