In order to get a weight reduction, a 15 cm diameter solid steel shaft rotating at 150 rpm. The shaft is 3 meters long and the d
ensity of steel is 8.05 g/cm3. This shaft is replaced by a Titanium (p = 4.506 g/cm3) tube with a 15 cm OD and an 10 cm ID. Determine the reduction in kinetic energy in this system.
1 answer:
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Given:
mass of water, m = 2000 kg
temperature, T = = 303 K
extacted mass of water = 100 kg
Atmospheric pressure, P = 101.325 kPa
Solution:
a) Using Ideal gas equation:
PV = mT (1)
where,
V = volume
m = mass of water
P = atmospheric pressure
R= Rydberg's constant = 8.314 KJ/K
M = molar mass of water = 18 g/ mol
Now, using eqn (1):
Therefore, the volume of the tank is
b) After extracting 100 kg of water, amount of water left, m' = m - 100
m' = 2000 - 100 = 1900 kg
The remaining water reaches thermal equilibrium with surrounding temperature at T' = = 303 K
At equilibrium, volume remain same
So,
P'V = m'T'
Therefore, the final pressure is P' = 96.258 kPa