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kicyunya [14]
3 years ago
5

1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con

stant heat addition from 100 kPa and 27∘C to 700 kPa 2-3 Isothermal expansion until V3 = 7v23-1 P = constant heat rejection to the initial state2. Assume air has constant properties with cv = 0.718 kJ/kg K, cp = 1.005 kJ/kg K, R = 0.287 kJ/kg K, and k = 1.4.(a) Sketch the P- and T-s diagrams for the cycle.(b) Determine the ratio of the compression work to the expansion work (the back work ratio).(c) Determine the cycle thermal efficiency.
Engineering
1 answer:
QveST [7]3 years ago
5 0

Answer:

Explanation: Here it is: 67 Hope that helps! :)

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1. Which type of fit implies that a piece will never fit? a. interference fit b. construction fit c. transition fit d. impeding
VladimirAG [237]

Answer:

d

Explanation:

interference fit: the dimensions are larger and high force like heating is required to fit in the object

transition fit: the dimensions are fractionally larger and slight force like hammering is required to fit in the object

impeding fit: the dimensions are larger and the object will not fit at all

construction fit: activities making a commercial space sutiable for tenant occupation.

6 0
3 years ago
Help please i will give brainlist
vlada-n [284]

Answer:

Open Circuit

Explanation:

Because the circuit isnt closed, the current wont flow

5 0
2 years ago
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You live on a street that runs East to West. You just had 2 inche of snow and you live on the North side of the street. You retu
RSB [31]

Answer:

The heat from the sun melted it

Explanation:

If the street runs east to west, houses on the south (across the street) will project shadows on their sidewalk, while the northern sidewalk will be illuminated. This is for the northern hemisphere, on the southern hemisphere it would be the other way around.

5 0
3 years ago
Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D
slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m   ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N)   --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

<u>F = 9.11 x 10³ N = 9.11 KN</u>

6 0
3 years ago
W²-5w+14
alexira [117]

Answer:w²-5w+14

4x²+11x+16

6x²+7x-10

x²-16x+56

Explanation:

7 0
3 years ago
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