Question

Hydrogen gas \(\left(\mathrm{H}_{2}\right)\) at \(35^{\circ} \mathrm{C}\) and pressure \(p\) enters an insulated control volume operating at steady state for which \(\dot{W}_{\mathrm{cv}}=0\). Half of the hydrogen exits the device at 2 bar and \(90^{\circ} \mathrm{C}\) and the other half exits at 2 bar and \(-20^{\circ} \mathrm{C}\). The effects of kinetic and potential energy are negligible. Employing the ideal gas model with constant \(c_{p}=14.3 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the minimum possible value for the inlet pressure \(p\), in bar.

Short answer

The energy balance is upheld; valid calculation conditions

Step-by-step solution

Identify Given Data

Given data includes:- Inlet temperature: 35°C- Exit temperatures: 90°C and -20°C- Exit pressures: 2 bar- Heat capacity at constant pressure: \(c_p = 14.3 \text{ kJ/kg·K}\)- Negligible kinetic and potential energy effects- Steady state and adiabatic conditions meaning \(\frac{dE}{dt} = 0\) and \(Q_{cv} = 0\)- \(\frac{dE}{dt} = \frac{\text{d}}{\text{d}t}(mc_vT) = m\frac{d}{dt}(c_vT)\), under constant mass, \(\frac{dE}{dt} = 0\)

Energy Balance Equation

Since the process is adiabatic and steady-state, write the energy balance equation for the control volume:\( \frac{dE_{cv}}{dt} = \frac{Q_{cv}}{dt} + \frac{W_{cv}}{dt} + \frac{m(h_in)_{in}} - \frac{m(h_out)_{out}} \ \frac{Q_{cv}}{dt} = 0, \frac{W_{cv}}{dt} = 0 \ Therefore: \ 0 = \frac{m(h_in)_{in}} - \frac{m(h_{90},h_{-20)}}\)

Calculate Enthalpy

Since we assume ideal gas behavior with constant \(c_p\), use the equation for enthalpy \(h = c_p T\). Convert temperatures to Kelvin (K = °C + 273.15): \(T_{\text{in}} = 35 + 273.15 = 308.15 K\)\(T_{\text{out1}} = 90 + 273.15 = 363.15 K\)\(T_{\text{out2}} = -20 + 273.15 = 253.15 K\)The enthalpies will be:\(h_{\text{in}} = c_p T_{\text{in}}\)\(h_{\text{out1}} = c_p T_{\text{out1}}\)\(h_{\text{out2}} = c_p T_{\text{out2}}\)

Apply Mass Flow Considerations

Assume equal mass flow rates for both exit streams, since total flow rate must be conserved:\((\text{mass flow in}) = \text{in}\)\((\text{mass flow out}) = \text{out}\)Mass flow rate: \(0.5\) each for two outflows.

Set Up Steady-State Energy Balance

At steady-state, the total enthalpy remains constant:\[m\frac{m(c_p (h)_{\text{in}}) - (h_{\text{out1}} + h_{\text{out2}}) = 0}\]\[(h_{\text{in}}) = \frac{1}{2}(h_{\text{out1}}+ h_{\text{out2}})\]

Solve for Inlet Enthalpy

Plug the values from Step 3 back into the equation:\[c_p (308.15 K) = \frac{1}{2} ( c_p (363.15 K) + c_p (253.15 K))\]Simplify, eliminating \(c_p\) from both sides:\[308.15 = (1/2)(363.15 + 253.15)\]Calculate:\[308.15 = \frac{1}{2}(616.3)\]\[308.15 = 308.15\]

Key concepts

Ideal Gas Model

The ideal gas model simplifies the behavior of gases by assuming no intermolecular forces and gas molecules occupying no volume. This simplification works well under standard conditions of temperature and pressure. For thermodynamic problems, we assume ideal gas behavior, aiding in straightforward calculations for properties like enthalpy and internal energy. The ideal gas law is represented as \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is the universal gas constant, and \(T\) is temperature.
With constant heat capacity at constant pressure (\(c_p\)), the enthalpy of an ideal gas can be expressed as \(h = c_pT\). Understanding the energy transformations in ideal gases is crucial for calculating enthalpy changes during various processes, such as heating or cooling.

Energy Balance Equation

The energy balance equation is critical for analyzing energy changes in a control volume system. Here, the general energy balance is given by:\r \[ \frac{dE_{cv}}{dt} = \frac{Q_{cv}}{dt} + \frac{W_{cv}}{dt} + \frac{m(h_{in})_{in}} - \frac{m(h_{out})_{out}} \]
In our problem, considering steady-state conditions (\r\(\frac{dE_{cv}}{dt} = 0\)), no heat transfer (adiabatic process), and no work done (\r\(W_{cv} = 0\)), the energy balance simplifies to:
\[ 0 = m(h_{in}) - m(h_{out}) \]
This equation asserts that the total enthalpy entering the control volume equals the total enthalpy exiting it. This principle helps in determining unknown parameters, such as the inlet pressure in our problem.

Enthalpy Calculation

Enthalpy, denoted as \(h\), is a measure of the total energy content in a system and includes internal energy plus the product of pressure and volume. For ideal gases with constant specific heat at constant pressure (\(c_p\)), enthalpy can be calculated using the formula: \r\[ h = c_p \times T \]
In this problem, we convert temperatures from degrees Celsius to Kelvin using \(T(K) = T(°C) + 273.15\). Then, we apply the above formula to compute the enthalpies for the inlet and outlet states:

• Inlet enthalpy: \(h_{in} = c_p \times 308.15 \text{ K}\)
• First outlet enthalpy: \(h_{out1} = c_p \times 363.15 \text{ K}\)
• Second outlet enthalpy: \(h_{out2} = c_p \times 253.15 \text{ K}\)

This computation is vital for comparing energy states within the system and confirming the steady-state energy balance.

Steady-State Conditions

Steady-state conditions imply that properties like mass and energy do not change with time within the control volume. Expressed mathematically, \r\( \frac{dE}{dt} = 0 \):
For our hydrogen gas system, this steady-state assumption simplifies the energy balance equation as it indicates no accumulation of energy. With constant flow rates in and out, the system ensures that the amount of energy entering is equal to the amount of energy leaving. We also assume equal mass flow rates for both exit streams, simplifying mass flow considerations. Therefore,
\(m_{in} = m_{out1} + m_{out2} = 0.5m + 0.5m\).
This assumption helps solve for unknown variables like inlet pressure.

Adiabatic Process

An adiabatic process occurs without any heat transfer between the system and its surroundings (\r\(Q_{cv} = 0\)). In thermodynamics, this is a key concept when dealing with insulated systems, like our control volume:
Given \(Q_{cv} = 0\), the energy change is purely due to work done and changes in enthalpy. For our problem where \(W_{cv} = 0\), the focus is entirely on enthalpy changes. Enthalpy change in adiabatic processes for ideal gases is solely a function of temperature differences, simplified by:

• Input: \(c_p \times T_{in}\)
• Output points: \(c_p \times T_{out1}\) and \(c_p \times T_{out2}\)

This helps establish energy equivalence and determine the required inlet conditions (pressure).