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zhenek [66]
2 years ago
14

A steam power plant is represented as a heat engine operating between two thermal reservoirs at 800 K and 300 K. The temperature

of the steam in the boiler is 550 K and the temperature of the saturated water in the condenser is 330 K. The rates of heat transfer in the boiler and condenser are 4 MW and 2 MW, respectively. Please answer the following.
Required:
a. Represent the devices present in the heat engine and include their name along with the data given in the problem.
b. If this steam power plant were to operate as a reversible heat engine, with the boiler and condenser temperatures, what would be its thermal efficiency?
c. Use your result from part (b) to determine if the actual steam power plant can achieve such thermal efficiency. Justify your answer.
d. For the actual steam power plant is the Clausius inequality satisfied? Show you computation and discuss whether your result agrees or disagrees with your answer in part (c).
Engineering
1 answer:
Sergeeva-Olga [200]2 years ago
4 0
Yeet is the answer .....
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5. Name two health problems that fume can cause?<br> a)....<br> b)......
Vlad [161]

Answer:

A) Cancer of the Lungs

B)Larynx and Urinary Tract, as well as nervous system and kidney damage

Explanation:

5 0
3 years ago
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Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the br
vagabundo [1.1K]

Answer:

Explanation:

Using the kinematics equation v = v_o + a_ct to determine the velocity of car B.

where;

v_o = initial velocity

a_c = constant deceleration

Assuming the constant deceleration is = -12 ft/s^2

Also, the kinematic equation that relates to the distance with the time is:

S = d + v_ot + \dfrac{1}{2}at^2

Then:

v_B = 60-12t

The distance traveled by car B in the given time (t) is expressed as:

S_B = d + 60 t - \dfrac{1}{2}(12t^2)

For car A, the needed time (t) to come to rest is:

v_A = 60 - 18(t-0.75)

Also, the distance traveled by car A in the given time (t) is expressed as:

S_A = 60  * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2

Relating both velocities:

v_B = v_A

60-12t = 60 - 18(t-0.75)

60-12t =73.5 - 18t

60- 73.5 = - 18t+ 12t

-13.5 =-6t

t = 2.25 s

At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars

i.e.

S_B = S_A

d + 60 t - \dfrac{1}{2}(12t^2) = 60  * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2

d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60  * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2

d + 104.625 = 114.75

d = 114.75 - 104.625

d = 10.125 ft

3 0
2 years ago
An inductor and resistor are connected in parallel to a 120-V, 60-Hz line. The resistor has a resistance of 50 ohms, and the ind
BARSIC [14]

Answer:

hi

Explanation:

the answer would be I dont know

6 0
3 years ago
"Design a sequential circuit with two T flip-flops A and B, and one input x. When x = 0, the circuit remains in the same state.
Harlamova29_29 [7]

Answer:

See attached images for the diagrams and tables

3 0
3 years ago
A cylinder is to be cast out of aluminum. The diameter of the disk is 500 mm and its thickness is 20 mm. The mold constant 2.0 s
Nezavi [6.7K]

Answer:

a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min

b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²

first we find the volume and Area;

Volume V = πD²t / 4

Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³

Area A = 2πD²/ 4 + πDt

Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}

Area A = 392,699.08 + 31,415.93

Area A = 424,115 mm²

a)

Chvorinov’s rule

T(aluminium) = Cm (V/A) ²

T(aluminium) =  2.0 × (3,926,991 / 424,115) ²

T(aluminium) = 171.5 s = 2.86 min

∴ the minimum time (minutes) for the casting to solidify is 2.86 min

b)

For cast iron

Cm (mold constant = 1.488 sec/mm²)

Chvorinov’s rule

T(iron) = Cm (V/A) ²

T(iron) = 1.488 × (3,926,991 / 424,115) ²

T(iron) = 127.5720s = 2.13 min

Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

6 0
3 years ago
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