Joanna Benson

2021-12-15

A power cycle operating at steady state receives energy by heat transfer at a rate ${\dot{Q}}_{H}\text{}at\text{}{T}_{H}=18000K$ and regects

energy by heat transfer to a cold reservoir at a rate${\dot{Q}}_{C}\text{}at\text{}{T}_{c}=600K$ . For each of the following cases, determine

whether the cycle operates reversibly, operates irreversibly, or is impossible.

a.${\dot{Q}}_{H}=500kW,{\dot{Q}}_{C}=100kW$

b.${\dot{Q}}_{H}=500kW,\dot{W}\text{}cyc\le =250kW,{\dot{Q}}_{C}=200kW$

c.$\dot{W}=350kW,{\dot{Q}}_{C}=150kW$

d.${\dot{Q}}_{H}=500kW,{\dot{Q}}_{C}=200kW$

energy by heat transfer to a cold reservoir at a rate

whether the cycle operates reversibly, operates irreversibly, or is impossible.

a.

b.

c.

d.

Thomas Nickerson

Beginner2021-12-16Added 32 answers

The maximum efficiency in the cycle is defined as;

$\eta}_{max}=1-\frac{{T}_{C}}{{T}_{H}$

In our case${T}_{C}=600K\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{T}_{H}=1800K$ , which is maximum efficiency of:

$\eta}_{max}=\frac{2}{3}\approx 66.7\mathrm{\%$

The actual efficiency cant

In our case

The actual efficiency cant

Elaine Verrett

Beginner2021-12-17Added 41 answers

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${\dot{W}}_{cyc\le}=250KW$

c)

Cycle work$\dot{{W}_{cyc\le}=350KW}$

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

$W}_{cyc\le}={Q}_{H}-{Q}_{C$

$\Rightarrow =500-100=400KW$

Maximum efficiency check

$\left\{{W}_{max}\right\}=1-\frac{{T}_{e}}{{T}_{H}}$

$\Rightarrow 1-\frac{600}{1800}=0.6667=66.67\mathrm{\%}$

$\left\{{W}_{avtual}\right\}=\frac{{W}_{cyc\le}}{{Q}_{H}}$

$\Rightarrow \frac{400}{500}=0.8=80\mathrm{\%}$

At actual efficiency is higher than maximum efficiency then the cycle is not possibe.

b)

Maximum work check.

$W}_{cyc\le}={Q}_{H}-{Q}_{C$

$\Rightarrow =500-200=300KW$

As actual work is higher than the maximum work the cycle is not possible.

Maximum efficiency check

$\left\{{W}_{max}\right\}=1-\frac{{T}_{e}}{{T}_{H}}$

$\Rightarrow 1-\frac{600}{1800}=0.6667=66.67\mathrm{\%}$

$\left\{{W}_{avtual}\right\}=\frac{{W}_{cyc\le}}{{Q}_{H}}$

$\Rightarrow \frac{250}{500}=0.5=50\mathrm{\%}$

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.

$Q}_{H}={W}_{cyc\le}+{Q}_{e$

$\Rightarrow =350+150=500KW$

Maximum efficiency check.

$\left\{{W}_{max}\right\}=1-\frac{{T}_{e}}{{T}_{H}}$

$\Rightarrow 1-\frac{600}{1800}=0.6667=66.67\mathrm{\%}$

$}$

Cold reservoir temperature

a)

Heat transfer at hot reservoir

Heat transfer at cold reservoir

b

Heat transfer at hot reservoir

Heat transfer at cold reservoir

Cycle work

c)

Cycle work

Heat transfer at cold reservoir

d)

Heat transfer at hot reservoir

Heat transfer at cold reservoir

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

Maximum efficiency check

At actual efficiency is higher than maximum efficiency then the cycle is not possibe.

b)

Maximum work check.

As actual work is higher than the maximum work the cycle is not possible.

Maximum efficiency check

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.

Maximum efficiency check.

0

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