Question

Four kg of water is placed in an enclosed volume of 1m3. Heat is added until the temperature is 150°C. Find ( a ) the pressure, ( b )the mass of vapor, and ( c ) the volume of the vapor.

Answer 1

Answer on Question #83797 – Engineering | Mechanical Engineering

Four kg of water is placed in an enclosed volume of $1\mathrm{m}^3$. Heat is added until the temperature is $150^{\circ}\mathrm{C}$. Find (a) the pressure, (b) the mass of vapor, and (c) the volume of the vapor.

Answer:

NOTE: Steam table is available at:

https://www.nist.gov/sites/default/files/documents/srd/NISTIR5078-Tab1.pdf

The density of the saturated steam at $150^{\circ}\mathrm{C}$ is $\rho_{V} = 2.5481\mathrm{kg / m}^{3}$. Thus, $4\mathrm{kg}$ of the saturated steam at $150^{\circ}\mathrm{C}$ occupies volume of

$$\frac {4 \mathrm {k g}}{2 . 5 4 8 1 \mathrm {k g} / \mathrm {m} ^ {3}} = 1. 5 7 \mathrm {m} ^ {3},$$

which is more than $1\mathrm{m}^3$. It shows that, the final state is the saturated water-steam mixture.

The pressure is equal to the saturation pressure 4.7616 bar.

The density of the saturated water $\rho_{L} = 917.01\mathrm{kg / m}^{3}$. Noting the mass of the water and steam in the mixture by $m_{L}$ and $m_{V}$, respectively, we can define the total mass $M$ and volume $V$ of the mixture by following equations

$$M = m _ {L} + m _ {V},$$ $$V = \frac {m _ {L}}{\rho_ {L}} + \frac {m _ {V}}{\rho_ {V}}.$$

Considering these equations together, yields

$$m _ {L} = M - m _ {V},$$ $$V = \frac {M - m _ {V}}{\rho_ {L}} + \frac {m _ {V}}{\rho_ {V}},$$ $$m _ {V} = \frac {V \rho_ {L} - M}{\frac {\rho_ {L}}{\rho_ {V}} - 1}.$$

Substitute to define the mass of the steam

$$m _ {V} = \frac {1 \cdot 9 1 7 . 0 1 - 4}{\frac {9 1 7 . 0 1}{2 . 5 4 8 1} - 1} = 2. 5 4 4 \mathrm {k g}.$$

The volume $V_{V}$ of the steam equals to

$$V _ {V} = \frac {m _ {V}}{\rho_ {V}} = \frac {2 . 5 4 4}{2 . 5 4 8 1} = 0. 9 9 8 \mathrm {m} ^ {3}.$$

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