Chapter 6.11, Problem 27P

To determine

The price of coal that will enable the IGCC plant to recover their cost difference from fuel saving in 5 year.

Answer to Problem 27P

The price of coal that will enable the IGCC plant to recover their cost difference from fuel saving in 5 year is $\underset{\_}{\text{\$85}\text{.2}/\text{ton}}$.

Explanation of Solution

Determine the construction costs of coal.

$\text{Construction}{\text{cost}}_{\text{coal}}=\left[\begin{array}{l}\left(\begin{array}{l}\text{the amount of elelctricity }\\ \text{generated}\text{by a plant}\end{array}\right)\\ \times \left(\begin{array}{l}\text{construction}\text{cost}\text{for}\\ \text{coal}\text{power}\text{plant}\end{array}\right)\end{array}\right]$ (I)

Determine the construction costs of IGCC.

$\text{Construction}{\text{cost}}_{\text{IGCC}}=\left[\begin{array}{l}\left(\begin{array}{l}\text{the amount of elelctricity}\\ \text{ generated}\text{by a plant}\end{array}\right)\\ \times \left(\begin{array}{l}\text{construction}\text{cost}\text{for}\\ \text{IGCC}\text{power}\text{plant}\end{array}\right)\end{array}\right]$ (II)

Determine the construction costs difference of coal and IGCC.

$\text{Construction}\text{cost diference}=\left[\begin{array}{l}\left(\text{Construction}{\text{cost}}_{\text{coal}}\right)\\ -\left(\text{Construction}{\text{cost}}_{\text{IGCC}}\right)\end{array}\right]$ (III)

Determine the amount of electricity produced by either plant in 5 years.

$W_e=\dot{W}\Delta t$ (IV)

Here, the power generation capacity of power plant is $\dot{W}$ and the rate of time is $\Delta t$.

Determine the amount of fuel needed to generate a specified amount of power.

$\begin{array}{l}\eta =\frac{W_e}{Q_{\text{in}}}\\ \eta =\frac{W_e}{\left(m_{\text{fuel}}\right)\times HV}\\ m_{\text{fuel}}=\frac{W_e}{\eta \times HV}\end{array}$ (V)

Here, the efficiency of plant is $\eta$, the mass of the fuel is $m_{\text{fuel}}$, and the heating value of coal is HV.

Determine the mass difference between plants.

$\Delta m_{\text{coal}}=m_{\text{coal,coal}\text{plant}}-m_{\text{coal},\text{IGCC}\text{plant}}$ (VI)

Here, the mass of the coal in coal plant is $m_{\text{coal,coal}\text{plant}}$ and the mass of coal in the IGCC plant is $m_{\text{coal},\text{IGCC}\text{plant}}$.

Determine the unit cost of coal per ton used in the power plant.

$\text{unit}{\text{cost}}_{\text{coal}}=\frac{\Delta \text{cost}}{\Delta {\text{m}}_{\text{coal}}}$ . (VII)

Here, the construction cost difference is $\Delta \text{cost}$.

Conclusion:

Substitute $150,000,000\text{kW}$ for the amount of electricity generated by a plant and $\text{\$1300}/\text{kW}$ for construction cost for coal power plant in Equation (I).

$\begin{array}{c}\text{Construction}{\text{cost}}_{\text{coal}}=\left(150,000,000\text{kW}\right)\times \left(\text{\$1300}/\text{kW}\right)\\ =1.95\times {10}^{11}\end{array}$

Substitute $150,000,000\text{kW}$ for the amount of electricity generated by a plant and $\text{\$1500}/\text{kW}$ for construction cost for coal power plant in Equation (II).

$\begin{array}{c}\text{Construction}{\text{cost}}_{\text{coal}}=\left(150,000,000\text{kW}\right)\times \left(\text{\$1500}/\text{kW}\right)\\ =2.25\times {10}^{11}\end{array}$

Substitute $2.25\times {10}^{11}$ for construction cost of coal and $1.95\times {10}^{11}$ for construction cost of IGCC in Equation (III).

$\begin{array}{c}\text{Construction}\text{cost diference}=\left(2.25\times {10}^{11}\right)-\left(1.95\times {10}^{11}\right)\\ =3\times {10}^{10}\end{array}$

Substitute $150,000,000\text{kW}$ for the amount of electricity generated by a plant and 5 year for $\Delta t$ in Equation (IV).

$\begin{array}{c}{W}_e=\left(150,000,000\text{kW}\right)\times \left(5\text{year}\right)\\ =\left(150,000,000\text{kW}\right)\times \left(5\text{year}\right)\times \left(\frac{356\text{days}}{1\text{year}}\right)\left(\frac{24\text{hours}}{1\text{day}}\right)\\ =6.570\times {10}^{12}\text{kWh}\end{array}$

For coal plant,

Substitute $6.570\times {10}^{12}\text{kWh}$ for $W_e$, 0.4 for ${\eta }_{\text{coal}}$, and $28\times {10}^6\text{kJ}/\text{ton}$ for HV in Equation (V)

$\begin{array}{c}{m}_{\text{coal}}=\frac{\left(6.570\times {10}^{12}\text{kWh}\right)}{\left(0.4\right)\times \left(28\times {10}^6\text{kJ}/\text{ton}\right)}\\ =\frac{\left(6.570\times {10}^{12}\text{kWh}\right)}{\left(0.4\right)\times \left(28\times {10}^6\text{kJ}/\text{ton}\right)}\times \left(\frac{3600\text{kJ}}{1\text{kWh}}\right)\\ =2.1117\times {10}^9\text{tons}\\ \cong 2.112\times {10}^9\text{tons}\end{array}$

For IGCC plant,

Substitute $6.570\times {10}^{12}\text{kWh}$ for $W_e$, 0.48 for ${\eta }_{\text{coal}}$, and $28\times {10}^6\text{kJ}/\text{ton}$ for HV in Equation (V)

$\begin{array}{c}{m}_{\text{IGCC}}=\frac{\left(6.570\times {10}^{12}\text{kWh}\right)}{\left(0.48\right)\times \left(28\times {10}^6\text{kJ}/\text{ton}\right)}\\ =\frac{\left(6.570\times {10}^{12}\text{kWh}\right)}{\left(0.48\right)\times \left(28\times {10}^6\text{kJ}/\text{ton}\right)}\times \left(\frac{3600\text{kJ}}{1\text{kWh}}\right)\\ =1.7598\times {10}^9\text{tons}\\ \cong 1.760\times {10}^9\text{tons}\end{array}$

Substitute $2.112\times {10}^9\text{tons}$ for $m_{\text{coal}}$ and $1.760\times {10}^9\text{tons}$ for $m_{\text{IGCC}}$ in Equation (VI).

$\begin{array}{c}\Delta m_{\text{coal}}=\left(2.112\times {10}^9\text{tons}\right)-\left(1.760\times {10}^9\text{tons}\right)\\ =0.352\times {10}^9\text{tons}\end{array}$

Substitute $3\times {10}^{10}$ for $\text{Δcost}$ and $0.352\times {10}^9\text{tons}$ for $\Delta m_{\text{coal}}$ in Equation (VII).

$\begin{array}{c}\text{unit}{\text{cost}}_{\text{coal}}=\frac{\left(3\times {10}^{10}\right)}{\left(0.352\times {10}^9\text{tons}\right)}\\ =\text{\$85}\text{.2}/\text{ton}\end{array}$

Thus, the price of coal that will enable the IGCC plant to recover their cost difference from fuel saving in 5 year is $\underset{\_}{\text{\$85}\text{.2}/\text{ton}}$.