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3 years ago

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11.A heat engine operates between two reservoirs at 800 and 20°C. One-half of the work output of the heat engine is used to driv

e a Carnot heat pump that removes heat from the cold surroundings at 2°C and transfers it to a house maintained at 22°C. If the house is losing heat at a rate of 62,000 kJ/h, determine the minimum rate of the heat supply to the heat engine required to keep the house at 22°C.

1 answer:

statuscvo [17]3 years ago

3 0

Answer: Meh account banned

Explanation:

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Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim workin

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4 years ago

At any given state the specific internal energy of a gas is always greater its specific enthalpy. a)True b) False

IgorC [24]

Answer:

(b)False

Explanation:

We know that specific Internal energy of gas u= $C_v$ T

and specific enthalpy of gas h= $C_p$ T

If we take the case of air we know that

$C_v$ =0.707 KJ/Kg=K , $C_p$ =1.005 KJ/Kg=K

If we take A fixed temperature T=300 K

so u=212.1 KJ/ kg ,h=301.5 KJ/kg

So we can say that specific enthaply of gas is always greater than its specific internal energy.

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3 years ago

Describe in your own words the three strengthening mechanisms discussed in this chapter (i.e., grain size reduction, solid-solut

Kobotan [32]

Explanation:

Strengthening by grain size reduction

It is based on the fact that dislocations will experience hindrances while trying to move from a grain into the next because of abrupt change in orientation of planes.

Hindrances can be two types: forcible change of slip direction, and discontinuous slip plane.
Smaller the grain size, often a dislocation encounters a hindrance. Yield strength of material will be increased.
Yield strength is related to grain size (diameter, d ) as Hall Petch relation:

$\sigma_{y}=\sigma_{i}+k d^{-1 / 2}$

Strengthening by Grain size reduction (contd..)

Grain size reduction improves not only strength, but also the toughness of many alloys.
If d is average grain diameter, $S_{v}$ is grain boundary area per unit volume, $N_{L}$ is mean number of intercepts of grain boundaries per unit length of test line, $N_{A}$ is number of grains per unit area on a polished surface:

$S_{v}=2 N_{L} \quad d=\frac{3}{S_{v}}=\frac{3}{2 N_{L}} \quad d=\sqrt{\frac{6}{\pi V_{A}}}$

Grain size can also be measured by comparing the grains at a fixed magnification with standard grain size charts.

Other method: Use of ASTM grain size number (Z). It is related to grain diameter, (in mm) as follows:

$D=\frac{1}{100} \sqrt{\frac{645}{2^{6-1}}}$

Solid solution strengthening

Impure foreign atoms in a single phase material produces lattice strains which can anchor the dislocations.

Effectiveness of this strengthening depends on two factors size difference and volume fraction of solute. Solute atoms interact with dislocations in many ways:

- elastic interaction

- modulus interaction

- stacking-fault interaction

- electrical interaction

- short-range order interaction

- long-range order interaction

3 0

4 years ago

What size discharge line ( pressure line ) would you select for a 20 gpm pump?

Alenkasestr [34]

Answer:

diameter= 1 in

Explanation:

Some authors have developed studies to find the recommended flow rate at the pump outlet, in order to avoid high pressure losses and cavitation.

Assuming that this pump carries water, the recommended speed is 2m / s, so knowing the flow rate = 20gpm we can find the diameter of the pipe.

Q=flow rate=20gpm=0.0012618m^3/s

V=speed=2m/s

A=Area

Q=VA

A=Q/V

$A=\frac{0.0012618m^3/s}{2m/s} =0.0006309m^2=630.2mm^2\\\\$

Taking into account that the flow is through a circular pipe we can use the equation to find the area of a circle, and then find the diameter

$A=\frac{\pi D^2 }{4} \\D=\sqrt[2]{\frac{4A}{\pi } } =\sqrt{\frac{4(630.8)}{\pi } } =23.34mm$

the result is 23.34mm, if we find the closest commercial diameter to this value, we find that it is 25.4mm = 1in

3 0

3 years ago

To inspect a 12,500 N car, it is raised with a hydraulic lift. If the radius of the small piston is 5.0 cm, and the radius of th

labwork [276]

Answer:

The force will be "125 N".

Explanation:

The given values are:

$F_1=12500 \ N$

$R_1 = 50 \ cm$

$R_2=5 \ cm$

As we know,

⇒ $A=\pi(H)^2$

then,

⇒ $A_2=\pi(5)^2$

⇒ $A_1=\pi(50)^2$

Since,

The pressure on both the pistons are equal, then

⇒ $\frac{F_1}{A_1} =\frac{F_2}{A_2}$

or,

⇒ $\frac{F_2}{F_1} =\frac{A_2}{A_1}$

By substituting the values, we get

⇒ $\frac{F_2}{12500} =\frac{\pi(5)^2}{\pi(50)^2}$

⇒ $\frac{F_2}{12500} =\frac{\pi(25)}{\pi(2500)}$

⇒ $F_2=\frac{25}{2500}\times 12500$

⇒ $=0.01\times 12500$

⇒ $=125 \ N$

5 0

3 years ago