Question

An air-standard diesel cycle, as shown in the following figure with a compression ratio of 16, has an initial pressure 0.9 bar and temperature 300 K. Assume \(\gamma = 1.4\) and \(C_p = 1.004\) kJ/kg-K. If the heat added during the constant pressure process is 900 kJ/kg, then the peak temperature during the cycle is .................... K (round off to the nearest integer) 

 

Hint:

For air-standard cycle problems, clearly identify the processes (e.g., isentropic, isobaric, isochoric). Remember the key relations for each process, especially the ideal gas law and the relations for temperature, pressure, and volume changes during an isentropic process.

Answer 1

Step 1: Understanding the Concept:
The problem asks for the maximum temperature in an air-standard Diesel cycle. In a Diesel cycle, the peak temperature is reached at the end of the constant pressure heat addition process (state 3 in the diagram). To find this temperature (\(T_3\)), we first need to find the temperature at the end of compression (\(T_2\)).
Step 2: Key Formula or Approach:
1. Use the isentropic relation for process 1-2 (compression) to find \(T_2\): \(T_2 = T_1 . r^{\gamma-1}\), where \(r\) is the compression ratio. 2. Use the first law for a constant pressure process 2-3 (heat addition) to find \(T_3\): \(q_{in} = C_p (T_3 - T_2)\).
Step 3: Detailed Explanation or Calculation:
Given values:
Compression ratio, \(r = v_1/v_2 = 16\)
Initial temperature, \(T_1 = 300\) K
Ratio of specific heats, \(\gamma = 1.4\)
Specific heat at constant pressure, \(C_p = 1.004\) kJ/kg-K
Heat added, \(q_{in} = 900\) kJ/kg
1. Calculate Temperature at the end of compression (\(T_2\)): The process 1-2 is isentropic compression. \[ T_2 = T_1 . r^{\gamma-1} = 300 \times (16)^{1.4-1} = 300 \times (16)^{0.4} \] Calculating \(16^{0.4}\): \[ \log(16^{0.4}) = 0.4 \log(16) \approx 0.4 \times 1.204 = 0.4816 \] \[ 16^{0.4} = 10^{0.4816} \approx 3.0314 \] \[ T_2 = 300 \times 3.0314 \approx 909.42 \text{ K} \] 2. Calculate Peak Temperature (\(T_3\)): The process 2-3 is constant pressure heat addition. \[ q_{in} = C_p (T_3 - T_2) \] Rearranging to solve for \(T_3\): \[ T_3 = T_2 + \frac{q_{in}}{C_p} \] \[ T_3 = 909.42 + \frac{900}{1.004} \] \[ \frac{900}{1.004} \approx 896.41 \] \[ T_3 = 909.42 + 896.41 = 1805.83 \text{ K} \] Step 4: Final Answer:
Rounding to the nearest integer, the peak temperature during the cycle is 1806 K.
Step 5: Why This is Correct:
The solution correctly applies the thermodynamic relations for the compression and heat addition processes of the Diesel cycle. The calculations are performed step-by-step, leading to a peak temperature of 1806 K, which is consistent with the provided answer range.