 # A power cycle operating at steady state receives energy by heat transf Joanna Benson 2021-12-15 Answered
A power cycle operating at steady state receives energy by heat transfer at a rate and regects
energy by heat transfer to a cold reservoir at a rate . For each of the following cases, determine
whether the cycle operates reversibly, operates irreversibly, or is impossible.
a. ${\stackrel{˙}{Q}}_{H}=500kW,{\stackrel{˙}{Q}}_{C}=100kW$
b.
c. $\stackrel{˙}{W}=350kW,{\stackrel{˙}{Q}}_{C}=150kW$
d. ${\stackrel{˙}{Q}}_{H}=500kW,{\stackrel{˙}{Q}}_{C}=200kW$
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The maximum efficiency in the cycle is defined as;
${\eta }_{max}=1-\frac{{T}_{C}}{{T}_{H}}$
In our case , which is maximum efficiency of:
${\eta }_{max}=\frac{2}{3}\approx 66.7\mathrm{%}$
The actual efficiency cant

We have step-by-step solutions for your answer! Elaine Verrett
Hot reservoir temperature ${T}_{H}=1800K$
Cold reservoir temperature ${T}_{e}=600K$
a)
Heat transfer at hot reservoir ${Q}_{H}=500KW$
Heat transfer at cold reservoir ${Q}_{C}=100KW$
b
Heat transfer at hot reservoir ${Q}_{H}=500KW$
Heat transfer at cold reservoir ${Q}_{C}=200KW$
Cycle work ${\stackrel{˙}{W}}_{cyc\le }=250KW$
c)
Cycle work $\stackrel{˙}{{W}_{cyc\le }=350KW}$
Heat transfer at cold reservoir ${Q}_{C}=150KW$
d)
Heat transfer at hot reservoir ${Q}_{H}=500KW$
Heat transfer at cold reservoir ${Q}_{C}=200KW$
Required
Determine whether the cycle operates reversibly, operates irreversibly, or is impossible
Assumption
Constant average values is used
a)
Maximum net work could be defined by
${W}_{cyc\le }={Q}_{H}-{Q}_{C}$
$⇒=500-100=400KW$
Maximum efficiency check
$\left\{{W}_{max}\right\}=1-\frac{{T}_{e}}{{T}_{H}}$
$⇒1-\frac{600}{1800}=0.6667=66.67\mathrm{%}$
$\left\{{W}_{avtual}\right\}=\frac{{W}_{cyc\le }}{{Q}_{H}}$
$⇒\frac{400}{500}=0.8=80\mathrm{%}$
At actual efficiency is higher than maximum efficiency then the cycle is not possibe.
b)
Maximum work check.
${W}_{cyc\le }={Q}_{H}-{Q}_{C}$
$⇒=500-200=300KW$
As actual work is higher than the maximum work the cycle is not possible.
Maximum efficiency check
$\left\{{W}_{max}\right\}=1-\frac{{T}_{e}}{{T}_{H}}$
$⇒1-\frac{600}{1800}=0.6667=66.67\mathrm{%}$
$\left\{{W}_{avtual}\right\}=\frac{{W}_{cyc\le }}{{Q}_{H}}$
$⇒\frac{250}{500}=0.5=50\mathrm{%}$
As actual efficiency is lower than maximum efficiency then the cycle operates irreversibly.
Then the cycle is not possible.
c)
Heat added from hot reservoir could be calculated as following.
${Q}_{H}={W}_{cyc\le }+{Q}_{e}$
$⇒=350+150=500KW$
Maximum efficiency check.
$\left\{{W}_{max}\right\}=1-\frac{{T}_{e}}{{T}_{H}}$
$⇒1-\frac{600}{1800}=0.6667=66.67\mathrm{%}$

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