Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²
x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K
K= 58.66 lb/ft
The damper coefficient for critically damped system
Cc= 12.7 lb.sec/ft
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci
1 answer:
You might be interested in
Answer:
The rate of work output = -396.17 kJ/s
Explanation:
Here we have the given parameters
Initial temperature, T₁ = 355°C = 628.15 K
Initial pressure, P₁ = 350 kPa
h₁ = 763.088 kJ/kg
s₁ = 4.287 kJ/(kg·K)
Assuming an isentropic system, from tables, we look for the saturation temperature of saturated air at 4.287 kJ/(kg·K) which is approximately
h₂ = 79.572 kJ/kg
The saturation temperature at the given
T₂ = 79°C
The rate of work output =
×
×(T₂ - T₁)
Where;
= The specific heat of air at constant pressure = 0.7177 kJ/(kg·K)
= The mass flow rate = 2.0 kg/s
Substituting the values, we have;
= 2.0 × 0.7177 × (79 - 355) = -396.17 kJ/s
= -396.17 kJ/s