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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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A closed vessel of volume 80 litres contain gas at a gauge pressure of 150 kPa. If the gas is compressed isothermally to half it
horsena [70]

Answer:

The resulting pressure is 300 kilopascals.

Explanation:

Let consider that gas within the closed vessel behaves ideally. By the equation of state for ideal gases, we construct the following relationship for the isothermal relationship:

P_{1}\cdot V_{1} = P_{2} \cdot V_{2} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kilopascals.

V_{1}, V_{2} - Initial and final volume, measured in litres.

If we know that \frac{V_{1}}{V_{2}} = 2 and P_{1} = 150\,kPa, then the resulting pressure is:

P_{2} = P_{1}\times \frac{V_{1}}{V_{2}}

P_{2} = 300\,kPa

The resulting pressure is 300 kilopascals.

6 0
3 years ago
A turbojet aircraft is flying with a velocity of 280 m/s at an altitude of 9150 m, where the ambient conditions are 32 kPa and -
artcher [175]

Answer:

(a) The velocity of the exhaust gases. is 832.7 m/s

(b) The rate of fuel consumption is 0.6243 kg/s

Explanation:

For the given turbojet engine operating on an ideal cycle, the pressure ,temperature, velocity, and specific enthalpy of air at i^{th} state are P_i , T_i , V_i , and h_i , respectively.

Use "ideal-gas specific heats of various common gases" to find the properties of air at room temperature.

Specific heat at constant pressure, c_p = 1.005 kJ/kg.K

Specific heat ratio, k = 1.4

3 0
3 years ago
A column in a building is subjected to the following load effects:
vagabundo [1.1K]

Answer:

attached below

Explanation:

6 0
3 years ago
Technician A says that to start a fuel-injected car, the accelerator should be depressed
Bumek [7]

Answer:

Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A​

Explanation:

Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A​

7 0
2 years ago
The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i
koban [17]

Answer:  (a) 9.00 Mega Newtons or 9.00 * 10^6 N

               (b)  17.1 m

Explanation:  The length of wall under the surface can be given by

                                            b=25m/sin(60)\\=28.867

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

F(resultant) = Pavg ( A) = (Patm +  \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N

Noting from the Bernoulli  equation that

Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0
3 years ago
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