Using the tables for water, determine the specified property data at the indicated states.
(a) At p = 3 bar, v = 0.35m ^ 3 / k * g find Tin °C and u in kJ/kg.
(b) At T = 320 deg * C v = 0.13m ^ 3 / k * g find p in MPa and u in kJ/kg.
(c) At , T = 400 deg * C find v in m³/kg and h in kJ/kg. p = 28MPi
(d) At T = 10 deg * C v = 100m ^ 3 / k * g find pin kPa and h in kJ/kg.
The Correct Answer and Explanation is :
To determine the specified thermodynamic properties of water at the given states, we refer to standard steam tables, which provide detailed data on water’s properties across various temperatures and pressures. Here’s the analysis for each part:
(a) At ( p = 3 \, \text{bar} ) and ( v = 0.35 \, \text{m}^3/\text{kg} ):
- Determine the Saturation Temperature:
- At ( p = 3 \, \text{bar} ) (which is equivalent to 300 kPa), the saturation temperature ( T_{\text{sat}} ) is approximately 133.5°C.
- Compare Specific Volume with Saturated Values:
- From steam tables:
- Specific volume of saturated liquid (( v_f )) at 3 bar: approximately ( 0.001017 \, \text{m}^3/\text{kg} ).
- Specific volume of saturated vapor (( v_g )) at 3 bar: approximately ( 0.6058 \, \text{m}^3/\text{kg} ).
- Given ( v = 0.35 \, \text{m}^3/\text{kg} ), which lies between ( v_f ) and ( v_g ), indicating a mixture of liquid and vapor (i.e., a saturated mixture).
- Calculate Quality (x):
- Quality ( x ) represents the mass fraction of vapor in the mixture and is calculated as:
[
x = \frac{v – v_f}{v_g – v_f} = \frac{0.35 – 0.001017}{0.6058 – 0.001017} \approx 0.577
]
- Determine Temperature and Specific Internal Energy (u):
- For a saturated mixture at 3 bar:
- Temperature ( T ) is ( T_{\text{sat}} = 133.5^\circ\text{C} ).
- Specific internal energy ( u ) is calculated as:
[
u = u_f + x \times (u_g – u_f)
]
where: - ( u_f ) (internal energy of saturated liquid) ≈ 561.7 kJ/kg.
- ( u_g ) (internal energy of saturated vapor) ≈ 2550.5 kJ/kg.
- Thus:
[
u = 561.7 + 0.577 \times (2550.5 – 561.7) \approx 1615.5 \, \text{kJ/kg}
]
(b) At ( T = 320^\circ\text{C} ) and ( v = 0.13 \, \text{m}^3/\text{kg} ):
- Determine the Saturation Pressure:
- At ( T = 320^\circ\text{C} ), the saturation pressure ( p_{\text{sat}} ) is approximately 11.29 MPa.
- Compare Specific Volume with Saturated Values:
- From steam tables:
- ( v_f ) at 320°C: approximately ( 0.001307 \, \text{m}^3/\text{kg} ).
- ( v_g ) at 320°C: approximately ( 0.01803 \, \text{m}^3/\text{kg} ).
- Given ( v = 0.13 \, \text{m}^3/\text{kg} ), which is significantly greater than ( v_g ), indicating a superheated vapor state.
- Determine Pressure and Specific Internal Energy:
- For superheated steam at 320°C and ( v = 0.13 \, \text{m}^3/\text{kg} ):
- Using superheated steam tables, interpolate to find the corresponding pressure and internal energy. However, since ( v ) is much larger than ( v_g ), this suggests a low-pressure superheated state.
- Estimating from available data, the pressure ( p ) is approximately 0.5 MPa (5 bar).
- Specific internal energy ( u ) at this state is approximately 2700 kJ/kg.
(c) At ( T = 400^\circ\text{C} ) and ( p = 28 \, \text{MPa} ):
- Compare with Critical Point:
- The critical pressure of water is 22.064 MPa. Since 28 MPa > 22.064 MPa, the water is in a compressed liquid state at 400°C.
- Determine Specific Volume and Enthalpy:
- Compressed liquid data at such high pressures are limited in standard steam tables. However, for approximation:
- Specific volume ( v ) is slightly less than that of saturated liquid at the same temperature, so ( v \approx v_f ) at 400°C, which is approximately ( 0.00146 \, \text{m}^3/\text{kg} ).
- Enthalpy ( h ) can be approximated as ( h_f ) at 400°C, which is approximately 1738 kJ/kg.
(d) At ( T = 10^\circ\text{C} ) and ( v = 100 \, \text{m}^3/\text{kg} ):
- Determine Saturation Pressure:
- At ( T = 10^\circ\text{C} ), the saturation pressure ( p_{\text{sat}} ) is approximately 1.23 kPa.
- Compare Specific Volume with Saturated Values:
- From steam tables:
- ( v_f ) at 10°C: approximately ( 0.001 \, \text{m}^3/\text{kg} ).
- ( v_g ) at 10°C: approximately ( 14.67 \, \text{m}^3/\text{kg} ).
- Given ( v = 100 \, \text{m}^3/\text{kg} ), which is much greater than ( v_g ), indicating a superheated vapor state.
- Determine Pressure and Enthalpy:
- For superheated steam at 10°C and ( v = 100 \, \text{m}^3/\text{kg} ):
- Using the ideal gas equation ( p = \frac{RT}{v} ):
- ( R ) (specific gas constant for water vapor) ≈ 0.4615 kJ/(kg·K).
- ( T ) in Kelvin = 10 + 273.15 = 283.15 K.
- Thus:
[
p = \frac{0.4615 \times 283.15}{100} \approx 1.31 \, \text{kPa}
] - This pressure is close to the saturation pressure, confirming the superheated state.