Question

A rigid tank holds 22 kg of 127 °C water. If 9 kg of that is liquid water what is the pressure in the tank and volume of the tank?

Answer 1

Answer:

The pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.

Explanation:

Volume is constant as the tank is rigid. Take the saturation condition of water from the steam table for pressure at 127°C.  

Given:  

Total mass of water is 22 kg.  

Mass of liquid water is 9 kg.  

Temperature of water is 127°C.  

From steam table at 127°C:  

The pressure in the tank is 246.878 Kpa.  

Specific volume of saturated water is 0.00106683 m³/kg.  

Specific volume of saturated steam is 0.72721 m³/kg.  

Calculation:  

Step1  

From steam table at 127°C:  

The pressure in the tank is 246.878 Kpa.  

Step2  

Dryness fraction is calculated as follows:  

$x=\frac{m_{v}}{m_{t}}$

Here, dyness fraction is x, mass of vapor is $m_{v}$and total mass is $m_{t}$.  

Substitute the values in the above equation as follows:  

$x=\frac{m_{v}}{m_{t}}$

$x=\frac{22-9}{22}$

x = 0.59  

Step3  

Specific volume of tank is calculated as follows:  

$v=v_{f}+x(v_{g}-v_{f})$

$v=0.00106683+0.59(0.72721-0.00106683)$

$v=0.00106683+0.42842447$

v=0.4295 m³/kg.  

Step4  

Volume is calculated as follows:  

$V=v\times m_{t}$

$V=0.4295 \times22$

V=9.449 m³.  

Thus, the pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.