Question

How much steam is required to evaporate 627kg of water, when steam is saturated steam @8BARG and 170C?

Answer 1

QUESTION:

How much steam is required to evaporate 627kg of water, when steam is saturated steam and 170C?

SOLUTION:

When steam is cooled down, it releases the heat

$$Q _ {s t e a m} = c _ {s t e a m} m _ {s t e a m} \left(T _ {s t e a m, i n i t i a l l} - T _ {0}\right)$$

Here $c_{\text{steam}}$ is specific heat of steam, $T_0 = 373.15 \, \text{K}$ ( $100{}^\circ \text{C}$ ).

When water absorbs this heat, its temperature decreases, and water boils.

The heat needed to evaporate water is

$$Q _ {w a t e r} = c _ {w a t e r} m _ {w a t e r} \left(T _ {0} - T _ {w a t e r, i n i t i a l}\right) + L _ {w a t e r} m _ {w a t e r}$$

Here $L_{water}$ is latent heat for evaporation of water. Let's assume, that the initial temperature of water is 293.15 K (20°C). Hence

$$Q _ {s t e a m} = Q _ {w a t e r}$$ $$c _ {s t e a m} m _ {s t e a m} \left(T _ {s t e a m, i n i t i a l l} - T _ {0}\right) = c _ {w a t e r} m _ {w a t e r} \left(T _ {0} - T _ {w a t e r, i n i t i a l}\right) + L _ {w a t e r} m _ {w a t e r}$$ $$m _ {s t e a m} = \frac {4 1 8 0 \cdot 6 2 7 \cdot (3 7 3 . 1 5 - 2 9 3 . 1 5) + 2 2 6 0 \cdot 1 0 ^ {3} \cdot 6 2 7}{2 0 8 0 \cdot (4 4 3 . 1 5 - 3 7 3 . 1 5)}$$ $$m _ {s t e a m} = 1 1 1 7 2 k g$$

ANSWER

$$m _ {s t e a m} = 1 1 1 7 2 k g$$