A large electrical power station generates 1100 MW of electricity with an efficiency of 35.0%.(a) Calculate the heat transfer (in J) to the power station, Q, in one day.(b)How much heat transfer Q. (in J) occurs to the environment in one day?(c) If the heat transfer in the cooling towers is from 35.0°C water into the local air mass, which increases in temperature from 18.0°C to 20.0°C, what is the total increase in entropy(in J/K) due to this heat transfer?J/K(d) How much energy (in J) becomes unavailable to do work because of this increase in entropy, assuming an 18.0°C lowest temperature? (Part of Q. could be utilized to operate heatengines or for simple space heating, but it rarely is.)

Given data: The electricity generated, P = 1100 MW. The efficiency, ƞ = 35 %. Calculate the power…

Answered: A large electrical power station…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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A large electrical power station generates 1100 MW of electricity with an efficiency of 35.0%.
(a) Calculate the heat transfer (in J) to the power station, Q, in one day.
(b)
How much heat transfer Q. (in J) occurs to the environment in one day?
(c) If the heat transfer in the cooling towers is from 35.0°C water into the local air mass, which increases in temperature from 18.0°C to 20.0°C, what is the total increase in entropy
(in J/K) due to this heat transfer?
J/K
(d) How much energy (in J) becomes unavailable to do work because of this increase in entropy, assuming an 18.0°C lowest temperature? (Part of Q. could be utilized to operate heat
engines or for simple space heating, but it rarely is.)

Expert Solution

Step 1

Given data:

The electricity generated, P = 1100 MW.
The efficiency, ƞ = 35 %.

Calculate the power generated in one day as follows:

$\begin{array}{rcl} E_{o u t} & = & P \times t \\ = & 1100 MW \times \frac{10^{6} W}{1 MW} \times 24 h \times \frac{3600 s}{1 h} \\ = & 95040000 \times 10^{6} W / s \times \frac{1 J}{1 W / s} \\ = & 9.504 \times 10^{13} J \end{array}$

Step 2

(a)

Calculate the heat transfer as follows:

$\begin{array}{rcl} η & = & \frac{E_{o u t}}{Q_{h}} \\ Q_{h} & = & \frac{E_{o u t}}{η} \\ = & \frac{9.504 \times 10^{13} J}{0.35} \\ = & 27.154 \times 10^{13} J \end{array}$

Hence, the heat transfer is $27.154 \times 10^{13} J$ .

Step 3

(b)

Calculate the heat transfer in environment in one day as follows:

$\begin{array}{rcl} Q_{c} & = & E_{o u t} - Q_{h} \\ = & 27.154 \times 10^{13} J - 9.504 \times 10^{13} J \\ = & 17.65 \times 10^{13} J \end{array}$

Hence, the heat transfer in the environment in one day is $17.65 \times 10^{13} J$ .

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