Question 9.2: An Air-Standard Cycle An air-standard cycle is executed in a...

An Air-Standard Cycle

An air-standard cycle is executed in a closed system and is composed of the following four processes:

1-2  Isentropic compression from 100 kPa and 27°C to 1 MPa

2-3  P = constant heat addition in amount of 2800 kJ/kg

3-4  v = constant heat rejection to 100 kPa

4-1  P = constant heat rejection to initial state

(a) Show the cycle on P-v and T-s diagrams.

(b) Calculate the maximum temperature in the cycle.

(c) Determine the thermal efficiency.

Assume constant specific heats at room temperature.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the maximum temperature in the cycle and the thermal efficiency are to be determined.

Assumptions     1 The air-standard assumptions are applicable. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

Properties     The properties of air at room temperature are c_{p}=1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, c_{v}=0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, and k=1.4 (Table A-2a).

Analysis    (a) The cycle is shown on P-v and T-s diagrams in Fig. 9-9 .

(b) From the ideal gas isentropic relations and energy balance,

\begin{aligned}T_{2}&=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(300  \mathrm{~K})\left(\frac{1000  \mathrm{kPa}}{100  \mathrm{kPa}}\right)^{0.4 / 1.4}=579.2  \mathrm{~K} \\\\q_{\mathrm{in}}&=h_{3}-h_{2}=c_{p}\left(T_{3}-T_{1}\right) \\\\2800  \mathrm{~kJ} / \mathrm{kg}&=(1.005  \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\left(T_{3}-579.2\right) \longrightarrow T_{\max }=T_{3}=3360  \mathrm{~K}\end{aligned}

(c) The temperature at state 4 is determined from the ideal gas relation for a fixed mass,

\frac{P_{3} v_{3}}{T_{3}}=\frac{P_{4} v_{4}}{T_{4}} \longrightarrow T_{4}=\frac{P_{4}}{P_{3}} T_{3}=\frac{100  \mathrm{kPa}}{1000  \mathrm{kPa}}(3360  \mathrm{~K})=336  \mathrm{~K}

The total amount of heat rejected from the cycle is

\begin{aligned}q_{\mathrm{out}} &=q_{34,  \mathrm{out}}+q_{41,  \mathrm{out}}=\left(u_{3}-u_{4}\right)+\left(h_{4}+h_{1}\right) \\&=c_{v}\left(T_{3}-T_{4}\right)+c_{p}\left(T_{4}-T_{1}\right) \\&=(0.718  \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(3360-336)  \mathrm{K}+(1.005  \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(336-300)  \mathrm{K} \\&=2212  \mathrm{~kJ} / \mathrm{kg}\end{aligned}

Then, the thermal efficiency is determined from its definition to be

\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{2212  \mathrm{~kJ} / \mathrm{kg}}{2800  \mathrm{~kJ} / \mathrm{kg}}=0.210 \text { or } 21.0 \%

Discussion     The assumption of constant specific heats at room temperature is not realistic in this case since the temperature changes involved are too large.

9.9

Related Answered Questions