Question

Problem 2.3 (Clausius Inequality Applications) 

A system executes a power cycle while receiving 1050 kJ by heat transfer at a temperature of 525 K and discharging 700 kJ by heat transfer at 350 K. There are no other heat transfers 

(a) Using eq.5.13, determine whether the cycle is internally reversible, irreversible, or impossible. 

(b) Determine the thermal efficiency using eq.5.4 and the given heat transfer data. Compare this value with the Carnot efficiency calculated using eq.5.9 and comment.

Answer 1

Answer 2

(a)

Determine whether the cycle is internally reversible, irreversible, or impossible

$$ \begin{aligned} \sigma_{\text {cycle }} &=-\Phi\left(\frac{\delta Q}{T}\right) \\ &=-\left(\frac{Q_{H}}{T_{H}}-\frac{Q_{C}}{T_{C}}\right) \\ &=-\left(\frac{1050}{525}-\frac{700}{350}\right) \\ &=0 \end{aligned} $$

Since \(\sigma_{\text {cycle }}=0\), the cycle is internally reversible.

(b)

Calculate of Thermal efficiency using the following equation:

$$ \begin{aligned} \eta &=1-\frac{Q_{C}}{Q_{H}} \\ &=1-\frac{700}{1050} \\ \eta &=0.333 \end{aligned} $$

Calculate the thermal efficiency of Carnot cycle,

$$ \begin{aligned} \eta_{\max } &=1-\frac{T_{C}}{T_{H}} \\ &=1-\frac{350}{525} \\ &=0.333 \end{aligned} $$

Comment:

Here \(\eta=\eta_{\max }\), this happens because of the internally reversible nature of the cycle.