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Marrrta [24]
3 years ago
5

Which of the following are all desirable properties of a hydraulic fluid? a. good heat transfer capability, low viscosity, high

density b. good lubricity, high viscosity, low density c. chemically stable, compatible with system materials, good heat insulative capability d, readily available, high density, large bulk modulus e. fire resistant, inexpensive, non-toxic.
Engineering
1 answer:
Vinvika [58]3 years ago
5 0

Answer:

e.Fire resistance,Inexpensive,Non-toxic.

Explanation:

Desirable hydraulic property of fluid as follows

1. Good chemical and environment stability

2. Low density

3. Ideal viscosity

4. Fire resistance

5. Better heat dissipation

6. Low flammability

7. Good lubrication capability

8. Low volatility

9. Foam resistance

10. Non-toxic

11. Inexpensive

12. Demulsibility

13. Incompressibility

So our option e is right.

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The cables of a power line are copper-coated steel wire. The overall diameter of the wire is 5/8 in. The steel core has a diamet
Jobisdone [24]

Answer:

The stresses in steel and the copper are, 128 ksi and 72.44 ksi, respectively.

Explanation:

In this case, we need to calculate the area of both portions separately.

Thus, we know that:

Inner Radius = Rin = (1/2 in)/2 = 0.25 in

Outer Radius = Rout = (5/8 in)/2 = 0.3125 in

FOR STEEL CORE:

Area of steel core = As = π(Rin)² = π(0.25 in)² = 0.0625 in²

Hence, the stress will be:

σs = Force / As

σs = 8000 lb/ 0.0625 in²

<u>σs = 128 ksi</u>

FOR COPPER:

Area of copper = Ac = π[(Rout)² - (Rin)²]

Ac = π[(0.3125 in)² - (0.25 in)²] = 0.11044 in²

Hence, the stress will be:

σc = Force / Ac

σc = 8000 lb/ 0.11044 in²

<u>σc = 72.44 ksi</u>

<u></u>

3 0
3 years ago
W<br>n só<br>i<br>Eo<br>E<br>find the transfer function​
zaharov [31]
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8 0
3 years ago
On a day in which the local atmospheric pressure is 99.5 kPa, answer each of the following: (a) Calculate the column height of m
horrorfan [7]

Answer:

C . . . . . . . . . . . . . . . . . . . . . .

8 0
2 years ago
A 2.2-kg model rocket is launched vertically and reaches an altitude of 70 m with a speed of 30 m/s at the end of powered flight
kirill115 [55]

Answer:

r_b= (30.8\hat{i} + 69.96 \hat{j}) m

Explanation:

given,

mass = 2.2 kg

altitude(r₀) = (70 j) m

speed = 30 m/s

m_a = 0.77 kg

m_b =1.43 kg

part A strike ground (r_a)= (80 i) m

t = 6 s

r = r_0 + v_ot-\dfrac{1}{2}gt^2

r = 60\hat{j} + (30\hat{j})\times 6-\dfrac{1}{2}\times 9.8 \times 6^2

r = 63.6 j m

by conservation of energy

mr = m_ar_a+m_br_b

2.2\times 63.6\hat{j} = 0.77\times (-80 \hat{i})+2\times r_b

r_b= (30.8\hat{i} + 69.96 \hat{j}) m

8 0
3 years ago
1. A cylindrical casting is 0.3 m in diameter and 0.5 m in length. Another casting has the same metal is rectangular in cross-se
Lorico [155]

Based on the Chvorinov's rule, the diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the prism casting.

<h3>How to apply the Chvorinov's rule for casting processes</h3>

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a <em>reference</em> time. This rule states that cooling time (<em>t</em>) is directly proportional to the square of the volume (<em>V</em>), in cubic meters, divided to the surface area (<em>A</em>), in square meters. Now we proceed to model each casting:

<h3>Cylindrical casting</h3>

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]²    (1)

<h3>Prism casting</h3>

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]²     (2)

<h3>Relationship between the cross sections of both castings</h3>

3 · T² = 0.25π · D²     (3)

Where:

  • <em>t</em> - Cooling time of the cylindrical casting, in time unit.
  • <em>t'</em> - Cooling time of the prism casting, in time unit.
  • <em>C</em> - Cooling factor, in time unit per square meter.
  • <em>D</em> - Diameter of the cylinder, in meters.
  • <em>L</em> - Length of the casting, in meters.
  • <em>T</em> - Width of the cross section of the prism casting, in meters.

If we know that <em>D =</em> <em>0.3 m</em>, then the thickness of the prism casting is:

T = \sqrt{\frac{\pi}{12} }\cdot D

<em>T ≈ 0.153 m</em>

<em />

And (1) and (2) simplified into these forms:

<h3>Cylindrical casting</h3>

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C     (1b)

<h3>Prism casting</h3>

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C     (2b)

Lastly we find the <em>percentual</em> difference in the solidification times of the two castings by using the following expression:

<em>r = (</em>1 <em>- t'/t) ×</em> 100 %

<em>r = (</em>1 <em>-</em> 0.00218<em>/</em>0.0329<em>) ×</em> 100 %

<em>r =</em> 93.374 %

The <em>cooling</em> time of the <em>prism</em> casting is 6.626 % of the <em>solidification</em> time of the <em>cylindrical</em> casting. The diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the <em>prism</em> casting. \blacksquare

To learn more on solidification times, we kindly invite to check this verified question: brainly.com/question/13536247

3 0
3 years ago
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