Answer:
Impossible.
Explanation:
The ideal Coefficient of Performance is:
The real Coefficient of Performance is:
Which leads to an absurds, since the real Coefficient of Performance must be equal to or lesser than ideal Coefficient of Performance. Then, the cycle is impossible, since it violates the Second Law of Thermodynamics.
Answer:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
Explanation:
Given:
- The diameter of the pipe d = 5-in
- The pipe is supported every L = 9 ft of pipe in length
- The weight if the pipe + contents W = 10 lb/ft
Find:
determine the magnitude and location of the maximum bending moment in members AC and BC.
Solution:
- The figure (missing) is given in the attachment.
- We will first determine the external forces acting on each member:
Section: 9-ft section of pipe.
Sum of forces perpendicular to member AC = 0
F_d - 0.8*W*L = 0
F_d = 0.8*10*9 = 72 lb
Sum of forces perpendicular to member BC = 0
F_e - 0.6*W*L = 0
F_e = 0.6*10*9 = 54 lb
F_d = 72 lb , F_e = 54 lb
- Then we will determine the support reactions for each member AC point A and BC point B.
Section: Entire Frame.
Sum of moments about point B = 0
-A_y*(18.75/12) + F_d*(d /2*12) + F_e*((11.25-2.5)/12) = 0
-A_y*(1.5625) + 15 + 39.375 = 0
A_y = 34.8 lb
Sum of forces in vertical direction = 0
A_y + B_y - 0.8*F_d - 0.6*F_e = 0
B_y = 0.8*(72) + 0.6*(54) - 34.8
B_y = 55.2 lb
Sum of forces in horizontal direction = 0
A_x + B_x - 0.6*F_d + 0.8*F_e = 0
A_x + B_x = 0
Section: Member AC
Sum of moments about point C = 0
F_d*(2.5/12) - A_y*(12/12) - A_x*(9/12) = 0
72*2.5 - 34.8*12 - 9*A_x = 0
A_x = -237.6 / 9 = - 26.4 lb
B_x = - A_x = 26.4 lb
A_x = -26.4 lb , B_x = 26.4 lb
- Now we will calculate bending moment for each member at different sections.
Member AC:
From point A till just before point D
-0.6*A_x*x - A_y*0.8*x + M = 0
15.84*x - 27.84*x + M = 0
M = 12*x ..... max value at D, x = 12.25 in
M_max = 12*12.25/12 = 12.25 lb-ft
Member BC:
From point B till just before point E
-0.8*B_x*x + B_y*0.6*x + M = 0
-21.12*x + 33.12*x + M = 0
M = -12*x ..... max value at E, x = 11.25 - 2.5 = 8.75 in
M_max = -12*8.75/12 = -8.75 lb-ft
- The maximum bending moments and their locations are:
AC: at D , M_max = 12.25 lb-ft
BC: at E , M_max = 8.75 lb-ft
Answer:
the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
Explanation:
Given the data in the question;
From the first law of thermodynamics;
dQ = dU + dW ------ let this be equation 1
where dQ is the heat transfer, dU is internal energy and dW is the work done.
from the question, the process is isothermal ( internally reversible process )
Thus, the change in internal energy is 0
dU = 0
given that; Air is compressed by a 40-kW compressor from P1 to P2
since it is compressed, dW = -40 kW
we substitute into equation 1
dQ = 0 + ( -40 kW )
dQ = -40 kW
Now, change in entropy of air is;
ΔS = dQ / T
given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K
so we substitute
ΔS = -40 kW / 298.15 K
ΔS = -0.13416 ≈ -0.1342 kW/K
Therefore, the rate of entropy change of the air is -0.1342 kW/K
the assumptions made in solving this problem
- Air is an ideal gas.
- the process is isothermal ( internally reversible process ). the change in internal energy is 0.
- It is a steady flow process
- Potential and Kinetic energy changes are negligible.
