A rigid tank contains 10 kg of water at 90°C. If 8 kg of water is in liquid form and the rest is in vapour form, what will be the volume of the tank? [Given: At 90°C, saturation pressure is 70 kPa , specific volume at saturated liquid state = 0.001 m3/kg , specific volume at saturated vapour state = 2.36 m3/kg.]
4.728 m3
In this problem, we are given a rigid tank containing a mixture of liquid water and water vapor at a specific temperature. To find the total volume of the tank, we need to calculate the volume occupied by each phase (liquid and vapor) and then sum them up.
Water State Analysis
The problem states that the water is at 90°C and is present in both liquid and vapor forms. This indicates that the water is in a saturated liquid-vapor mixture state. We are provided with the total mass of water, the mass of the liquid phase, and the specific volumes for both saturated liquid and saturated vapor at 90°C.
- Total mass of water ($m_{total}$) = 10 kg
- Mass of liquid water ($m_f$) = 8 kg
- Mass of vapor water ($m_g$) = $m_{total} - m_f$ = 10 kg - 8 kg = 2 kg
- Specific volume of saturated liquid ($v_f$) at 90°C = 0.001 m3/kg
- Specific volume of saturated vapor ($v_g$) at 90°C = 2.36 m3/kg
Liquid Volume Calculation
The volume occupied by the liquid water can be calculated by multiplying its mass by its specific volume. The fundamental formula used is:
$$V_{liquid} = m_f \times v_f$$
Substituting the given values into the formula:
$$V_{liquid} = 8 \text{ kg} \times 0.001 \text{ m}^3/\text{kg}$$
$$V_{liquid} = 0.008 \text{ m}^3$$
Vapor Volume Calculation
Similarly, the volume occupied by the water vapor can be calculated by multiplying its mass by its specific volume. First, we determine the mass of the vapor, then apply the volume formula:
The mass of the vapor is found by subtracting the liquid mass from the total mass:
$$m_g = m_{total} - m_f$$
$$m_g = 10 \text{ kg} - 8 \text{ kg} = 2 \text{ kg}$$
Now, calculate the vapor volume using its specific volume:
$$V_{vapor} = m_g \times v_g$$
Substituting the values:
$$V_{vapor} = 2 \text{ kg} \times 2.36 \text{ m}^3/\text{kg}$$
$$V_{vapor} = 4.72 \text{ m}^3$$
Tank Volume Determination
Since the tank is rigid and contains both liquid and vapor phases of water, its total volume must be equal to the sum of the volumes occupied by the liquid and vapor components. The rigid tank's volume does not change, and it accommodates both phases.
$$V_{tank} = V_{liquid} + V_{vapor}$$
Substituting the calculated volumes for the liquid and vapor phases:
$$V_{tank} = 0.008 \text{ m}^3 + 4.72 \text{ m}^3$$
$$V_{tank} = 4.728 \text{ m}^3$$
Therefore, the volume of the rigid tank containing 10 kg of water, with 8 kg in liquid form and the rest in vapor form at 90°C, is 4.728 m3.
Important Questions from Phase Change of a Pure Substance
A state where all the phase of water can simultaneously co-exist is called _________.
Which of the following statements is true about sensible heat?
What is the lowest pressure at which water can exist in liquid phase in stable equilibrium?
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