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ladessa [460]
3 years ago
5

A steam power plant with a power output of 230 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,0

00 kJ/kg, determine the overall efficiency of this plant.
Engineering
1 answer:
NARA [144]3 years ago
7 0

Answer:

\eta =46\%

Explanation:

Hello!

In this case, we compute the heat output from coal, given its heating value and the mass flow:

Q_H=60\frac{tons}{h}*\frac{1000kg}{1ton}*\frac{1h}{3600s}*\frac{30,000kJ}{kg}\\\\Q_H=500,000\frac{kJ}{s}*\frac{1MJ}{1000J} =500MW

Next, since the work done by the power plant is 230 MW, we compute the efficiency as shown below:

\eta =\frac{230MW}{500MW}*100\% \\\\\eta =46\%

Best regards!

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Which type of system is being researched to deliver power to several motors to drive multiple systems in vehicles?
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Answer:

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Explanation:

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8 0
2 years ago
You have designed a treatment system for contaminant Z. The treatment system consists of a pipe that feeds into a CSTR. The pipe
IRISSAK [1]

Answer:

0.667 per day.

Explanation:

Our values here are

Q=10m^3/dV_p=15m^3\\V_{cstr}=60m^3\\c_p=2500mg/L\\c_{cstr}=500mg/L

Degradation constant=k and is unknown.

We calculate the concentration through the formula,

cc_{cstr} =\frac{c_{in}}{1+K(V/Q)} \\cc_{cstr}=\frac{c_p}{1+K*\frac{V_{csrt}}{Q}}

Replacing values we have

1+k(\frac{60}{10})=\frac{2500}{500}\\1+k=5\\K(6)=5-1\\K(6)=4\\K=2/3\\K=0.667/day

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3 0
3 years ago
On diesel engines, data from ________ sensors are commonly used to adjust exhaust gas recirculation (EGR) rates.
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Explanation:

3 0
2 years ago
A belt drive was designed to transmit the power of P=7.5 kW with the velocity v=10m/s. The tensile load of the tight side is twi
Leviafan [203]

Answer:

F₁ = 1500 N

F₂ = 750 N

F_{e} = 500 N

Explanation:

Given :

Power transmission, P = 7.5 kW

                                      = 7.5 x 1000 W

                                      = 7500 W

Belt velocity, V = 10 m/s

F₁ = 2 F₂

Now we know from power transmission equation

P = ( F₁ - F₂ ) x V

7500 = ( F₁ - F₂ ) x 10

750 =  F₁ - F₂

750 = 2 F₂ - F₂      ( ∵F₁ = 2 F₂ )

∴F₂  = 750 N

Now F₁ = 2 F₂

        F₁ = 2 x F₂

        F₁ = 2 x 750

        F₁ = 1500 N   ,   this is the maximum force.

Therefore we know,

F_{max} = 3 x F_{e}

where F_{e} is centrifugal force

 F_{e} = F_{max} / 3

                          = 1500 / 3

                         = 500 N

8 0
3 years ago
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