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sdas [7]
3 years ago
5

A 5 MW gas turbine power plant is reported to have a thermodynamic efficiency of 35%. Assume products of the combustion reaction

exit at 750K with water in the gaseous phase. a. What is the corresponding heat rate of the power cycle
Engineering
1 answer:
sineoko [7]3 years ago
4 0

Answer:

The corresponding heat rate of the power cycle is  9749 BTU/kWh

Explanation:

Heat rate is the common measure of system efficiency in a steam power plant. It is defined as "the energy input to a system, typically in Btu/kWh, divided by the electricity generated, in kW."

Heat rate (BTU/kWh) = (Input Energy. Btu/hr)/(output power. kW)

Given;

Electrical energy output of gas turbine power plant = 5 MW = 5000 KW

Chemical energy input of the turbine = ?

Also, Efficiency = output power/input power

Given, efficiency = 35% = 0.35

0.35 = 5000 kW/input power

Input Power (kW) = 5000/0.35

Input Power (kW) = 14,285.7 KW

1 KW = 3412.142 BTU/hr

14,285.7 KW =   48,744,836.97 BTU/hr

Heat rate (BTU/kWh) = (Input Energy. Btu/hr)/(output power. kW)

Heat rate (BTU/kWh) = (48,744,836.97 BTU/hr)/(5000 KW)

Heat rate (BTU/kWh) = 9748.97 BTU/kWh = 9749 BTU/kWh

Therefore, the corresponding heat rate of the power cycle is  9749 BTU/kWh

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3 years ago
Read 2 more answers
A 15.00 mL sample of a solution of H2SO4 of unknown concentration was titrated with 0.3200M NaOH. the titration required 21.30 m
natima [27]

Answer:

a. 0.4544 N

b. 5.112 \times 10^{-5 M}

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O

NaOH\ Mass = Normality \times equivalent\ weight \times\ volume

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= 0.27264 g

NaOH\ mass = \frac{mass}{molecular\ weight}

= \frac{0.27264\ g}{40g/mol}

= 0.006816 mol

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= 0.003408 mol

Mass\ of\ H_2SO_4 = moles \times molecular\ weight

= 0.003408\ mol \times 98g/mol

= 0.333984 g

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a. Normality of acid is

= \frac{acid\ mass}{equivalent\ weight \times volume}

= \frac{0.333984 g}{49 \times 0.015}

= 0.4544 N

b. And, the acid solution molarity is

= \frac{moles}{Volume}

= \frac{0.003408 mol}{15\ mL \times  1L/1000\ mL}

= 0.00005112

=5.112 \times 10^{-5 M}

We simply applied the above formulas

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