At p = 3 bar, v = 0.2 m³/kg, find T in °C and u in kJ/kg.
Understand the Problem
The question requires finding the temperature (T) in degrees Celsius and the specific internal energy (u) in kJ/kg, given the pressure (p) of 3 bar and specific volume (v) of 0.2 m³/kg. This will likely require consulting thermodynamic tables or using appropriate equations of state, such as the ideal gas law.
Answer
$T = 133.52 \, ^\circ\text{C}$ $u = 1212.95 \, \text{kJ/kg}$
Answer for screen readers
$T = 133.52 , ^\circ\text{C}$ $u = 1212.95 , \text{kJ/kg}$
Steps to Solve
- Determine the substance
Since the substance is not specified, we will assume it is water.
- Convert pressure to Pascals
Convert the pressure from bar to Pascals:
$p = 3 , \text{bar} = 3 \times 10^5 , \text{Pa}$
- Consult saturated water tables
Check the saturated water tables to determine if the water is in a compressed liquid, saturated mixture, or superheated vapor state. At $p = 3 , \text{bar} = 0.3 , \text{MPa}$, the saturation temperature $T_{sat}$ is $133.52^\circ\text{C}$, the specific volume of saturated liquid $v_f$ is $0.001073 , \text{m}^3/\text{kg}$ and the specific volume of saturated vapor $v_g$ is $0.6058 , \text{m}^3/\text{kg}$.
- Determine the state of the water
Since $v_f < v < v_g$ ($0.001073 < 0.2 < 0.6058$), the water is in a saturated mixture state. Therefore, $T = T_{sat} = 133.52^\circ\text{C}$.
- Calculate the quality (x)
Calculate the quality $x$ of the saturated mixture:
$x = \frac{v - v_f}{v_g - v_f} = \frac{0.2 - 0.001073}{0.6058 - 0.001073} = \frac{0.198927}{0.604727} \approx 0.3288$
- Determine the specific internal energy
Obtain the specific internal energy of saturated liquid $u_f$ and saturated vapor $u_g$ at $0.3 , \text{MPa}$ from the saturated water tables: $u_f = 561.11 , \text{kJ/kg}$ and $u_g = 2543.6 , \text{kJ/kg}$.
- Calculate the specific internal energy of the mixture
Calculate the specific internal energy $u$ of the saturated mixture:
$u = u_f + x(u_g - u_f) = 561.11 + 0.3288(2543.6 - 561.11) = 561.11 + 0.3288(1982.49) = 561.11 + 651.84 \approx 1212.95 , \text{kJ/kg}$
$T = 133.52 , ^\circ\text{C}$ $u = 1212.95 , \text{kJ/kg}$
More Information
The solution involves using steam tables to identify the temperature and specific internal energy of water at a given pressure and specific volume. Since the specific volume falls between the saturated liquid and saturated vapor values at the given pressure, the state is a saturated mixture, and the temperature is equal to the saturation temperature. The specific internal energy is calculated using the quality and the saturated liquid and vapor internal energies.
Tips
A common mistake is to assume the substance is an ideal gas and apply the ideal gas law, which is inappropriate for water near saturation conditions. Another mistake is to not check the saturation conditions and incorrectly assume the water is in a compressed liquid or superheated vapor state.
AI-generated content may contain errors. Please verify critical information
