The output voltage of a power supply is normally distributed with mean 12 V and standard deviation 0.11 V. If the upper and lowe
1 answer:
Answer:
82.62%
Explanation:
The z score is a score used in statistics to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
From the normal distribution table, P(11.85 < x < 12.15) = P(-1.36 < z < 1.36) = P(z < 1.36) - P(z < -1.36) = 0.9131-0.0869 = 0.8262 = 82.62%
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Answer:
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
Now, we will find (enthalpy at the outlet for the isentropic process):
Now, the isentropic efficiency of the turbine can be given as follows:
Answer:
T₂ =93.77 °C
Explanation:
Initial temperature ,T₁ =27°C= 273 +27 = 300 K
We know that
Absolute pressure = Gauge pressure + Atmospheric pressure
Initial pressure ,P₁ = 300+1=301 kPa
Final pressure ,P₂= 367+1 = 368 kPa
Lets take temperature=T₂
We know that ,If the volume of the gas is constant ,then we can say that
Now by putting the values in the above equation we get
The temperature in °C
T₂ = 366.77 - 273 °C
T₂ =93.77 °C