Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

Answer 1

Complete Question

A large power plant heats 1917 kg of water per second to high-temperature steam to run its electrical generators.

(a) How much heat transfer is needed each second to raise the water temperature from 35.0°C to 100°C, boil it, and then raise the resulting steam from 100°C to 450°C? Specific heat of water is 4184 J/(kg · °C), the latent heat of vaporization of water is 2256 kJ/kg, and the specific heat of steam is 1520 J/(kg · °C).

J

(b) How much power is needed in megawatts? (Note: In real power plants, this process occurs under high pressure, which alters the boiling point. The results of this problem are only approximate.)

MW

Answer:

The heat transferred is  $Q = 5.866 * 10^9 J$

The power is  $P = 5866\ MW$

Explanation:

From the question we are told that

      Mass of the water per second is $m = 1917 \ kg$

      The initial temperature of the water is $T_i = 35^oC$

      The boiling point of water is  $T_b = 100^oC$

      The final temperature $T_f = 450^oC$

      The latent heat of vapourization of water is  $c__{L}} = 2256*10^3 J/kg$

      The specific heat of water $c_w = 4184 J/kg^oC$

      The specific heat of stem is $C_s =1520 \ J/kg ^oC$

Generally the heat needed each second is mathematically represented as

         $Q = m[c_w (T_i - T_b) + m* c__{L}} + m* c__{S}} (T_f - T_b)]$

Then substituting the value

        $Q = m[c_w [T_i - T_b] + c__{L}} + C__{S}} [T_f - T_b]]$

         $Q = 1917 [(4184) [100 - 35] + [2256 * 10^3] +[1520] [450 - 100]]$

         $Q = 1917 * [3.05996 * 10^6]$

         $Q = 5.866 * 10^9 J$

The power required is mathematically represented as

         $P = \frac{Q}{t}$

From the question $t = 1\ s$

So  

        $P = \frac{5.866 *10^9}{1}$

        $P = 5866*10^6 \ W$

        $P = 5866\ MW$