Helium is to be compressed from 120 kPa and 320 K to 800 kPa, and 450 K., A heat loss of 20 kJ/kg,
occurs during the compression process, as shown in Figure 1 below. Neglecting kinetic energy changes,
determine the power input required for a mass flow rate of 90 kg/min.

To determine the power input required for the compression process, we can use the First Law of Thermodynamics for a control volume, which for a steady-state, steady-flow process (neglecting changes in kinetic and potential energy) is:

\[
\dot{Q} - \dot{W} = \dot{m} \cdot (\Delta h)
\]

Where:
- \(\dot{Q}\) is the heat transfer rate (negative since it's heat loss),
- \(\dot{W}\) is the power input (what we need to determine),
- \(\dot{m}\) is the mass flow rate,
- \(\Delta h\) is the change in specific enthalpy of helium.

### Step 1: Mass Flow Rate
We are given:
\[
\dot{m} = 90 \, \text{kg/min} = \frac{90}{60} = 1.5 \, \text{kg/s}
\]

### Step 2: Heat Transfer Rate
The heat loss is given as \(20 \, \text{kJ/kg}\), so for the mass flow rate of \(1.5 \, \text{kg/s}\), the total heat transfer rate is:

\[
\dot{Q} = \dot{m} \cdot q = 1.5 \, \text{kg/s} \times (-20 \, \text{kJ/kg}) = -30 \, \text{kJ/s} = -30 \, \text{kW}
\]
(Note: Heat loss is negative because it's leaving the system.)

### Step 3: Change in Enthalpy (\(\Delta h\))
We need to calculate the change in specific enthalpy (\(\Delta h\)) between the initial and final states. For helium, we can use the specific heat at constant pressure (\(c_p\)) to estimate the change in enthalpy:

\[
\Delta h = c_p \cdot (T_2 - T_1)
\]

Where:
- \(c_p\) for helium is approximately \(5.1926 \, \text{kJ/kg} \cdot \text{K}\),
- \(T_1 = 320 \, \text{K}\) (initial temperature),
- \(T_2 = 450 \, \text{K}\) (final temperature).

Now, calculate \(\Delta h\):

\[
\Delta h = 5.1926 \, \text{kJ/kg} \cdot (450 \, \text{K} - 320 \, \text{K}) = 5.1926 \times 130 = 674.04 \, \text{kJ/kg}
\]

### Step 4: Apply the First Law
Now, substitute the values into the First Law equation:

\[
-30 \, \text{kW} - \dot{W} = 1.5 \, \text{kg/s} \cdot (674.04 \, \text{kJ/kg})
\]

This simplifies to:

\[
-30 - \dot{W} = 1.5 \times 674.04
\]

\[
-30 - \dot{W} = 1011.06
\]

\[
\dot{W} = -30 - 1011.06 = -1041.06 \, \text{kW}
\]

The power input required is:

\[
\dot{W} = 1041.06 \, \text{kW}
\]

Thus, the power input required for the compression process is **1041.06 kW**.

### Any Questions or Details?
Would you like more details or clarifications about any step?

### Related Questions:
1. How does specific heat capacity (\(c_p\)) vary with temperature for gases like helium?
2. Why is the heat transfer rate negative in this process?
3. How would the power input change if the mass flow rate was different?
4. What assumptions are made when neglecting kinetic and potential energy changes in this process?
5. How would using a different gas, such as nitrogen, change the calculation?

### Tip:
When using the First Law of Thermodynamics in steady-state flow processes, always ensure that all energy forms (work, heat, and internal energy changes) are properly accounted for to avoid calculation errors.

Helium is to be compressed from 120 kPa and 320 K to 800 kPa, and 450 K. A heat loss of 20 kJ/kg occurs during the compression process. Neglecting kinetic energy changes, determine the power input required for a mass flow rate of 90 kg/min.

Calculate the power input required for compressing helium from 120 kPa and 320 K to 800 kPa and 450 K, with a heat loss of 20 kJ/kg. This solution uses the First Law of Thermodynamics and involves enthalpy change calculations, suitable for undergraduate engineering students.

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