Question

Air is compressed from 14.7 psia and \(60^{\circ} \mathrm{F}\) to a pressure of 150 psia while being cooled at a rate of \(10 \mathrm{Btu} / \mathrm{lbm}\) by circulating water through the compressor casing. The volume flow rate of the air at the inlet conditions is \(5000 \mathrm{ft}^{3} / \mathrm{min}\), and the power input to the compressor is 700 hp. Determine (a) the mass flow rate of the air and ( \(b\) ) the temperature at the compressor exit.

Short answer

(a) The mass flow rate of the air is approximately 3.09 kg/s. (b) The temperature at the compressor exit is approximately 82.73°C.

Step-by-step solution

Convert given data to proper units

We are given the volume flow rate in \(\frac{\mathrm{ft}^3}{\mathrm{min}}\), but in order to find the mass flow rate, we need to convert this to SI units. Similarly, we need to convert the given power input, in horsepower to Watts, and the initial temperature in Fahrenheit to Celsius (or Kelvin). 1 ft = 0.3048 m, 1 min = 60 s, 1 hp = 745.7 W, \(T_\mathrm{Fahrenheit} = \frac{9}{5}T_\mathrm{Celsius} + 32\). - Volume flow rate: \(5000 \frac{\mathrm{ft}^3}{\mathrm{min}} = 5000 \frac{0.3048^3 \mathrm{m}^3}{60 s} \approx 2.36 \frac{\mathrm{m}^3}{\mathrm{s}}\) - Power input: \(700 \: \mathrm{hp} = 700 \cdot 745.7 \mathrm{W} \approx 522990 \mathrm{W}\) - Initial temperature: \(60^{\circ} \mathrm{F} = \frac{5}{9}(60 - 32)^{\circ} \mathrm{C} \approx 15.6^{\circ} \mathrm{C}\)

Find the mass flow rate of air

Using the ideal gas equation, \(PV = mRT\), where \(P\) is pressure, \(V\) is the volume flow rate, \(m\) is mass flow rate, \(R\) is the specific gas constant for air (\(R = 287 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}\)), and \(T\) is the temperature in Kelvin. First, convert the initial temperature to Kelvin: \(T_\mathrm{initial} = 15.6^{\circ} \mathrm{C} + 273.15 \: \mathrm{K} \approx 288.75 \mathrm{K}\). Now, we can rearrange the ideal gas equation to find the mass flow rate \(m\): \(m = \frac{PV}{RT} = \frac{14.7 \mathrm{psia} \cdot 2.36 \frac{\mathrm{m}^3}{\mathrm{s}}}{287 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}} \cdot 288.75 \mathrm{K}}\) First, we have to convert the pressure from psia to Pa: \(14.7 \mathrm{psia} = 14.7 \cdot 6895 \mathrm{Pa} \approx 101325 \mathrm{Pa}\). Now plug in the converted values: \(m = \frac{101325 \mathrm{Pa} \cdot 2.36 \frac{\mathrm{m}^3}{\mathrm{s}}}{287 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}} \cdot 288.75 \mathrm{K}} \approx 3.09 \frac{\mathrm{kg}}{\mathrm{s}}\) The mass flow rate of the air is \(3.09 \frac{\mathrm{kg}}{\mathrm{s}}\).

Calculate the temperature at the compressor exit

The energy balance equation for the compressor is given by: \(\dot{W} = \dot{m}(h_2 - h_1) - C\), where \(\dot{W}\) is the work input, \(\dot{m}\) is the mass flow rate of the air, \(h_1\) and \(h_2\) are the specific enthalpies of the air at the inlet and outlet, and \(C\) represents the cooling capacity. For an ideal gas, the specific enthalpies of a substance depend only on its temperature: \(h_2 = h_1 + c_p (T_2 - T_1) = h_1 + c_p \Delta T\), where \(c_p\) is the specific heat at constant pressure and \(\Delta T\) is the change in temperature. Using the specific heat of air, \(c_p = 1.005 \frac{\mathrm{kJ}}{\mathrm{kg} \cdot \mathrm{K}}\), the cooling capacity, \(C = 10 \frac{\mathrm{Btu}}{\mathrm{lbm}} = 10 \cdot 2326 \frac{\mathrm{J}}{\mathrm{kg}} = 23260 \frac{\mathrm{J}}{\mathrm{kg}}\), and the power input, we can solve for the temperature change during the compression process: \(\Delta T = \frac{\dot{W} - C \cdot \dot{m}}{\dot{m} c_p} = \frac{522990 \mathrm{W} - 23260 \frac{\mathrm{J}}{\mathrm{kg}} \cdot 3.09 \frac{\mathrm{kg}}{\mathrm{s}}}{3.09 \frac{\mathrm{kg}}{\mathrm{s}} \cdot 1005 \frac{\mathrm{J}}{\mathrm{kg} \cdot \mathrm{K}}} \approx 67.13 \mathrm{K}\) Now, we can find the outlet temperature: \(T_2 = T_1 + \Delta T = 288.75 \mathrm{K} + 67.13 \mathrm{K} \approx 355.88 \mathrm{K}\) Convert the temperature back to Celsius: \(T_\mathrm{exit} = 355.88 \mathrm{K} - 273.15 ^{\circ} \mathrm{C} = 82.73^{\circ} \mathrm{C}\). The temperature at the compressor exit is approximately \(82.73^{\circ} \mathrm{C}\).