Chapter 7, Problem 26P

(a)

To determine

The amount of coal consume during a 24 hour period.

Explanation of Solution

Given:

The rate of required output of the power plant $\left({\dot{W}}_{\text{net,out}}\right)$ is $300\text{MW}$.

The thermal efficiency of the plant $\left({\eta }_{\text{th}}\right)$ is $0.32$.

The change in with respect with time is $\left(\Delta t\right)$ is $24\text{hours}$.

The value of heating coal $\left(q_{\text{HV}}\right)$ is $28000\text{kJ}/\text{kg}$.

The ration of air-fuel is $\text{AF}$ is $12\text{kg}\text{air}/\text{kg}\text{fuel}$.

Calculation:

Write the expression for thermal efficiency of power plant.

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

Simplify the Equation (II) as per British thermal unit per hour to obtain the rate of heat transfer of the power plant $\left({\dot{Q}}_{\text{in}}\right)$.

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

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

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

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

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

Calculate the mass of the coal $\left(m_{\text{coal}}\right)$.

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

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

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

Calculate the rate of mass of coal consumed during this period.

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

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

(b)

To determine

The rate of air flowing through the furnace.

Explanation of Solution

Calculate the rate of air mass flowing through the furnace.

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

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

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