Math Problem Statement

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

Solution

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=1500kJ/kgh = 1500 \, \text{kJ/kg}, and pressure P=10kPaP = 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 TsatT_{\text{sat}} at 10 kPa: 45.81°C
  • Saturation enthalpy for liquid hfh_f at 10 kPa: 191.81 kJ/kg
  • Saturation enthalpy for vapor hgh_g at 10 kPa: 2583.9 kJ/kg

Step 2: Determine the quality (x)

If the given enthalpy is within the range between hfh_f and hgh_g, the water is in the saturated mixture region, and we can find the quality xx using:

h=hf+x(hghf)h = h_f + x \cdot (h_g - h_f)

Rearranging to solve for xx:

x=hhfhghfx = \frac{h - h_f}{h_g - h_f}

Substitute the known values:

x=1500191.812583.9191.81=1308.192392.090.547x = \frac{1500 - 191.81}{2583.9 - 191.81} = \frac{1308.19}{2392.09} \approx 0.547

Thus, the quality x0.547x \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=Tsat=45.81°CT = T_{\text{sat}} = 45.81°C

Step 4: Determine the specific entropy (s)

Now that we know the quality xx, we can calculate the specific entropy ss using:

s=sf+x(sgsf)s = s_f + x \cdot (s_g - s_f)

From steam tables at 10 kPa:

  • Saturation entropy for liquid sf=0.6492kJ/kg\cdotpKs_f = 0.6492 \, \text{kJ/kg·K}
  • Saturation entropy for vapor sg=8.1508kJ/kg\cdotpKs_g = 8.1508 \, \text{kJ/kg·K}

Substitute the values:

s=0.6492+0.547(8.15080.6492)s = 0.6492 + 0.547 \cdot (8.1508 - 0.6492)

s=0.6492+0.5477.5016=0.6492+4.1004=4.7496kJ/kg\cdotpKs = 0.6492 + 0.547 \cdot 7.5016 = 0.6492 + 4.1004 = 4.7496 \, \text{kJ/kg·K}

Thus, the specific entropy s4.75kJ/kg\cdotpKs \approx 4.75 \, \text{kJ/kg·K}.

Final Answers:

  • Quality (x): 0.5470.547
  • Temperature (T): 45.81°C45.81°C
  • Specific entropy (s): 4.75kJ/kg\cdotpK4.75 \, \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 xx 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).

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Enthalpy
Quality of a Mixture
Entropy

Formulas

x = (h - h_f) / (h_g - h_f)
T = T_sat
s = s_f + x * (s_g - s_f)

Theorems

Saturation Properties of Water
Thermodynamic Property Relations

Suitable Grade Level

College Level (Engineering or Physics)