Answer:

Explanation:
For pressure gage we can determine this by saying:
The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

Answer: 0.95 inches
Explanation:
A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.
The length= 64 inches
Ends are fixed Le= 64/2 = 32 inches
Factor Of Safety (FOS) = 3. 0
E= 10.6× 10^6 ps
σy= 4000ps
The square cross-section= ia^4/12
PE= π^2EI/Le^2
6500= 3.142^2 × 10^6 × a^4/12×32^2
a^4= 0.81 => a=0.81 inches => a=0.95 inches
Given σy= 4000ps
σallowable= σy/3= 40000/3= 13333. 33psi
Load acting= 6500
Area= a^2= 0.95 ×0.95= 0.9025
σactual=6500/0.9025
σ actual < σallowable
The dimension a= 0.95 inches
Answer:
Explained
Explanation:
This situation can occur because of various factors such as:
- Gradual deterioration of lubrication and coolant.
- change of environmental condition such as temperature, humidity, moisture, etc.
- Change in the properties of incoming raw material
- An increase or decrease in the temperature of the heat treating operation
- Debris interfering with the manufacturing process.
The exit temperature is 586.18K and compressor input power is 14973.53kW
Data;
- Mass = 50kg/s
- T = 288.2K
- P1 = 1atm
- P2 = 12 atm
<h3>Exit Temperature </h3>
The exit temperature of the gas can be calculated isentropically as

Let's substitute the values into the formula

The exit temperature is 586.18K
<h3>The Compressor input power</h3>
The compressor input power is calculated as

The compressor input power is 14973.53kW
Learn more on exit temperature and compressor input power here;
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