Answer
a) $\approx111\;\rm kg/s$
Work Step by Step
a)
We are given that the efficiency of the two engines is 0.65 of the Carnot efficiency.
So we need first to find the engines' efficiencies.
$$e_C=\dfrac{T_H-T_L}{T_H}\tag{$C$ referrs to Carnot}$$
Hence, the efficiency of the first engine;
$$e_1=0.65e_{1C}=0.65\cdot \dfrac{T_{H1}-T_{L1}}{T_{H1}}$$
Plugging the given;
$$e_1 =0.65\cdot \dfrac{(750+ 273)-(440+273)}{(750+273)}=\color{blue}{\bf 0.197}$$
The efficiency of the second engine;
$$e_2=0.65e_{2C}=0.65\cdot \dfrac{T_{H2}-T_{L2}}{T_{H2}}$$
Plugging the given;
$$e_2 =0.65\cdot \dfrac{(415+ 273)-(270+273)}{(415+273)}=\color{blue}{\bf 0.137}$$
Now we know that the input heat is from the coal.
Thus, $Q_{H1}=Q_{coal}$, and hence,
$$e_1=\dfrac{W_1}{Q_{coal}}$$
Solving for $W_1$;
$$W_1=e_1Q_{coal} \tag 1$$
We know that the input heat for the second engine is the output heat from the first one.
So, $Q_{H2}=Q_{L1}$, and we know that,
$$Q_{L1}=Q_{H1}-W_1=Q_{coal}-W_1$$
Plugging from (1);
$$Q_{L1}=Q_{H1}-e_1Q_{coal} =Q_{coal}(1-e_1) $$
Hence,
$$e_2=\dfrac{W_2}{Q_{L1}}=\dfrac{W_2}{Q_{coal}(1-e_1)}$$
Solving for $W_2$;
$$W_2=e_2Q_{coal}(1-e_1)\tag 2$$
We know that the total power of the plant is given by
$$P=\dfrac{W_{tot}}{t}=\dfrac{W_1+W_2}{t}$$
Plugging from (1) and (2);
$$P= \dfrac{e_1Q_{coal}+e_2Q_{coal}(1-e_1)}{t}$$
Solvig for $Q_{coal}/t$ and note that we need to the plant to put out 950 MW.
So,
$$950\times10^6=\dfrac{Q_{coal}}{t}\left[ e_1+e_2 (1-e_1) \right]$$
$$\dfrac{Q_{coal}}{t}=\dfrac{950\times10^6}{e_1+e_2 (1-e_1)}=\dfrac{950\times10^6}{0.197+0.137(1-0.197)}$$
$$\dfrac{Q_{coal}}{t}=\color{blue}{\bf 3.1\times10^9}\;\rm J/s\tag 3$$
And since heat of combustion of coal is $2.8\times10^7\;\rm J/kg$, the amount of coal needed each second is
$$\dfrac{m}{t}=\dfrac{Q_{coal}}{t} \cdot \dfrac{1}{2.8\times10^7}= 3.1\times10^9 \cdot \dfrac{1}{2.8\times10^7}$$
$$\boxed{\dfrac{m}{t}=\color{red}{\bf110.7}\;\rm kg/s}$$
b)
We know that the exahusted heat will make the water gain temperature.
And we also know that the exahusted heat is given by
$$Q_{L}=Q_{H}-W=Q_{coal}-W$$
This $Q_L$ will raise the water temperature, so
$$Q_L=m_{water}c_{water}\Delta T_{water}$$
From the last two equations,
$$Q_{coal}-W=m_{water}c_{water}\Delta T_{water}$$
Solving for $m_{water}$;
$$m_{water}=\dfrac{Q_{coal}-W}{c_{water}\Delta T_{water}}$$
And since we need to find the rate of water passed through plant, so we need to divide both sides by $t$;
$$\dfrac{m_{water}}{t}=\dfrac{\frac{1}{t}\left[Q_{coal}-W\right]}{c_{water}\Delta T_{water}}=\left[\dfrac{ Q_{coal}}{t}-\dfrac{W}{t}\right]\cdot \dfrac{1}{{c_{water}\Delta T_{water}}}$$
Thus,
$$\dfrac{m_{water}}{t}= \left[\dfrac{ Q_{coal}}{t}-P\right]\cdot \dfrac{1}{{c_{water}\Delta T_{water}}}$$
Plugging the know and plug from (3);
$$\dfrac{m_{water}}{t}= \left[ ( 3.1\times10^9)-(950\times10^6)\right]\cdot \dfrac{1}{{4186\cdot 4.5}}$$
We need to find the rate in $\rm m^3/h$ which means we need to the volume of the water that passes through the plant each hour.
And we know from the density law that $\rho =m/V$ and hence, $m=\rho V$
$$\dfrac{m_{water}}{t}=\dfrac{\rho_{water}V_{water}}{t}=\rm\dfrac{1.14\times10^5\; kg}{1\;s}\cdot \dfrac{3600\;s}{1\;h}$$
$$\dfrac{V_{water}}{t}=\rm\dfrac{4.104\times10^8\; kg}{1\;h}\cdot \dfrac{1}{\rho_{water}}$$
$$\dfrac{V_{water}}{t}=\rm\dfrac{4.104\times10^8\; kg}{1\;h}\cdot \dfrac{1\;m^3}{1000\;kg}$$
$$\boxed{\dfrac{V_{water}}{t}=\color{red}{\bf4.104\times10^5}\;\rm m^3/h}$$
