Answer:
So, the evaporating pressure of the R410A = 118 psig
Explanation:
Solution:
For R410A system:
Data Given:
Evaporator Outlet Temperature = 50°F
Evaporator Superheat = 10°F
Required:
Evaporating Pressure in the system = ?
For this, first of all, we need to calculate inlet temperature on R410A system from the given value of outlet temperature.
Evaporator inlet temperature is the difference of outlet temperature and evaporator superheat.
Evaporator inlet temperature = Outlet Temperature - Evaporator Superheat
Evaporator inlet Temperature = 50°F - 10°F
Evaporator inlet Temperature = 40°F
Now, as we have the inlet temperature and the R410A system. We can consult the pressure temperature chart or PT chart, which I have attached and highlighted the value of evaporating pressure for 40°F inlet temperature.
So, the evaporating pressure of the R410A = 118 psig
Answer:
The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.
Explanation:
ω1= 1.72x10^4 rad/sec
ω2= 5.42x10^4 rad/sec
ωmax= 8.42x10^4 rad/sec
θ= 1.72x10^4 rad
α=7.67 x10^4 rad/sec²
t= ωmax / α
t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²
t=1.097 sec