Question

0.05Kg of steam at 15bar is contained in a rigid vessel of volume 0.0076m^3. what is the temperature of the steam

Answer 1

Answer on Question #58786-Engineering-Other

$m = 0.05\, \text{Kg}$ of steam at $P = 15\, \text{bar}$ is contained in a rigid vessel of volume $V = 0.0076\, \text{m}^3$. What is the temperature of the steam?

Solution

The steam is superheated

$$V = m v_1$$ $$0.0076 = 0.05 v_1$$ $$v_1 = 0.152\, \frac{\text{m}^3}{\text{kg}}.$$

From superheated steam table, corresponding to a steam press of 15 bar and volume $0.152\, \frac{\text{m}^3}{\text{kg}}$, the temperature of steam is $250^\circ\text{C}$.

Answer: $250^\circ\text{C}$.

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