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kap26 [50]
2 years ago
3

At steady state, a power cycle having a thermal efficiency of 38% generates 100 MW of electricity while discharging energy by he

at transfer to cooling water at an average temperature of 70 F. The average temperature of the steam passing through the boiler is 900 F.
Determine the rate at which is discharged to the cooling water, in BTU/h
Engineering
1 answer:
Tresset [83]2 years ago
5 0

Answer:

Heat rejected = 556,724,951 Btu/hr

Explanation:

given data

thermal efficiency = 38%

electricity = 100 MW

average temperature = 70 F

average temperature = 900 F

solution

Heat input = Output ÷ efficiency   ........................1

Heat input = 100 ÷ 0.38

Heat input = 263.16 MW  

and

Heat rejected = Heat input - Output    ........................2

Heat rejected = 263.16 - 100

Heat rejected = 163.16 MW

Heat rejected = 556,724,951 Btu/hr

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Which of the following is NOT a characteristic of hazardous waste
lbvjy [14]

Answer:

A RCRA characteristic hazardous waste is a solid waste that exhibits at least one of four characteristics defined in 40 CFR Part 261 subpart C — ignitability (D001), corrosivity (D002), reactivity (D003), and toxicity (D004 - D043). combustible, or have a flash point less than 60 °C (140 °F).

8 0
2 years ago
The density of a liquid is to be determined by an old 1-cm-diameter cylindrical hydrometer whose division marks are completely w
vodka [1.7K]

Answer:

The density of the unknown liquid is 1.025 kg/m³ (considering the density of the water as 1.000 kg/m³)

Explanation:

The hydrometer works by the Archimedes principle. The cylinder floats in the liquid because the hydrostatic thrust is equal to the weight force. This means:

Tr-W=0N\\Tr=W\\\delta_{fl} \cdot Vol \cdot g =W_{hydr}

If we measure 2 fluids, the weight of the hydrometer is the same, so:

\delta_{fl1} \cdot Vol_1 \cdot g =W_{hydr}=\delta_{fl2} \cdot Vol_2 \cdot g\\\delta_{fl1} \cdot Vol_1 =\delta_{fl2} \cdot Vol_2\\\delta_{fl1} H_1 \pi R^2 =\delta_{fl2} H_2 \pi R^2\\\delta_{fl1}\frac{H_1}{H_2}=\delta_{fl2}

If the original watermark height is 12.3cm (H₁) and the mark for water has risen 0.3 cm above the unknown liquid–air interface, the height of the unknown liquid mark is 12cm (H₂). Therefore:

delta_{fl1}\frac{H_1}{H_2}=\delta_{fl2}\\1.000\frac{kg}{m^3} \frac{12.3cm}{12cm}=\delta_{fl2}=1.025\frac{kg}{m^3}

8 0
2 years ago
A paint company produces glow in the dark paint with an advertised glow time of 15 min. A painter is interested in finding out i
andrew11 [14]

Answer:

p_v = P(z

Now we can decide based on the significance level \alpha. If p_v we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

\alpha=0.05 we see that p_v< \alpha so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

\alpha=0.01 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

\alpha=0.001 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

Explanation:

For this case they conduct the following system of hypothesis for the ture mean of interest:

Null hypothesis: \mu \leq 15

Alternative hypothesis: \mu >15

The statistic for this hypothesis is:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And on this case the value is given z = -2.30

For this case in order to take a decision based on the significance level we need to calculate the p value first.

Since we have a lower tailed test the p value would be:

p_v = P(z

Now we can decide based on the significance level \alpha. If p_v we reject the null hypothesis and in other case we FAIL to reject the null hypothesis.

\alpha=0.05 we see that p_v< \alpha so then we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly less than 15

\alpha=0.01 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

\alpha=0.001 we see that p_v> \alpha so then we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is NOT significantly less than 15

3 0
2 years ago
You want to know your grade in Computer Science, so write a program that continuously takes grades between 0 and 100 to standard
Margarita [4]

Answer:

The code is given below in python

# Code Block 1

count = 0  # count variable

total = 0  # total variable

enter = '' # input variable

while enter != 'stop':

   enter = input('Enter a grade:' )

   if enter != 'stop' and enter.isdigit():

       total += int(enter) # add to total value

       count = count + 1   # then to the count

print float(total) / count

# Code Block 2

numbers = []

while enter != 'stop':

   enter = input('Enter a grade:' )

   if enter != 'stop':

       numbers.append(int(enter))

print sum(numbers) / float(len(numbers))

5 0
2 years ago
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Leno4ka [110]
Manuel is a Gas compressor
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