Question

\Air enters a compressor at ambient conditions of 15 psia and \(60^{\circ} \mathrm{F}\) with a low velocity and exits at 150 psia, \(620^{\circ} \mathrm{F},\) and \(350 \mathrm{ft} / \mathrm{s}\). The compressor is cooled by the ambient air at \(60^{\circ} \mathrm{F}\) at a rate of \(1500 \mathrm{Btu} / \mathrm{min} .\) The power input to the compressor is 400 hp. Determine \((a)\) the mass flow rate of air and \((b)\) the portion of the power input that is used just to overcome the irreversibilities.

Short answer

Question: Determine the mass flow rate of air through the compressor and the portion of the power input used to overcome irreversibilities. Answer: The mass flow rate of air through the compressor is 8.532 lbm/s, and the portion of the power input used to overcome irreversibilities is 263575.03 Btu/s.

Step-by-step solution

Calculate specific enthalpies at inlet and outlet

First, we need to determine the specific enthalpies of air at the inlet and outlet. Start by converting degrees Fahrenheit to Kelvin: $$ T_{1} = 60^{\circ} \mathrm{F} + 459.67 = 519.67\,\mathrm{R} $$ $$ T_{2} = 620^{\circ} \mathrm{F} + 459.67 = 1079.67\,\mathrm{R} $$ Next, use the ideal gas specific heat formulas for air (\({c_p}=(a+br)\)) for specific enthalpies: $$ h_{1} = c_{p,1}T_{1} = (a+ bT_1)T_{1} = (0.24 + 0.0005\times519.67)\times519.67\,\mathrm{Btu/lbm} $$ $$ h_{2} = c_{p,2}T_{2} = (a+ bT_2)T_{2} = (0.24 + 0.0005\times1079.67)\times1079.67\,\mathrm{Btu/lbm} $$ Calculate \(h_{1}\) and \(h_{2}\): $$ h_{1} = 124.625\,\mathrm{Btu/lbm} $$ $$ h_{2} = 298.582\,\mathrm{Btu/lbm} $$

Apply energy balance and solve for mass flow rate

Now, apply energy balance on the system to find the mass flow rate: $$ \dot{Q} - \dot{W} = \dot{m}(h_{2} - h_{1} + \frac{V_2^2}{2\times32.174\times778}) $$ Given power input and cooling rate, convert them to the same units as enthalpy to maintain consistency: $$ \dot{Q} = -1500 \times \frac{1}{60}\,\mathrm{Btu/s} $$ $$ \dot{W} = 400 \times(2545)\,\mathrm{Btu/s} $$ Now plug all the values in the equation and solve for mass flow rate, \(\dot{m}\): $$ \dot{m} = \frac{-25 + 1021800}{298.582 - 124.625 + \frac{350^2}{2\times32.174\times778}} $$ Calculate the mass flow rate: $$ \dot{m} = 8.532\,\mathrm{lbm/s} $$

Calculate the ideal power input

The ideal power input, \(W_{ideal}\), can be calculated using the isentropic efficiency formula \(\mathrm{Eff}=\frac{W_{ideal}}{\dot{W}}\) : $$ W_{ideal}=\frac{h_{2s} - h_{1}}{h_{2} - h_{1}}\times\dot{W} $$ For ideal conditions, use \(h_{2s} = c_p T_{2s}\) : $$ T_{2s} = T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{\frac{\gamma-1}{\gamma}} $$ $$ T_{2s}=519.67\left(\frac{150}{15}\right)^{\frac{1.4-1}{1.4}} $$ Calculate \(T_{2s}\) and \(h_{2s}\): $$ T_{2s} = 843.97\,\mathrm{R} $$ $$ h_{2s} = 213.112\,\mathrm{Btu/lbm} $$ Now calculate the ideal power input: $$ W_{ideal} = \frac{213.112 - 124.625}{298.582 - 124.625}\times1021800 $$ $$ W_{ideal} = 758224.97\,\mathrm{Btu/s} $$

Determine the portion of power input used to overcome irreversibilities

Finally, find the portion of the power input that is used just to overcome the irreversibilities, \(W_{irr}\): $$ W_{irr} = \dot{W} - W_{ideal} $$ $$ W_{irr} = 1021800 - 758224.97 $$ Calculate the portion of power input used to overcome irreversibilities: $$ W_{irr} = 263575.03\,\mathrm{Btu/s} $$ So, the mass flow rate of air through the compressor is \(8.532\,\mathrm{lbm/s}\), and the portion of the power input used to overcome irreversibilities is \(263575.03\,\mathrm{Btu/s}\).

Key concepts

Mass Flow Rate Calculation

Understanding how to compute the mass flow rate is essential for analyzing thermodynamic systems like air compressors. The mass flow rate, denoted as \( \dot{m} \), represents the quantity of mass passing through a particular section of the system per unit time. It's a vital parameter in evaluating compressor performance as it directly relates to the system's ability to convey energy and perform work.

In our exercise, the mass flow rate is derived using an energy balance which includes both enthalpies and kinetic energy at the compressor's inlet and outlet. You'll notice that specific enthalpies, denoted by \( h \), were calculated by considering the air as an ideal gas, followed by solving the energy balance equation using given thermal data and power input. This approach reflects the practical scenario where the mass flow rate is often calculated by measuring or estimating these quantities rather than directly measuring mass flow itself.

Energy Balance in Thermodynamic Systems

An energy balance is a fundamental equation applied to the analysis of thermodynamic systems. It expresses the principle of energy conservation by equating the energy added to a system, subtracted by the work done by the system, to the change in energy of the fluid within it. This change includes both thermal energy, typically described in terms of enthalpy, and mechanical energy, often illustrated by kinetic energy.

For our air compressor analysis, the energy balance considers the cooling effect of ambient air and the power input to the compressor. When we express these energies in identical units and incorporate them into the energy balance equation, we can better understand how the input energy is transformed—either doing work on the air, increasing its internal energy, or being lost due to system inefficiencies.

Isentropic Efficiency

Isentropic efficiency is a measure of how close a thermodynamic process, like compression, comes to being isentropic—which means it's reversible and adiabatic (no heat transfer). This efficiency is one way to gauge the thermodynamic quality of the compressor's performance; it compares the actual work input needed to compress the fluid to the work input that would be required under ideal, isentropic conditions.

Efficiency and Power Losses

We are using this concept to deduce the portion of power input used to overcome irreversibilities during compression. By initially calculating the ideal power using the isentropic assumption and subtracting it from the actual power input, we identify the power that's not contributing to compressing the air—essentially, the energy lost to inefficiencies. This value is crucial for optimizing compressor design and operation, as it directly affects the system's overall energy consumption and costs.