H2O at 10 kPa, with h = 1500 kJ/kg; find x, T and s

To solve this problem, we are dealing with **water (H₂O)** under a pressure of **10 kPa** and with a specific enthalpy (**h**) of **1500 kJ/kg**. We need to find the **quality (x)**, **temperature (T)**, and **specific entropy (s)**.

### Step 1: Locate the state
We know the enthalpy \( h = 1500 \, \text{kJ/kg} \), and pressure \( P = 10 \, \text{kPa} \). This is low pressure, so it’s likely a saturated or superheated condition. First, let’s check the saturation properties of water at 10 kPa.

From steam tables:
- Saturation temperature \( T_{\text{sat}} \) at 10 kPa: **45.81°C**
- Saturation enthalpy for liquid \( h_f \) at 10 kPa: **191.81 kJ/kg**
- Saturation enthalpy for vapor \( h_g \) at 10 kPa: **2583.9 kJ/kg**

### Step 2: Determine the quality (x)
If the given enthalpy is within the range between \( h_f \) and \( h_g \), the water is in the **saturated mixture region**, and we can find the quality \( x \) using:

\[
h = h_f + x \cdot (h_g - h_f)
\]

Rearranging to solve for \( x \):

\[
x = \frac{h - h_f}{h_g - h_f}
\]

Substitute the known values:

\[
x = \frac{1500 - 191.81}{2583.9 - 191.81} = \frac{1308.19}{2392.09} \approx 0.547
\]

Thus, the quality \( x \approx 0.547 \), meaning the water is about 54.7% vapor and 45.3% liquid.

### Step 3: Determine the temperature (T)
Since the water is a saturated mixture, the temperature is equal to the saturation temperature at the given pressure. From the steam tables:

\[
T = T_{\text{sat}} = 45.81°C
\]

### Step 4: Determine the specific entropy (s)
Now that we know the quality \( x \), we can calculate the specific entropy \( s \) using:

\[
s = s_f + x \cdot (s_g - s_f)
\]

From steam tables at 10 kPa:
- Saturation entropy for liquid \( s_f = 0.6492 \, \text{kJ/kg·K} \)
- Saturation entropy for vapor \( s_g = 8.1508 \, \text{kJ/kg·K} \)

Substitute the values:

\[
s = 0.6492 + 0.547 \cdot (8.1508 - 0.6492)
\]

\[
s = 0.6492 + 0.547 \cdot 7.5016 = 0.6492 + 4.1004 = 4.7496 \, \text{kJ/kg·K}
\]

Thus, the specific entropy \( s \approx 4.75 \, \text{kJ/kg·K} \).

### Final Answers:
- **Quality (x)**: \( 0.547 \)
- **Temperature (T)**: \( 45.81°C \)
- **Specific entropy (s)**: \( 4.75 \, \text{kJ/kg·K} \)

Would you like further clarification on any step or details?

---

Here are 5 questions to expand your understanding:
1. What happens to the quality \( x \) when the enthalpy increases at the same pressure?
2. How would the state change if the pressure were decreased further?
3. How do you determine if water is in the superheated region?
4. Why is the specific entropy important in thermodynamic processes?
5. How would the temperature change if the quality of the mixture were increased?

**Tip:** Always check if the given enthalpy lies within the saturation region to determine the state of the water (liquid, mixture, or vapor).

Learn how to calculate the quality (x), temperature (T), and specific entropy (s) of water at 10 kPa with an enthalpy of 1500 kJ/kg. This thermodynamic problem involves using steam tables and key formulas for enthalpy and entropy. Suitable for college-level engineering or physics students.

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