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`

Step-by-step 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 = (Q)/(t)`

From the question
`t = 1\ s`

So

`P = (5.866 *10^9)/(1)`

`P = 5866*10^6 \ W`

`P = 5866\ MW`