a worker cleans a railroad car with steam of 15000ft^3 car on inside, t=240F P=14.7psia of steam. the car is closed. the steam inside cools until P=12 psia, at which time salad oil is added to tank. find a. heat transfer from steam during cooling, b. amount of liquid water in bottom of tank car that will dilute the salad oil.
I found the T2 using ideal gas law, using dU=dQ=dW, dw=0 cause no volume change. so would I just look up the values of internal energy change in a table, then what about part b?
For Part (a), to do it accurately you need to use "Steam Tables". You go from superheated steam at 240F (800 R) and 14.7 psia in a closed volume to a saturated mixture at 12 psia. dW = 0 and dQ = dU. This is a rather tedius process; my Keenan and Kayes tables has only specific enthalpy (h) and specific volume data, not u.
Here is an approximate way to get the heat release when the steam condenses:
When the steam has cooled enough to lower the pressure to 12 psia, some of the water will have condensed and the temperature becomes 202 F (762 R), the corresponding pressure on the P vs T saturation curve. In the gas phase, the number of moles of H2O is reduced by a factor (P2/P1)*(T1/T2) = 0.857. This assumes the volume available for gas does not change.
Therefore 14.3% of the steam in the tank condensed. Calculate this mass and multipy by 1000 btu/lbm (an approximate heat of condensation value) for the heat release.
(b) The "14.3% condensed" figure and the original mass of H2O in the steam will tell you the amount of liquid water available todilute salad oil.
To find the heat transfer from steam during cooling, you can use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat transferred to the system minus the work done by the system. In this case, since there is no work done (dw=0), the change in internal energy is equal to the heat transfer.
To calculate the change in internal energy, we can use the equation:
ΔU = m·C·ΔT
Where:
ΔU = change in internal energy (heat transfer)
m = mass of the steam
C = specific heat capacity of steam
ΔT = change in temperature
First, we need to find the mass of the steam using the ideal gas law equation:
PV = nRT
Given:
V = 15000 ft^3
T = 240°F
P = 14.7 psia
Convert the volume to cubic meters:
15000 ft^3 * 0.02832 m^3/ft^3 = 424.8 m^3
Convert temperature to Kelvin:
T = (240 + 459.67)°F * (5/9) = 388.08 K
Convert pressure to Pascals:
P = 14.7 psia * 6894.76 Pa/psia = 101.3 kPa
The equation can now be rearranged to solve for the mass (m):
m = PV / RT
Put the values into the equation:
m = (101.3 kPa)(424.8 m^3) / (8.314 J/(mol·K))(388.08 K)
Next, to find the change in temperature (ΔT), note that the steam cools until the pressure reaches 12 psia. The change in pressure (ΔP) can be obtained as follows:
ΔP = 14.7 psia - 12 psia = 2.7 psia
Now, we need to look up the specific heat capacity of steam at constant volume (Cv) from a table or use the equation:
Cv = R / (γ - 1)
Where:
R = specific gas constant for steam (approximated as 8.314 J/(mol·K))
γ = specific heat ratio for steam (approximated as 1.33)
Calculate Cv using the equation:
Cv = 8.314 J/(mol·K) / (1.33 - 1)
Finally, substitute the values into the equation to find the heat transfer:
ΔU = m·C·ΔT
For part b, to calculate the amount of liquid water in the bottom of the tank car that will dilute the salad oil, we need to know the composition of the steam, such as the initial mass fraction of water in the steam. Without that information, it is not possible to determine the amount of liquid water in the bottom of the tank car.