The pressure of water is 7.3851 kPa
<u>Explanation:</u>
Given data,
V = 150×
m = 1 Kg
= 2 MPa
= 40°C
The waters specific volume is calculated:
= V/m
Here, the waters specific volume at initial condition is , the containers volume is V, waters mass is m.
= 150× /1
= 0.15 / Kg
The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 / Kg and 0.13 / Kg.
= 350+(400-350) 
= 395.17°C
Hence, the initial temperature is 395.17°C.
The volume is constant in the rigid container.
= = 0.15 / Kg
In saturated water labels for = 40°C.
= 0.001008 / Kg
= 19.515 / Kg
The final state is two phase region < < .
In saturated water labels for = 40°C.
= 
= 7.3851 kPa
= 7.3851 kPa
78950W the answer
Explanation:
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
Answer:
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Explanation:
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Based on the percent moisture content of the dried product, the mass of dried casein produced os 852.3 kg.
<h3>What is the mass of casein in wet casein?</h3>
The mass of casein in 1000 Kg of wet casein is 75% 1000 kg = 750 Kg
Mass of water 250 kg
The mass of casein is constant while the moisture content can be changed.
At 12% moisture content;
750 kg = 88%%
100 % = 100 ×750/88 = 852.27 kg
Therefore, the mass of dried casein produced os 852.3 kg.
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