a. The work done by the assembly can be calculated using the formula:
w = P(V2 - V1)
where P is the constant pressure, V2 is the final specific volume, and V1 is the initial specific volume.
From the steam tables, we can find the final specific volume at 435 °C and the initial specific volume at 350 °C. The values are:
V2 = 0.00988 m³/kg
V1 = 0.00699 m³/kg
Substituting these values into the formula, we get:
w = P(V2 - V1)
w = 1 bar (0.00988 - 0.00699) m³/kg
w = 0.00289 kJ/kg
Therefore, the work done by the assembly is 0.00289 kJ/kg.
b. The heat transfer to the assembly can be calculated using the first law of thermodynamics:
Q = ΔU + w
where Q is the heat transfer, ΔU is the change in internal energy, and w is the work done by the assembly.
At constant pressure, the change in internal energy is given by:
ΔU = C_p ΔT
where C_p is the specific heat capacity at constant pressure and ΔT is the change in temperature.
From the steam tables, we can find the specific heat capacity at constant pressure for steam at 350 °C and 435 °C. The values are:
C_p = 2.06 kJ/kg·K (at 350 °C)
C_p = 2.19 kJ/kg·K (at 435 °C)
Substituting these values and the temperature difference into the formula, we get:
ΔU = C_p ΔT
ΔU = 2.06 kJ/kg·K (435 - 350) °C
ΔU = 17.56 kJ/kg
Substituting the values of w and ΔU into the first law of thermodynamics, we get:
Q = ΔU + w
Q = 17.56 kJ/kg + 0.00289 kJ/kg
Q = 17.56 kJ/kg
Therefore, the heat transfer to the assembly is 17.56 kJ/kg.
c. The process in the p-v and T-v diagrams can be sketched as follows:
In the p-v diagram, the process is a vertical line from the initial specific volume to the final specific