Answer:
Q=36444.11 Btu
Explanation:
Given that
Initial temperature = 60° F
Final temperature = 110° F
Specific heat of water = 0.999 Btu/lbm.R
Volume of water = 90 gallon
Mass = Volume x density
Mass ,m= 90 x 0.13 x 62.36 lbm
m=729.62 lbm
We know that sensible heat given as
Q= m Cp ΔT
Now by putting the values
Q= 729.62 x 0.999 x (110-60) Btu
Q=36444.11 Btu
Answer:
D. Both pull-in and hold-in windings are energized.
Explanation:
The instant the ignition switch is turned to the start position, "Both pull-in and hold-in windings are energized." This is because the moment the ignition switch is turned to the start position, voltage passes through to the S terminal of the solenoid.
The hold-in winding is attached to the case of the solenoid. Similarly, the pull-in winding is also attached to the starter motor. Thereby, the current will move across both windings by getting energized to generate a strong magnetic field.
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>
