Answer:
For
- 5.556 lb/s
For
- 7.4047 lb/s
Solution:
As per the question:
System Load = 96000 Btuh
Temperature, T = 
Temperature rise, T' =
Now,
The system load is taken to be at constant pressure, then:
Specific heat of air, 
Now, for a rise of
in temeprature:


Now, for
:


Complete Question
The complete question is shown on the first uploaded image.
Answer:
The answer is shown on the second uploaded image
Explanation:
The explanation is also shown on the second uploaded image
Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
Answer: THERE IS NO GRAVITY IN SPACE SO ROCKS SPIN
Explanation: