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Marrrta [24]
2 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]2 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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For an Otto cycle, plot the cycle efficiency as a function of compression ratio from 4 to 16.
Elza [17]

Assumptions:

  • Steady state.
  • Air as working fluid.
  • Ideal gas.
  • Reversible process.
  • Ideal Otto Cycle.

Explanation:

Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):

  • Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
  • Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).

        r =\frac{V_1}{V_2}

  • Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
  • Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.

         r = \frac{V_4}{V_3} = \frac{V_1}{V_2}

  • Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
  • Exhaust 1-0: the working fluid is vented to the atmosphere.

If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:

           \eta = 1-(\frac{1}{r^{\gamma - 1} } )

where:

           \gamma = \frac{C_{p} }{C_{v}} : specific heat ratio

Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.

           \gamma = 1.4

Answer:

See image attached.

5 0
3 years ago
Design a program that calculates the area and circumstance of rectangle?​
Phantasy [73]
Let “w” and “L” be the width and length of the rectangle. “p” and “a” are perimeter and area
For python,
w=int(input(“width”))
l=int(input(“length”))
a= w*l
p=2*w+2*l
print(str(a), str(p)
4 0
2 years ago
Is Tesla French, American, German, or Russian?
ozzi

Answer:

I am gonna say it is AMERICAN

6 0
3 years ago
Read 2 more answers
Ignoring any losses, estimate how much energy (in units of Btu) is required to raise the temperature of water in a 90-gallon hot
Rudik [331]

Answer:

Q=36444.11 Btu

Explanation:

Given that

Initial temperature = 60° F

Final temperature = 110° F

Specific heat of water = 0.999 Btu/lbm.R

Volume of water = 90 gallon

Mass = Volume x density

1\ gallon = 0.13ft^3

Mass ,m= 90 x 0.13 x 62.36 lbm

m=729.62 lbm

We know that sensible heat given as

Q= m Cp ΔT

Now by putting the values

Q= 729.62 x 0.999 x (110-60) Btu

Q=36444.11 Btu

5 0
3 years ago
What is the metal removal rate when a 2 in-diameter hole 3.5 in deep is drilled in 1020 steel at cutting speed of 120 fpm with a
Studentka2010 [4]

Answer:

a) the metal removal rate is 14.4 in³/min

b) the cutting time is 0.98 min

Explanation:

Given the data from the question

first we find the rpm for the spindle of the drilling tool, using the equation

Ns = 12V/πD

V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)

so we substitute

Ns = 12 × 120 / π2

Ns = 1440 / 6.2831

Ns = 229.18 rmp

Now we find the metal removal rate using the equation

MRR = (πD²/4) Fr × Ns

Fr is the feed rate( 0.02 ipr ),

so we substitute

MRR = ((π × 2²)/4) × 0.02 × 229.18

MRR = 14.3998 ≈ 14.4 in³/min

Therefore the metal removal rate is 14.4 in³/min

Next we find the allowance for approach of the tip of the drill

A = D/2

A = 2/2

= 1 in

now find the time required to drill the hole

Tm = (L + A) / (Fr × Ns)

Lis the the depth of the hole( 3.5 in)

so we substitute our values

Tm = (3.5 + 1) / (0.02 × 229.18  )

Tm = 4.5 / 4.5836

Tm = 0.98 min

Therefore the cutting time is 0.98 min

8 0
2 years ago
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