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Oksi-84 [34.3K]
2 years ago
5

Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constan

t pressure to a final state. If the heat transfer for the process is 2960 kJ, determine the temperature at the final state, in °C, and the work, in kJ. Kinetic and potential energy effects are negligible.
Engineering
1 answer:
Digiron [165]2 years ago
7 0

Answer:

The final temperature of water is 381.39  °C.

Explanation:

Given that

Mass of water = 5 kg

Heat transfer at constant pressure Q = 2960 KJ

Initial temperature = 240 °C

We know that heat transfer at constant pressure given as follows

Q=mC_p\Delta T

We know that for water

C_p=4.187\ \frac{KJ}{kg.K}

Lets take final temperature of water is T

So

Q=mC_p\Delta T

2960=5\times 4.187(T-240)

T=381.39  °C

So the final temperature of water is 381.39  °C.

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3 years ago
Unlike expendable molds, permanent molds do not collapse, so the mold must be opened before appreciable cooling contraction occu
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Answer: True

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3 years ago
I don’t understand this
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Answer:

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7 0
3 years ago
7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21
Irina18 [472]

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

In this problem, the first 3 timing stages are given as:

200, 187, and 210 veh/h.

We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

g = (\frac{v}{s}) (\frac{C}{X})

Let's first find the fourth stage critical lane group ratio \frac{v}{s} , using the formula:

C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}

60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}

Solving for (\frac{v}{s}), we have:

(\frac{v}{s}) = 0.185

Let's also calculate the volume capacity ratio X,

X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

g_4 = (\frac{v}{s}) (\frac{C}{X})

Substituting figures in the equation, we now have:

g_4 = (0.185) (\frac{60}{0.704})

g_4 = 15.78 seconds

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

8 0
2 years ago
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