Chapter 6.11, Problem 26P

A coal-burning steam power plant produces a net power of 300 MW with an overall thermal efficiency of 32 percent. The actual gravimetric air–fuel ratio in the furnace is calculated to be 12 kg air/kg fuel. The heating value of the coal is 28,000 kJ/kg. Determine (a) the amount of coal consumed during a 24-hour period and (b) the rate of air flowing through the furnace.

To determine

The amount of coal consume during a 24 hour period.

Answer to Problem 26P

The amount of coal consume during a 24 hour period is $\underset{\_}{\text{2}\text{.89}\times {\text{10}}^6\text{kg}}$.

Explanation of Solution

Write the expression for thermal efficiency of power plant.

${\eta }_{\text{th}}=\frac{W_{\text{net,out}}}{Q_{\text{in}}}$ (I)

Here, the net work output of the power plant is $W_{\text{net,out}}$ and the amount of the heat supplied to the power plant is $Q_{\text{in}}$.

Simplify the Equation (II) as per British thermal unit per hour.

$\begin{array}{l}{\eta }_{\text{th}}=\frac{{\dot{W}}_{\text{net,out}}}{{\dot{Q}}_{\text{in}}}\\ {\dot{Q}}_{\text{in}}=\frac{{\dot{W}}_{\text{net,out}}}{{\eta }_{\text{th}}}\end{array}$ (II)

Here, the rate of heat transfer of the power plant is ${\dot{Q}}_{\text{in}}$ and the rate of required output of the power plant is ${\dot{W}}_{\text{net,out}}$.

Determine the amount heat input to the power plant with respect to time.

$Q_{\text{in}}={\dot{Q}}_{\text{in}}\Delta t$ (III)

Here, the change in with respect with time is $\Delta t$.

Determine the heat supplied to the power plant.

$\begin{array}{l}{Q}_{\text{in}}=m_{\text{coal}}\times q_{\text{HV}}\\ m_{\text{coal}}=\frac{Q_{\text{in}}}{q_{\text{HV}}}\end{array}$ (IV)

Here, the mass of the coal is $m_{\text{coal}}$ and the value of heating coal is $q_{\text{HV}}$.

Determine the rate of mass of coal consumed during this period.\

${\dot{m}}_{\text{coal}}=\frac{m_{\text{coal}}}{\Delta t}$ (V)

Conclusion:

Substitute $300\text{MW}$ for ${\dot{W}}_{\text{net,out}}$ and $0.32$ for ${\eta }_{\text{th}}$ in Equation (II).

$\begin{array}{c}{\dot{Q}}_{\text{in}}=\frac{\left(300\text{MW}\right)}{\left(0.32\right)}\\ =\frac{\left(300\text{MW}\right)}{\left(0.32\right)}\\ =937.5\text{MW}\end{array}$

Substitute $937.5\text{MJ}/\text{s}$ for ${\dot{Q}}_{\text{in}}$ and $24\text{hours}$ for $\Delta t$ in Equation (III).

$\begin{array}{c}{Q}_{\text{in}}=\left(937.5\text{MJ}/\text{s}\right)\left(24\text{h}\right)\\ =\left(937.5\text{MJ}/\text{s}\right)\left(24\text{h}\times \frac{3600\sec }{1\text{h}}\right)\\ =81000000\text{MJ}\\ =8.1\times {10}^7\text{MJ}\end{array}$

Substitute $8.1\times {10}^7\text{MJ}$ for $Q_{\text{in}}$ and $28000\text{kJ}/\text{kg}$ for $q_{\text{HV}}$ in Equation (IV).

$\begin{array}{c}{m}_{\text{coal}}=\frac{\left(8.1\times {10}^7\text{MJ}\right)}{\left(28000\text{kJ}/\text{kg}\right)}\\ =\frac{\left(8.1\times {10}^7\text{MJ}\right)}{\left(28000\text{kJ}/\text{kg}\times \left(\frac{{10}^{-3}\text{MJ}/\text{kg}}{1\text{kJ}/\text{kg}}\right)\right)}\\ =2892857\text{kg}\\ \cong \text{2}\text{.89}\times {\text{10}}^6\text{kg}\end{array}$

Thus, the amount of coal consume during a 24 hour period is $\underset{\_}{\text{2}\text{.89}\times {\text{10}}^6\text{kg}}$.

Substitute $\text{2}\text{.89}\times {\text{10}}^6\text{kg}$ for $m_{\text{coal}}$ and $24\text{hours}$ for $\Delta t$ in Equation (V).

$\begin{array}{c}{\dot{m}}_{\text{coal}}=\frac{\left(\text{2}\text{.89}\times {\text{10}}^6\text{kg}\right)}{\left(24\text{h}\right)}\\ =\frac{\left(\text{2}\text{.89}\times {\text{10}}^6\text{kg}\right)}{\left(24\text{h}\times \left(\frac{3600\sec }{1\text{h}}\right)\right)}\\ =33.48\text{kg}/\text{s}\end{array}$

To determine

The rate of air flowing through the furnace.

Answer to Problem 26P

The rate of air flowing through the furnace is $\underset{\_}{401.8\text{kg}/\text{s}}$.

Explanation of Solution

Determine the rate of air mass flowing through the furnace.

${\dot{m}}_{\text{air}}=\left(\text{AF}\right)\times {\dot{m}}_{\text{coal}}$ (VI)

Here, the ration of air-fuel is $\text{AF}$.

Conclusion:

Substitute $12\text{kg}\text{air}/\text{kg}\text{fuel}$ for $\text{AF}$ and $33.48\text{kg}/\text{s}$ for ${\dot{m}}_{\text{coal}}$ in Equation (VI).

$\begin{array}{c}{\dot{m}}_{\text{air}}=\left(12\text{kg}\text{air}/\text{kg}\text{fuel}\right)\times \left(33.48\text{kg}/\text{s}\right)\\ =401.8\text{kg}/\text{s}\end{array}$

Thus, the rate of air flowing through the furnace is $\underset{\_}{401.8\text{kg}/\text{s}}$.