Q. 4.26 - An Introduction to Thermal Physics

Question

A coal-fired power plant, with parameters similar to those used in the text above, is to deliver $1 \text{GW }\left({10}^9 \text{watts}\right)$ of power. Estimate the amount of steam (in kilograms) that must pass through the turbine(s) each second.

Step-by-Step Solution

Answer

When a coal-fired plant is supposed to deliver $1 \text{GW}$ of power, then the amount of steam that pass through the turbine per second is $620.03 \text{kg}$.

Step 1. Given information

Power delivered by coal-fired plant$=1 {\text{GW=10}}^9\text{Watt}$.

Step 2. Explanation

The efficiency of steam engine$=48\%$.

As, $$\begin{gathered} \text{e=1-}\frac{{\text{Q}}_{\text{c}}}{{\text{Q}}_{\text{h}}}\text{=}\frac{\text{W}}{{\text{Q}}_{\text{h}}} \\ \text{where,} \\ {\text{W=Q}}_{\text{h}}{\text{-Q}}_c \\ {\text{Q}}_{\text{h}}\text{=Heat of hot reservoir} \\ {\text{Q}}_{\text{c}}\text{=Heat of cold reservoir} \end{gathered}$$

So,

$\begin{array}{rcl}& & {\text{Q}}_{\text{h}}\text{=}\frac{\text{W}}{\text{e}}\\ & & \text{=}\frac{{10}^9}{0.48}\\ & & {\text{=2.0833×10}}^9\text{ watt}\\ & & \text{=2.0833 GW}\end{array}$

Step 3. Amount of steam

As some water is passing the turbine every second and let the mass of steam be $\text{M}$ given by:

$$\begin{gathered} \frac{{\text{Q}}_{\text{h}}}{\text{M}}={\text{H}}_3{\text{-H}}_1 \\ =\left(3444-84\right)\raisebox{1ex}{$\text{KJ}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right. \\ =3360 \raisebox{1ex}{$\text{KJ}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right. \end{gathered}$$

So,

$$\begin{gathered} \text{M=}\frac{{\text{Q}}_{\text{h}}}{3360{\displaystyle \raisebox{1ex}{$\text{KJ}$}\!\left/ \!\raisebox{-1ex}{$\text{kg}$}\right.}} \\ \text{=}\frac{2.0833\times {10}^9}{3360}\text{ kg} \\ =620.029 \text{kg} \end{gathered}$$

Step 4. Conclusion

When a coal-fired plant is supposed to deliver $1\text{ KW}$ of power, then the amount of steam that pass through the turbine per second is $620.03 \text{kg}$.