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

View Answer

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) m_{s} \(\{1-\frac{v_g}{v_a}\}\)

b) m_{s} \(\{1+\frac{v_g}{v_a}\}\)

c) m_{s} \(\{1-\frac{v_a}{v_g}\}\)

d) m_{s} \(\{1+\frac{v_a}{v_g}\}\)

View Answer

Explanation: The correct expression for calculating the total mass of mixture (of gas and vapour) present inside a container is m

_{s}\(\{1+\frac{v_g}{v_a}\}\).

where, m

_{s}= Mass of vapour inside the container

v

_{g}= Specific volume of saturated water vapour

v

_{a}= Specific volume of air

v

_{g}and v

_{a}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) m_{a}\(\{1+\frac{v_g}{v_a}\}\)

b) m_{a}\(\{1-\frac{v_g}{v_a}\}\)

c) m_{a}\(\{1+\frac{v_a}{v_g}\}\)

d) m_{a}\(\{1-\frac{v_a}{v_g}\}\)

View Answer

Explanation: The correct expression for calculating the total mass of mixture (of gas and vapour) present inside a container is m

_{a}\(\{1+\frac{v_a}{v_g}\}\).

where, m

_{a}= Mass of air inside the container

v

_{g}= Specific volume of saturated water vapour

v

_{a}= Specific volume of air

v

_{g}and v

_{a}at a particular temperature ‘t’.

4. Given specific volume of steam as 54.25 m^{3}/kg and volume of the container as 190 m^{3}, 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

View Answer

Explanation: Given, v = 54.25 m

^{3}/kg, V = 190 m

^{3}

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 m^{3}/kg and 55 m^{3}/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

View Answer

Explanation: Given, v

_{a}= 42 m

^{3}/kg, v

_{g}= 55 m

^{3}/kg

We know that,

Total mass of mixtue = m

_{s}\(\{1+\frac{v_g}{v_a}\}\) = 4.23\(\{1+\frac{55}{42}\}\) = 9.77 kg.

**Sanfoundry Global Education & Learning Series – Thermal Engineering**

To practice all areas of Thermal Engineering Problems, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

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