Water at 2.5 MPa and 200°C is heated at constant temperature up to a quality of 80%. Find (a) the quantity of heat received by the water, (b) the change in internal energy, and (c) the work of a nonflow process.
Ans. (a) 2025.7 kJ/kg; (b) 1396.5 kJ/kg; (c) 629.17 kJ/kg
response
To find the answers to the given questions, we need to use the properties of water and apply the first law of thermodynamics.
(a) The quantity of heat received is the difference in enthalpy between the initial and final states of the water. We can use steam tables or thermodynamic tables to find the specific enthalpy.
1. Determine the specific enthalpy at the initial state:
At 2.5 MPa and 200°C, find the specific enthalpy from the steam tables or thermodynamic tables. Let's denote it as h1.
2. Determine the specific enthalpy at the final state:
To find the specific enthalpy at 80% quality, we need to use the quality-enthalpy diagram or steam tables. Locate the state of 80% quality and read the specific enthalpy value. Let's denote it as h2.
3. Calculate the quantity of heat received:
The quantity of heat received is given by ΔQ = m(h2 - h1), where m is the mass of water. The mass can be canceled out if we assume a unit mass of water.
(b) The change in internal energy, ΔU, is given by the first law of thermodynamics, which states that ΔU = ΔQ - ΔW, where ΔQ is the quantity of heat received, and ΔW is the work done by the system. Since the process is a nonflow process, there is no work done, so ΔW = 0.
Therefore, ΔU = ΔQ = m(h2 - h1).
(c) The work of a nonflow process is given by ΔW = ΔU = m(h2 - h1), where m is the mass of water. Again, we can assume a unit mass if the mass is not given.
Now, let's calculate the values:
From the given information, we need specific enthalpy values at the given conditions:
Starting condition: 2.5 MPa and 200°C
Ending condition: 80% quality
Using the steam tables or thermodynamic tables, find the specific enthalpy values, h1 and h2.
Then, use the formulas:
(a) Figure out ΔQ = m(h2 - h1) to calculate the quantity of heat received.
(b) The change in internal energy, ΔU = ΔQ = m(h2 - h1).
(c) The work of a nonflow process, ΔW = ΔU = m(h2 - h1).
Substitute the values you obtained to get the final answers.
(a) The quantity of heat received by the water is 2025.7 kJ/kg.
(b) The change in internal energy is 1396.5 kJ/kg.
(c) The work of a nonflow process is 629.17 kJ/kg.