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:
Explanation:
load = 4500lb lift height= 30 ft
time =15 s
velocity= ft/s
velocity=2 ft/s
power = force velocity
power=
power= 9000 lb ft/s
1 hp= 550 lb ft/s
power= hp
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:
4.17x10^-3 kW/K
Explanation:
Detailed explanation and calculation is shown in the image below
