Bogata, Columbia is 8500 ft above sea level (2500 m) where the pressure is appro
ID: 718086 • Letter: B
Question
Bogata, Columbia is 8500 ft above sea level (2500 m) where the pressure is approximately 75 kPa. If water is boiled there in a cylindrical pot with a diameter of 0.25 m, answer the following questions. 2. a. At what temperature will the water boil in G and what is the specific volume in m? (5 pts) 91.76 °G and 0.001037 m'/kg If water is filled in the pot to a height of 0.1 m, what is the mass of water in kg? Assume b. saturation conditions for this calculation. (5 pts) 4.7336 kg Assuming the water is boiled in the pot with a lid, if the height of the liquid decreases by 0.02 m, what mass of water becomes steam in kg and what is the quality, x, of the saturated mixture? (5 pts) c. 0.9467 kg and x-0.20 d. What is the total volume of the mixture under the above scenario in m3? This is the volume that would be required to avoid verpressurizatien if the pot were truly covered and sealed. (5 pts) 2.10299 m3 (Vg-2.099 m3 and V0.003927 m3)Explanation / Answer
Saturation pressure P = 75 kPa
Part a
From the steam table
Boiling temperature of water = 91.76°C
Specific volume of water vf = 0.001037 m3/kg
Part b
Height of water in pot h = 0.1 m
Radius of pot r = 0.25/2 = 0.125 m
Volume of water filled in the pot = x r^2 x h
= 3.14 x 0.125^2 x 0.1
V = 4.91 x 10^-3 m3
Mass of water = V/vf = (4.91 x 10^-3 m3) / (0.001037 m3/kg)
= 4.7336 kg
Part c
Liquid decreased by 0.02 m.
0.02 m height of liquid becomes water.
Volume of water filled in the pot of height 0.02 m = x r^2 x h
= 3.14 x 0.125^2 x 0.02
V = 9.82 x 10^-4 m3
Mass of water becomes steam = V/vf = (9.82 x 10^-4 m3) / (0.001037 m3/kg)
= 0.9467 kg
Steam quality = mass of steam / total mass of water
= 0.9467/4.7336
= 0.20
Part d
At 75 kPa
Volume of liquid water vf = specific volume x mass of liquid water remain in pot
= 0.001037 m3/kg x (4.7336 - 0.9467) kg
= 0.003927 m3
Volume of steam vg = specific volume x mass of steam in pot
= 2.2172 m3/kg x (0.9467) kg
= 2.099 m3
Total volume = vf + vg
= 0.003927 + 2.099
= 2.10299 m3