91Ó°ÊÓ

A diesel engine intakes atmospheric air at \(60^{\circ} \mathrm{F}\) and adds \(800 \mathrm{Btu} / \mathrm{lbm}\) of energy. If the maximum pressure is 1200 psia, calculate \((a)\) the cutoff ratio, \((b)\) the thermal efficiency, and \((c)\) the power output for an airflow of \(0.2 \mathrm{lbm} / \mathrm{s}\). (a) The compression process is isentropic. The temperature at state 2 (see Fig. 8.9) is calculated to be $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(520)\left(\frac{1200}{14.7}\right)^{0.2857}=1829^{\circ} \mathrm{R} $$ The temperature at state 3 is found from the first law as follows: $$ q_{\text {in }}=C_{p}\left(T_{3}-T_{2}\right) \quad 800=(0.24)\left(T_{3}-1829\right) \quad \therefore T_{3}=5162^{\circ} \mathrm{R} $$ The specific volumes of the three states are $$ v_{1}=\frac{R T_{1}}{P_{1}}=\frac{(53.3)(520)}{(14.7)(144)}=13.09 \mathrm{ft}^{3} / \mathrm{lbm} \quad v_{2}=\frac{R T_{2}}{P_{2}}=\frac{(53.3)(1829)}{(1200)(144)}=0.5642 \mathrm{ft}^{3} / 1 \mathrm{bm} $$ $$ v_{3}=\frac{R T_{3}}{P_{3}}=\frac{(53.3)(5162)}{(1200)(144)}=1.592 \mathrm{ft}^{3} / \mathrm{lbm} $$ The cutoff ratio is then \(r_{c}=v_{3} / v_{2}=1.592 / 0.5642=2.822\). (b) The compression ratio is \(r=v_{1} / v_{2}=13.09 / 0.5642=23.20\). The thermal efficiency can now be calculated, using (8.25): $$ \eta=1-\frac{1}{r^{k-1}} \frac{r_{c}^{k}-1}{k\left(r_{c}-1\right)}=1-\frac{1}{(23.2)^{0.4}} \frac{(2.822)^{1.4}-1}{(1.4)(2.822-1)}=0.6351 \quad \text { or } \quad 63.51 \% $$ (c) \(\dot{W}_{\text {out }}=\eta \dot{Q}_{\text {in }}=\eta \dot{m} q_{\text {in }}=[(0.6351)(0.2)(800)](778)=79,060 \mathrm{ft}-\mathrm{lbf} / \mathrm{s} \quad\) or \(\quad 143.7 \mathrm{hp}\).

Step by step solution

01

- Calculate State 2 Temperature

Use the isentropic relation to find the temperature at state 2: \[ T_{2} = T_{1} \left( \frac{P_{2}}{P_{1}} \right)^{(k-1)/k} = (520) \left( \frac{1200}{14.7} \right)^{0.2857} = 1829^{\circ} \mathrm{R} \]

02

- Calculate State 3 Temperature

Apply the first law of thermodynamics to find the temperature at state 3: \[ q_{\text{in}} = C_{p} (T_{3} - T_{2}) \quad 800 = (0.24)(T_{3} - 1829) \ \therefore T_{3} = 5162^{\circ} \mathrm{R} \]

03

- Calculate Specific Volumes

Determine the specific volumes at states 1, 2, and 3 using the gas law: \[ v_{1} = \frac{R T_{1}}{P_{1}} = \frac{(53.3)(520)}{(14.7)(144)} = 13.09 \mathrm{ft}^{3} / \mathrm{lbm} \]\[ v_{2} = \frac{R T_{2}}{P_{2}} = \frac{(53.3)(1829)}{(1200)(144)} = 0.5642 \mathrm{ft}^{3} / \mathrm{lbm} \]\[ v_{3} = \frac{R T_{3}}{P_{3}} = \frac{(53.3)(5162)}{(1200)(144)} = 1.592 \mathrm{ft}^{3} / \mathrm{lbm} \]

04

- Calculate the Cutoff Ratio

Use the specific volumes to find the cutoff ratio:\[ r_{c} = \frac{v_{3}}{v_{2}} = \frac{1.592}{0.5642} = 2.822 \]

05

- Calculate the Compression Ratio

Determine the compression ratio: \[ r = \frac{v_{1}}{v_{2}} = \frac{13.09}{0.5642} = 23.20 \]

06

- Calculate Thermal Efficiency

Using the compression ratio and cutoff ratio, calculate the thermal efficiency: \[ \eta = 1 - \frac{1}{r^{k-1}} \frac{r_{c}^{k} - 1}{k (r_{c} - 1)} = 1 - \frac{1}{(23.2)^{0.4}} \frac{(2.822)^{1.4} - 1}{(1.4)(2.822 - 1)} = 0.6351 \quad \text{or} \quad 63.51\% \]

07

- Calculate Power Output

Find the power output given the thermal efficiency and airflow rate: \[ \dot{W}_{\text{out}} = \eta \dot{Q}_{\text{in}} = \eta \dot{m} q_{\text{in}} = [(0.6351)(0.2)(800)](778) = 79,060 \mathrm{ft}-\mathrm{lbf} / \mathrm{s} \quad \text{or} \quad 143.7 \mathrm{hp} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isentropic Process

An isentropic process is a thermodynamic process that occurs at constant entropy. In simpler terms, during an isentropic process, there is no heat exchange with the surrounding environment, and the process is both adiabatic (no heat transfer) and reversible.
In the context of a diesel engine, the compression process is typically considered isentropic. This assumption simplifies calculations and is a good approximation for ideal engine cycles. The temperature at the end of the isentropic compression (state 2) can be calculated using the relation:
\ T_{2} = T_{1} \left( \frac{P_{2}}{P_{1}} \right)^{(k-1)/k}
\(T_1\) is the initial temperature, \(P_1\) and \(P_2\) are the initial and final pressures, respectively, and \(k\) is the specific heat ratio (usually around 1.4 for air).
As seen in the exercise, the process from state 1 to state 2 is calculated to determine the temperature at the end of compression, which is essential for further calculations.

Thermal Efficiency

Thermal efficiency of an engine measures how well it converts heat from fuel into work. Higher thermal efficiency means better performance and fuel economy. In a diesel engine, thermal efficiency can be greatly influenced by its compression ratio and cutoff ratio.
The thermal efficiency of a diesel cycle can be calculated using the formula:
\'\eta = 1 - \frac{1}{r^{k-1}} \frac{r_{c}^{k} - 1}{k (r_c - 1)}\'
Where:

• \(\eta\): Thermal efficiency
• \(r\): Compression ratio \( (v_1/v_2)\)
• \(r_c\): Cutoff ratio \( (v_3/v_2)\)
• \(k\): Specific heat ratio

This formula indicates that increasing the compression ratio and optimizing the cutoff ratio can improve the engine's thermal efficiency. In the exercise, the thermal efficiency was calculated to be approximately 63.51%, which is quite efficient for a diesel engine.

Cutoff Ratio

The cutoff ratio in a diesel engine is defined as the ratio of the volume after combustion (state 3) to the volume after compression (state 2). It is a crucial parameter in understanding the performance of a diesel engine.
Mathematically, it is given by:
\r_c = \frac{v_3}{v_2}
Where:

• \(v_3\): Specific volume after combustion
• \(v_2\): Specific volume after compression

The cutoff ratio determines the duration of the fuel injection process and directly affects the thermal efficiency of the engine. A lower cutoff ratio usually implies a shorter combustion process, which can lead to higher efficiency but might limit the engine's power output.
In the given problem, the cutoff ratio was found to be 2.822. This value indicates the extent of the volume expansion during the combustion process and is used in calculating the thermal efficiency.