Cos of error of special characters
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an adiabatic compressor receives 1.5 meter cube per second of air at 30 degrees celsius and 101 kpa. The discharge pressure is 5
05 kpa and the power supplied is 325 kW, what is the discharge temperature
1 answer:
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Answer:
M_AD = 1359.17 lb-in
Explanation:
Given:
- T_ef = 46 lb
Find:
- Moment of that force T_ef about the line joining points A and D.
Solution:
- Find the position of point E:
mag(BC) = sqrt ( 48^2 + 36^2) = 60 in
BE / BC = 45 / 60 = 0.75
Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in
- Find unit vector EF:
mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in
vec(EF) = < -15 , -110 , 30 >
unit(EF) = (1/115) * < -15 , -110 , 30 >
- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb
- Find unit vector AD:
mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in
vec(AD) = < 48 , -12 , 36 >
unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >
unit (AD) = <0.7845 , -0.19612 , 0.58835 >
Next:
M_AD = unit(AD) . ( E x T_EF)
M_AD = 1835.73 + 116.49528 - 593.0568
M_AD = 1359.17 lb-in
Answer:
a)σ₁ = 265.2 MPa
b)σ₂ = -172.8 MPa
c)
d)Range = 438 MPa
Explanation:
Given that
Mean stress ,σm= 46.2 MPa
Stress amplitude ,σa= 219 MPa
Lets take
Maximum stress level = σ₁
Minimum stress level =σ₂
The mean stress given as
2 x 46.2 = σ₁ + σ₂
σ₁ + σ₂ = 92.4 MPa --------1
The amplitude stress given as
2 x 219 = σ₁ - σ₂
σ₁ - σ₂ = 438 MPa --------2
By adding the above equation
2 σ₁ = 530.4
σ₁ = 265.2 MPa
-σ₂ = 438 -265.2 MPa
σ₂ = -172.8 MPa
Stress ratio
Range = 265.2 MPa - ( -172.8 MPa)
Range = 438 MPa