Question

Two reversible heat engines a and b are arranged in series, a rejecting heat directly to b. Engine a receives 200 kj at a temperature of 421c from a hot source, while engine b is in communication with a cold sink at a temperature of 4.4c. If the work output of a is twice that of b, find (a) the intermediate temperature between a and b

Answer 1

Explanation:

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Answer 2

The intermediate temperature between two reversible heat engine  is 143.3 C

Step by step explanation:

Given:

Engine receives 200 kj at a temperature of 421 c.

Communication with cold sink at a temperature is 4.4 c

To find: Temperature between A and B

                             $T_{1}$  (421 c)

                              ↓  ($Q_{1} = 200 kj)$

                            Heat engine A (WA)

                               ↓  $T_{2} = ?$

                             Heat engine B(WB)

                                 ↓ $Q_{3}$

                               $T_{3} = 4.4 c$

$W_{A} = 2Wx_{B}$

Now,

$\frac{W_{A}}{Q_{1} } = \frac{T_{2} -T_{1}}{T_{1} }$      

$W_{A}= [\frac{T_{2} T_{1} }{T_{1} } ](Q_{1} )$      

$\frac{W_{B}}{Q_{2} } = \frac{T_{2} -T_{3}}{T_{2} }$

$W_{B}= [\frac{T_{2} T_{3} }{T_{2} } ](Q_{1} - W_{A} )$    

Now substitute the values in the equation

$W_{A} = 2 [ \frac{(T_{2} - T_{3} }{T_{2} }](Q_{1} - W_{A} )$

⇒ $[\frac{T_{2} }{2(T_{2} - T_{3}) } + 1] W_{A} = Q_{1}$

Substitute the values then we get,

⇒ $[\frac{T_{2} }{2(T_{2} - T_{3}) } + 1] [\frac{T_{T} - T_{2} }{T_{1} } ]Q_{1} = Q_{1}$

⇒ $[\frac{T_{2} }{2T_{2} - 2T_{3}) } ]$ = $\frac{T_{1} }{T_{1} T_{2} } = \frac{T_{2} }{T_{1} - T_{2} }$

$T_{1} - T_{2} = 2T_{2} - 2T_{3}$

$T_{2} = \frac{T_{1} + 2T_{3} }{3}$

⇒ $(694 +2 (277.4))/3$

⇒ 416.267 K

⇒ 143.3 c

Final answer:

The intermediate temperature between A and B is 143.3 C

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