Answer:
the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Explanation:
Given that;
depth 1 = 71 ft
depth 2 = 10 ft
pressure p = 17 psi = 2448 lb/ft²
depth h = 71 ft - 10 ft = 61 ft
we know that;
p = P_air + yh
where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )
so we substitute;
p = 2448 + ( 49.3 × 61 )
= 2448 + 3007.3
= 5455.3 lb/ft³
= 37.88 psi
Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Answer:
The specific weight of unknown liquid is found to be 15 KN/m³
Explanation:
The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.
Total Pressure = Pressure of oil + Pressure of unknown liquid
65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)
65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)
(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²
(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m
<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>
Answer:
Explanation:
Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:
The initial pressure is:
The speed at outlet is:
![v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B%28250%5C%2C%5Cfrac%7Bgal%7D%7Bmin%7D%20%29%5Ccdot%20%28%5Cfrac%7B3.785%5Ctimes%2010%5E%7B-3%7D%5C%2Cm%5E%7B3%7D%7D%7B1%5C%2Cgal%7D%20%29%5Ccdot%28%5Cfrac%7B1%5C%2Cmin%7D%7B60%5C%2Cs%7D%20%29%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%5Ccdot%20%5B%281.125%5C%2Cin%29%5Ccdot%28%5Cfrac%7B0.0254%5C%2Cm%7D%7B1%5C%2Cin%7D%20%29%5D%5E%7B2%7D%20%7D)
The initial pressure is:
