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Andreyy89
3 years ago
6

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?

Engineering
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

See the attached pictures for detailed answer.

Explanation:

See the attached pictures for step by step explanation.

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can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.​
marishachu [46]

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

6 0
4 years ago
Các đặc điểm chính của đường dây dài siêu cao áp .
rodikova [14]

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

8 0
3 years ago
Which of the following has special properties that allow forces and pressure to be distributed evenly?
Thepotemich [5.8K]

Answer:

Fluids

Explanation:

Fluids has special properties that allow forces and pressure to be distributed evenly within them.

  • Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
  • Therefore, when pressure is applied to them, it permeates evenly on all parts.
  • Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.
7 0
3 years ago
2. A well of 0.1 m radius is installed in the aquifer of the preceding exercise and is pumped at a rate averaging 80 liter/min.
hodyreva [135]

Question:

The question is not complete. See the complete question and the answer below.

A well that pumps at a constant rate of 0.5m3/s fully penetrates a confined aquifer of 34 m thickness. After a long period of pumping, near steady state conditions, the measured drawdowns at two observation wells 50m and 100m from the pumping well are 0.9 and 0.4 m respectively. (a) Calculate the hydraulic conductivity and transmissivity of the aquifer (b) estimate the radius of influence of the pumping well, and (c) calculate the expected drawdown in the pumping well if the radius of the well is 0.4m.

Answer:

T = 0.11029m²/sec

Radius of influence = 93.304m

expected drawdown = 3.9336m

Explanation:

See the attached file for the explanation.

8 0
3 years ago
The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc
Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy  

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.  

Given Data  

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist  

∅=4°

∅=4*\pi/180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist  

∅=TL/GJ

∅=TL/G*\pi/2*c^4

∅=2TL/G*\pi*c^4

c=\sqrt[4]{2TL/\pi G }∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J*\pi/2*c^4)]*c

г =[2T/(J*\pi*c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required  

We will use larger of the two values  

c= 18.0569 x 10^-3 m  

c = 18.0569 mm  

3 0
4 years ago
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