Summarize the difference in hydraulic and pneumatic systems.
1 answer:
You might be interested in
Answer:
0.5 kW
Explanation:
The given parameters are;
Volume of tank = 1 m³
Pressure of air entering tank = 1 bar
Temperature of air = 27°C = 300.15 K
Temperature after heating = 477 °C = 750.15 K
V₂ = 1 m³
P₁V₁/T₁ = P₂V₂/T₂
P₁ = P₂
V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³
For ideal gas, = 5/2×R = 5/2*0.287 = 0.7175 kJ
PV = NKT
N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)
N = 9.66×10²⁴
Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles
The average mass of one mole of air = 28.8 g
Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg
∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ
The power input required = The rate of heat transfer = 149.211/(60*5)
The power input required = 0.49737 kW ≈ 0.5 kW.
Answer:
The answer is given in the explanation.
Explanation:
The circuit is as indicated in the attached figure.
From the analytical description the zener voltage is given as
Here
Vzo is the voltage at which the slope of 1/rz intersects the voltage axis it is equal to knee voltage.
The equivalent model is shown in the attached figure.
From the above equation, Vzo is calculated as
Here Vz is given as 7.5 V
Iz is given as 10 mA
rz is given as 30 Ω
Thus the Vzo is given as
The value of I_L is given as 5 mA
Now the expression of current is as
Now the resistance is calculated as
So the value of resistance is 186.66 Ω.
Considering the supply voltage is increased by 10%
V is 10-10%*10=10+1=11 so the
Considering the supply voltage is decreased by 10%
V is 10-10%*10=10-1=9 so the
Now if the supply voltage is 10% high and the value of Load is removed i.e I=Iz only which is 10mA
so
Now the largest load current thus that the supply voltage is 10% low and the current of zener is knee current thus
The load voltage is 7.485 V
