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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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An ant starts at one edge of a long strip of paper that is 34.2 cm wide. She travels at 1.3 cm/s at an angle of 61◦ with the lon
ankoles [38]

Answer:

t = 30.1 sec

Explanation:

If the ant is moving at a constant speed, the velocity vector will have the same magnitude at any point, and can be decomposed in two vectors, along directions perpendicular each other.

If we choose these directions coincident with the long edge of the paper, and the other perpendicular to it, the components of the velocity vector, along these axes, can be calculated as the projections of this vector along these axes.

We are only interested in the component of the velocity across the paper, that can be calculated as follows:

vₓ = v* sin θ, where v is the magnitude of the velocity, and θ the angle that forms v with the long edge.

We know that v= 1.3 cm/s, and θ = 61º, so we can find vₓ as follows:

vₓ = 1.3 cm/s * sin 61º = 1.3 cm/s * 0.875 = 1.14 cm/s

Applying the definition of average velocity, we can solve for t:

t =\frac{x}{vx} = \frac{34.2 cm}{1.14 cm/s} =30.1 sec

⇒ t = 30.1 sec

7 0
3 years ago
Air enters an adiabatic gas turbine at 1590 oF, 40 psia and leaves at 15 psia. The turbine efficiency is 80%, and the mass flow
melamori03 [73]

Answer:

a) 158.4 HP.

b) 1235.6 °F.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to set up an energy balance for the turbine's inlets and outlets:

m_{in}h_{in}=W_{out}+m_{out}h_{out}

Whereas the mass flow is just the same, which means we have:

W_{out}=m_{out}(h_{out}-h_{in})

And the enthalpy and entropy of the inlet stream is obtained from steam tables:

h_{in}=1860.7BTU/lbm\\\\s_{in}= 2.2096BTU/lbm-R

Now, since we assume the 80% accounts for the isentropic efficiency for this adiabatic gas turbine, we assume the entropy is constant so that:

s_{out}= 2.2096BTU/lbm-R

Which means we can find the temperature at which this entropy is exhibited at 15 psia, which gives values of temperature of 1200 °F (s=2.1986 BTU/lbm-K) and 1400 °F (s=2.2604 BTU/lbm-K), and thus, we interpolate for s=2.2096 to obtain a temperature of 1235.6 °F.

Moreover, the enthalpy at the turbine's outlet can be also interpolated by knowing that at 1200 °F h=1639.8 BTU/lbm and at 1400 °F h=174.5 BTU/lbm, to obtain:

h_{out}=1659.15BUT/lbm

Then, the isentropic work (negative due to convention) is:

W_{out}=2500lbm/h(1659.15BUT/lbm-1860.7BUT/lbm)\\\\W_{out}=-503,875BTU

And the real produced work is:

W_{real}=0.8*-503875BTU\\\\W_{real}=-403100BTU

Finally, in horsepower:

W_{real}=-403100BTU/hr*\frac{1HP}{2544.4336BTU/hr} \\\\W_{real}=158.4HP

Regards!

6 0
3 years ago
Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an
fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

8 0
3 years ago
g Design of a spindle present in an existing design needs to be reviewed for use under new loading needs. It is currently design
ss7ja [257]

Answer: its c

Explanation:

6 0
3 years ago
In the combination of resistors above, consider the 1.50 µΩ and 0.75 µΩ. How can you classify the connection between these two r
Airida [17]

Answer: they are connected in series.

Explanation:

3 0
4 years ago
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