The substance is steam (H2O). NOTE: The purpose of this problem is to illustrate that there are conditions where water vapor is
far from ideal-gas behavior. For this reason, in this course, we will always use tables or NIST to determine properties of H2O for all conditions to avoid misuse of the ideal-gas EOS.
a. For the four conditions below, calculate the reduced pressure and the reduced temperature.
HINT: See Appendix E. Based on these values, decide if ideal-gas behavior is reasonable, i.e., that Pv = RiT within +/- 5%.
i. P = 15 MPa T = 1200 K
ii. P = 15 MPa T = 800 K
iii. P = 10 MPa T = 600 K
iv. P = 0.10 MPa T = 600 K
b. Use the generalized compressibility chart (Fig. 2.18 p. 92) to estimate a value of the compressibility factor Z for these conditions. How do these values compare with your thinking in part 1?
c. Use the NIST webbook to determine the compressibility factor Z for steam at the conditions above. How "ideal" is steam at these various conditions? How do these Z-values compare with your estimates from part 2?
a. For the four conditions below, calculate the reduced pressure and the reduced temperature.
HINT: See Appendix E. Based on these values, decide if ideal-gas behavior is reasonable, i.e., that Pv = RiT within +/- 5%.
i. P = 15 MPa T = 1200 K
ii. P = 15 MPa T = 800 K
iii. P = 10 MPa T = 600 K
iv. P = 0.10 MPa T = 600 K
b. Use the generalized compressibility chart (Fig. 2.18 p. 92) to estimate a value of the compressibility factor Z for these conditions. How do these values compare with your thinking in part 1?
c. Use the NIST webbook to determine the compressibility factor Z for steam at the conditions above. How "ideal" is steam at these various conditions? How do these Z-values compare with your estimates from part 2?
1 answer:
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Answer:
stress_ac = 5.333 MPa
shear stress_c = 1.763 MPa
Explanation:
Given:
- The missing figure is in the attachment.
- The dimensions of member AC = ( 6 x 25 ) mm x 2
- The diameter of the pin d = 19 mm
- Load at point A is P = 2 kN
Find:
- Find the axial stress in AE and the shear stress in pin C.
Solution:
- The stress in member AE can be calculated using component of force P along the member AE as follows:
stress_ac = P*cos(Q) / A_ae
Where, Angle Q: A_E_B and A_ac: cross sectional area of member AE.
cos(Q) = 4 / 5 ..... From figure ( trigonometry )
A_ae = 0.006*0.025*2 = 3*10^-4 m^2
Hence,
stress_ae = 2*(4/5) / 3*10^-4
stress_ae = 5.333 MPa
- The force at pin C can be evaluated by taking moments about C equal zero:
(M)_c = P*6 - F_eb*3
0 = P*6 - F_eb*3
F_eb = 0.5*P
- Sum of horizontal forces for member AC is zero:
P - F_eb - F_c = 0
F_c = 0.5*P
- The shear stress of double shear bolt is given by an expression:
shear stress = shear force / 2*A_pin
Where, The area of the pin C is:
A_pin = pi*d^2 / 4
A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2
Hence,
shear stress = 0.5*P / 2*A_pin
shear stress = 0.5*2 / 2*2.8353*10^-4
shear stress = 1.763 MPa
