Question

How much error would one expect in determining the specific enthalpy by applying the incompressible-liquid approximation to water at 3000 psia and \(400^{\circ} \mathrm{F} ?\)

Short answer

Answer: The error in determining the specific enthalpy of water at 3000 psia and 400°F using the incompressible-liquid approximation is 2.56%.

Step-by-step solution

Gather the data

We are given: - Pressure P = 3000 psia - Temperature T = \(400^{\circ} \mathrm{F}\)

Convert units to SI system

Convert Pressure to the SI unit (Pa) and Temperature to the SI unit (K): 1 psia = 6894.76 Pa 1\(^{\circ}\mathrm{F}\) = \(\frac{5}{9}(T_{\circ F} - 32) + 273.15\) K P = 3000 psia * 6894.76 Pa/psia = 20684080 Pa T = \(\frac{5}{9}(400 - 32) + 273.15\) K = 477.59 K

Check the water phase at given conditions

Check whether the water is in the liquid or vapor phase at given conditions. For water at 3000 psia and \(400^{\circ} \mathrm{F}\), it is in the compressed liquid (subcooled) phase. To confirm this, use a steam table or software to find the saturation temperature at 3000 psia. The saturation temperature at 3000 psia (20.686 MPa) is approximately 344.61\(^{\circ} \mathrm{F}\) (639.72 K). Since the given temperature (400\(^{\circ} \mathrm{F}\)) is greater than the saturation temperature, the water is in the compressed liquid phase (subcooled).

Find the specific enthalpy using incompressible liquid approximation

Use the following approximation for incompressible liquid to find specific enthalpy: \(h_{approx} = h_{ref} + C_p (T-T_{ref})\) Let's use the reference state as water at its normal boiling point (P = 1 atm, T = 100\(^{\circ} \mathrm{F}\) or 373.15 K). From the steam table, the specific enthalpy at the reference state is: \(h_{ref} = 419.17 \mathrm{kJ/kg}\). The specific heat capacity (Cp) for liquid water can be assumed to be constant and equal to 4.18 kJ/(kg K). Calculate the specific enthalpy using incompressible liquid approximation: \(h_{approx} = 419.17 + 4.18 (477.59 - 373.15) \mathrm{kJ/kg} = 838.17 \mathrm{kJ/kg}\)

Find the actual specific enthalpy using steam table

Using a steam table or software, find the specific enthalpy for water at 3000 psia and \(400^{\circ}\mathrm{F}\) (20.686 MPa and 477.59 K). The actual specific enthalpy, \(h_{actual}\) is approximately 860.19 kJ/kg.

Calculate the error

Determine the error between the approximation and the actual specific enthalpy: Error = \(\frac{h_{actual} - h_{approx}}{h_{actual}} \times 100\) Error = \(\frac{860.19 - 838.17}{860.19} \times 100 = 2.56 \%\) So, one would expect an error of 2.56% in determining the specific enthalpy of water at 3000 psia and \(400^{\circ}\mathrm{F}\) using the incompressible-liquid approximation.