91Ó°ÊÓ

Carbon dioxide gas is compressed at steady state from a pressure of \(20 \mathrm{lbf} /\) in. \({ }^{2}\) and a temperature of \(32^{\circ} \mathrm{F}\) to a pressure of \(50 \mathrm{lbf} / \mathrm{in}^{2}\) and a temperature of \(120^{\circ} \mathrm{F}\). The gas enters the compressor with a velocity of \(30 \mathrm{ft} / \mathrm{s}\) and exits with a velocity of \(80 \mathrm{ft} / \mathrm{s}\). The mass flow rate is \(0.98 \mathrm{lb} / \mathrm{s}\). The magnitude of the heat transfer rate from the compressor to its surroundings is \(5 \%\) of the compressor power input. Using the ideal gas model with \(c_{p}=0.21 \mathrm{Btu} / \mathrm{bb} \cdot{ }^{\circ} \mathrm{R}\) and neglecting potential energy effects, determine the compressor power input, in horsepower.

Step by step solution

01

- Convert Units

Convert all given temperatures to Rankine. Use the formula: \(T(°R) = T(°F) + 459.67\).So, the inlet temperature is \(T_1 = 32 + 459.67 = 491.67 \text{ °R}\).The outlet temperature is \(T_2 = 120 + 459.67 = 579.67 \text{ °R}\).

02

- Apply the First Law of Thermodynamics

Using the First Law of Thermodynamics for a control volume in steady state, neglecting potential energy effects: \[ \frac{\text{d} E_{\text{cv}}}{\text{d}t} = \frac{\text{d} m}{\text{d}t} (h_2 + \frac{V_2^2}{2} + gz_2 - h_1 - \frac{V_1^2}{2} - gz_1) + \text{Q}_{\text{in}} - \text{W}_{\text{cv}} = 0 \]Simplify to: \[ \text{W}_{\text{cv}} = \text{m} (h_2 - h_1) + \frac{\text{m}}{2}(V_2^2 - V_1^2) + \text{Q}_{\text{in}} \]

03

- Determine Enthalpy Change

For an ideal gas, the enthalpy change can be determined using the specific heat capacity at constant pressure: \[ \text{h}_2 - \text{h}_1 = \text{c}_p (T_2 - T_1) \]Given: \(T_1 = 491.67 \text{ °R}\), \(T_2 = 579.67 \text{ °R}\), and \(\text{c}_p = 0.21 \text{ Btu/lb·°R}\) So, \( \text{h}_2 - \text{h}_1 = 0.21 \times (579.67 - 491.67) = 0.21 \times 88 = 18.48 \text{ Btu/lb}\)

04

- Calculate Kinetic Energy Difference

Use the velocities to find the difference in kinetic energy: \[ \frac{V_2^2}{2} - \frac{V_1^2}{2} = \frac{80^2}{2} - \frac{30^2}{2} = 3200 - 450 = 2750 \text{(ft^2/s^2)} \]Convert to consistent units where \(1 \text{ Btu} = 778.17 \text{ ft-lb} \), so \( \frac{2750 \text{ ft}^2/\text{s}^2}{778.17 \text{ ft-lb/Btu}} ≈ 3.53 \text{ Btu/lb} \)

05

- Determine Heat Transfer

The heat transfer rate is given as 5% of the compressor power input. Using a heat transfer rate of \( \text{Q} = 0.05 \times \text{W} \).Therefore, \[ \text{W}_{\text{input}} = \frac{1}{0.95} (\text{W}_{\text{cv}}) \]

06

- Plug in Known Values and Solve for Power Input

Now, substitute the known values into the equation: \[ \text{W}_{\text{cv}} = \text{m}(\text{h}_2 - \text{h}_1 + \frac{V_2^2}{2} - \frac{V_1^2}{2}) = 0.98 \text{ lb/s} \times (18.48 \text{ Btu/lb} + 3.53 \text{ Btu/lb}) = 0.98 \times 22.01 \text{ Btu/s} = 21.57 \text{ Btu/s} \]Convert to horsepower: \[ 1 \text{ Btu/s} = 0.7079 \text{ hp} \]So, \[ 21.57 \times 0.7079 = 15.28 \text{ hp} \]Now, considering the heat loss: \[ \text{W}_{\text{input}} = \frac{15.28}{0.95} \text{ hp} ≈ 16.08 \text{ hp} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

steady state compression

Steady state compression refers to a process where the conditions (such as pressure, temperature, and mass flow rate) at the inlet and outlet of the compressor remain constant over time. This simplifies the analysis because the energy changes can be measured consistently. In our exercise, carbon dioxide gas is compressed steadily from a lower to a higher pressure and temperature, with specific velocities at the entry and exit points. Steady state ensures that while the gas properties change during compression, they do so predictably and can be accounted for using established thermodynamic principles.

First Law of Thermodynamics

The First Law of Thermodynamics, often referred to as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transformed. For a control volume like a compressor, this can be expressed in the form of an energy balance:
\[ \dot{Q} - \dot{W} = \dot{m} \left( h_2 + \frac{V_2^2}{2} - h_1 - \frac{V_1^2}{2} \right) \]

• \(\dot{Q}\): Heat transfer rate
• \(\dot{W}\): Work done by the compressor
• \(\dot{m}\): Mass flow rate
• \(h_1, h_2\): Enthalpy at the inlet and outlet
• \(V_1, V_2\): Velocity at the inlet and outlet

In our problem, we adapt this equation to neglect potential energy changes since they are minimal compared to other energy terms. By analyzing the work done and heat transfer, we can calculate the compressor's power input accurately.

ideal gas

An ideal gas is a theoretical gas that adheres perfectly to the ideal gas law (PV = nRT). In this model, gas molecules do not interact with each other, and their individual volume is negligible compared to the container's volume.

• PV: Pressure (P) times volume (V)
• nRT: Number of moles (n) times the gas constant (R) and temperature (T)

In our exercise, carbon dioxide is treated as an ideal gas, meaning we can use simple gas laws and properties (like specific heat capacity at constant pressure, \(c_p\)) to determine other thermodynamic properties, such as enthalpy and temperature relationships.

enthalpy change

Enthalpy (denoted as \(H\)) is a measure of the total energy of a thermodynamic system, including internal energy and the product of pressure and volume. The change in enthalpy (\(\Delta H\)) for an ideal gas during a compression process can be calculated as:
\[ \Delta H = c_p \Delta T \]
This simplifies to:

• \(c_p\): Specific heat capacity at constant pressure
• \(\Delta T\): Temperature change

In our exercise, this means we can determine the change in enthalpy of the carbon dioxide gas as it compresses from the initial to the final state, using the provided specific heat capacity and temperature values.

kinetic energy

Kinetic energy (KE) is the energy possessed by a body due to its motion and is given by the formula: \[ KE = \frac{1}{2}mv^2 \]
In a compressor, the difference in kinetic energy between the inlet and outlet states must be accounted for. The change in kinetic energy (\(\Delta KE\)) is:
\[ \Delta KE = \frac{1}{2}m(V_2^2 - V_1^2) \]

• m: Mass flow rate
• V_1: Velocity at the inlet
• V_2: Velocity at the outlet

In the given problem, the velocity change from the entry to exit significantly contributes to the total energy change. By converting these differences into consistent energy units, we can incorporate them into the overall energy balance to find the compressor's power input.