Question

Consider the turbocharger of an internal combustion engine. The exhaust gases enter the turbine at \(450^{\circ} \mathrm{C}\) at a rate of \(0.02 \mathrm{kg} / \mathrm{s}\) and leave at \(400^{\circ} \mathrm{C}\). Air enters the compressor at \(70^{\circ} \mathrm{C}\) and \(95 \mathrm{kPa}\) at a rate of \(0.018 \mathrm{kg} / \mathrm{s}\) and leaves at 135 kPa. The mechanical efficiency between the turbine and the compressor is 95 percent ( 5 percent of turbine work is lost during its transmission to the compressor). Using air properties for the exhaust gases, determine ( \(a\) ) the air temperature at the compressor exit and ( \(b\) ) the isentropic efficiency of the compressor.

Short answer

Question: Determine (a) the air temperature at the compressor exit and (b) the isentropic efficiency of the compressor. Answer: (a) The air temperature at the compressor exit is approximately 216.03°C. (b) The isentropic efficiency of the compressor is approximately 71.8%.

Step-by-step solution

Calculate the turbine work

First, let's find the work done by the turbine which can be found using the formula: Work (turbine) = Mass flow rate * Specific heat capacity * (Temperature at turbine inlet - Temperature at turbine outlet) The mass flow rate of exhaust gases is given as 0.02 kg/s, the specific heat capacity for air can be found in air property tables and is approximately 1.005 kJ/kg·K, and the temperatures at the turbine inlet and outlet are given as 450°C and 400°C, respectively. We calculate the work done by the turbine: Work (turbine) = \(0.02\,\text{kg/s} * 1.005\,\text{kJ/kg·K} * (450 - 400)\,^{\circ}\text{C}\) Work (turbine) = \(1.005\,\text{kJ/s}\)

Calculate mechanical work transmitted to the compressor

Now, let's find the mechanical work transmitted to the compressor after accounting for the 5% mechanical loss. We are given that the mechanical efficiency is 95%, so the work transmitted to the compressor is: Work (compressor) = 0.95 * Work (turbine) Work (compressor) = \(0.95 * 1.005\,\text{kJ/s}\) Work (compressor) = \(0.95475\,\text{kJ/s}\)

Find the temperature at the compressor exit

To find the air temperature at the compressor exit, we can use the following relationship for work done by the compressor: Work (compressor) = Mass flow rate (air) * Specific heat capacity * (Temperature at compressor exit - Temperature at compressor inlet) The mass flow rate of air is given as 0.018 kg/s, and the temperature at the compressor inlet is 70°C. We can now find the temperature at the compressor exit (\(T_2\)): \(0.95475\,\text{kJ/s} = 0.018\,\text{kg/s} * 1.005\,\text{kJ/kg·K} * (T_2 - 70)\,^{\circ}\text{C}\) \(T_2\) = \(\frac{0.95475\,\text{kJ/s}}{0.018\,\text{kg/s} * 1.005\,\text{kJ/kg·K}} + 70\) \(T_2\) ≈ \(216.03\,^{\circ}\text{C}\) So, the air temperature at the compressor exit is approximately \(216.03\,^{\circ}\text{C}\).

Determine the isentropic efficiency of the compressor

To determine the isentropic efficiency of the compressor, we need to know the actual work done by the compressor (\(W_{actual}\)) and the isentropic work done by the compressor (\(W_{isentropic}\)). The isentropic efficiency is given by the following formula: Isentropic efficiency = \(\frac{W_{isentropic}}{W_{actual}}\) We know the actual work done by the compressor from Step 2. Let's find the isentropic work done by the compressor, using the isentropic relation for ideal gases and air properties: \(W_{isentropic}\) = Specific heat capacity * (Temperature at compressor inlet - Isentropic temperature at compressor exit) We are given that the pressure ratio across the compressor (135 / 95) and the specific heat ratios for air (\(C_p=1.005\,\text{kJ/kg·K}\) and \(C_v=0.718\,\text{kJ/kg·K}\), which gives a specific heat ratio, \(\gamma=\frac{C_p}{C_v}\approx 1.4\)). Using the isentropic relation, we can find the isentropic temperature at the compressor exit (\(T_{2s}\)): \(T_{2s} = T_1 * \left(\frac{P_2}{P_1}\right)^{\frac{\gamma - 1}{\gamma}}\) \(T_{2s} = 70 * \left(\frac{135}{95}\right)^{\frac{1.4 - 1}{1.4}}\) \(T_{2s}\) ≈ \(145.86\,^{\circ}\text{C}\) Now, we can calculate the isentropic work done by the compressor: \(W_{isentropic}\) = \(0.018\,\text{kg/s} * 1.005\,\text{kJ/kg·K} * (145.86 - 70)\,^{\circ}\text{C}\) \(W_{isentropic}\) ≈ \(0.68571\,\text{kJ/s}\) Finally, we can determine the isentropic efficiency of the compressor: Isentropic efficiency = \(\frac{0.68571\,\text{kJ/s}}{0.95475\,\text{kJ/s}}\) Isentropic efficiency ≈ \(0.718\) So, the isentropic efficiency of the compressor is approximately 71.8%.