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ra1l [238]
4 years ago
7

A Carnot power cycle is executed in a piston-cylinder system using 1 kg of water. The cycle consists of an isobaric expansion of

a saturated liquid at 160 °C (State 1) to a volume of 0.30 m^3 (State 2) followed by a reversible adiabatic expansion to 20 °C (State 3). The water is then compressed isothermally to State 4, and finally compressed reversibly and adiabatically back to the original State 1.
(a) Sketch the cycle on a P-v diagram.
(b) Determine the heat transfer into the cycle and the net work for the cycle, in kJ.
(c) Calculate the thermal efficiency of this power cycle using the values found in part (b). Compare this value with the value based on the maximum possible value for a reversible cycle operating between the absolute temperatures of the high-temperature and low-temperature thermal reservoirs.
Engineering
1 answer:
jeka944 years ago
5 0

Answer:

b) Determine the heat transfer into the cycle and the net work for the cycle, in kJ.

Explanation:

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5 0
3 years ago
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Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
myrzilka [38]

Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

Inlet and outlet velocities are found:

v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }

v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

Now, the head associated with the pump is finally computed:

h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

h_{pump} = 7.705\,m

The power that pump adds to the fluid is:

\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

4 0
3 years ago
the hoop is cast on the rough surface such that it has an angular velocity w=4rad/s and an angular acceleration a=5rad/s^2. also
IgorLugansk [536]

Given Information:

Angular velocity = ω = 4 rad/s

Angular acceleration = α = 5 rad/s²

Center deceleration = a₀ = 2 m/s

Required Information:

Acceleration of point A at this instant = ?

Answer:

Acceleration of point A at this instant = 5.94 m/s²

Explanation:

Refer to the attached diagram of the question,

The acceleration of point A is given by

a = a₀ + rα - rω²

Where r is the radial distance between the center and point A, a₀ is the deceleration of center, α is the angular acceleration and ω is the angular velocity.

a = -2i + 0.3j*5k - 0.3j*4²

a = -2i + 1.5(j*k) - 0.3j*16

a = -2i + 1.5(-i) - 4.8j

a = -2i - 1.5i - 4.8j

a = -3.5i - 4.8j

The magnitude of acceleration vector is

a = √(-3.5)² + (-4.8)²

a = √35.29

a = 5.94 m/s²

Therefore, the acceleration of point A is 5.94 m/s²

The angle is given by

θ = tan⁻¹(y/x)

θ = tan⁻¹(-4.8/-3.5)

θ = 53.9°

7 0
4 years ago
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Ugo [173]

Answer: English please!

Explanation:

3 0
3 years ago
Is a street the same as a avenue
-BARSIC- [3]

they're essentially the same thing so i'd say yes

5 0
3 years ago
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