A 2.15-m3 rigid tank initially contains air whose density is 1.18 kg/m3. The tank is connected to a high-pressure supply line th
2 answers:
Answer:
The mass of air that has entered the tank is 8.86Kg
Explanation:
We are given:
We first calculate the intial and final masses,
Therefore:
= 1.18 * 2.15 = 2.54
= 5.30 * 2.15 = 11.4
Therefore to find the mass of air inside the tank, we use:
= 11.4 - 2.54 = 8.86Kg
Therefore the mass of air inside the tank is 8.86Kg
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Answer:
The maximum theoretical height that the pump can be placed above liquid level is
Explanation:
To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:
( stands here for density,
for height)
Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:
This means that pressure drop is proportional to the suction lift's height.
We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.
That means:
We insert that into our last equation and get:
And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.
Answer:
The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero
Explanation:
The expression for the maximum shear stress is given:
Where
σx = stress in vertical plane = 20 ksi
σy = stress in horizontal plane = -30 ksi
τM = 32 ksi
Replacing:
Solving for τxy:
τxy = ±19.98 ksi
The principal stress is:
Where
σp1 = 20 ksi
σp2 = -30 ksi
(equation 1)
equation 2
Solving both equations:
σp1 = 27 ksi
σp2 = -37 ksi
The shear stress on the vertical plane is zero