# Thermal Engineering Questions and Answers – Steam Condensers – Dalton’s Law of Partial Pressure

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This set of Thermal Engineering Problems focuses on “Steam Condensers – Dalton’s Law of Partial Pressure”.

1. Dalton’s law of partial pressure states that the total pressure exerted by a mixture of gas and vapour is the sum of partial pressure of the gas and partial pressure of the vapour at the common temperature.
a) True
b) False

Explanation: The given statement is the correct one for Dalton’s law of partial pressure.
Therefore, at a particular temperature ‘t’ –
Mathematically,
P = Pa + Ps
Where, P = Total pressure inside the container
Pa = Partial pressure of the gas (at temperature t)
Ps = Saturation pressure of water at temperature t

2. Which of the following expression is the correct one for calculating total mass of mixture present in a container? (In terms of mass of vapour)
a) ms $$\{1-\frac{v_g}{v_a}\}$$
b) ms $$\{1+\frac{v_g}{v_a}\}$$
c) ms $$\{1-\frac{v_a}{v_g}\}$$
d) ms $$\{1+\frac{v_a}{v_g}\}$$

Explanation: The correct expression for calculating the total mass of mixture (of gas and vapour) present inside a container is ms $$\{1+\frac{v_g}{v_a}\}$$.
where, ms = Mass of vapour inside the container
vg = Specific volume of saturated water vapour
va = Specific volume of air
vg and va at a particular temperature ‘t’.

3. Which of the following expression is the correct one for calculating total mass of mixture present in a container? (In terms of mass of air)
a) ma$$\{1+\frac{v_g}{v_a}\}$$
b) ma$$\{1-\frac{v_g}{v_a}\}$$
c) ma$$\{1+\frac{v_a}{v_g}\}$$
d) ma$$\{1-\frac{v_a}{v_g}\}$$

Explanation: The correct expression for calculating the total mass of mixture (of gas and vapour) present inside a container is ma$$\{1+\frac{v_a}{v_g}\}$$.
where, ma = Mass of air inside the container
vg = Specific volume of saturated water vapour
va = Specific volume of air
vg and va at a particular temperature ‘t’.

4. Given specific volume of steam as 54.25 m3/kg and volume of the container as 190 m3, calculate the mass of steam present inside the container.
a) 3.0 kg
b) 3.5 kg
c) 4.0 kg
d) 4.5 kg

Explanation: Given, v = 54.25 m3/kg, V = 190 m3
Mass of steam = $$\frac{V}{v} = \frac{190}{54.25}$$ = 3.5 kg.

5. In a conatainer, having a mixture of steam and air, the specific volumes of air and steam are 42 m3/kg and 55 m3/kg. Calculate the total mass of the mixture present inside the container if the mass of steam inside the container is 4.23 kg.
a) 9.77 kg
b) 10.23 kg
c) 12.85 kg
d) 15.64 kg

Explanation: Given, va = 42 m3/kg, vg = 55 m3/kg
We know that,
Total mass of mixtue = ms$$\{1+\frac{v_g}{v_a}\}$$ = 4.23$$\{1+\frac{55}{42}\}$$ = 9.77 kg. 