Problem 2 For the following we will utilize the tables for w

Problem 2: For the following we will utilize the tables for water, Appendix D. to include both u, (sat. liquid) and v, (sat. vapor) for each pressure data point. energy and the specific enthalpy (Linear interpolation between table values should be used for part ii) a) Plot the p-u diagram (using 0.04, 0.5, 1, 10, 50, 100, 200, 220.64 bar pressure data points and table D-2) on a logarithmic scale. (Both axes should be in natural log scale.) Make sure b) For saturated water vapor at 420K, determine the specific volume, the specific internal c For superheated water vapor determine the specific volume, the specific internal energy [10] and the specific enthalpy for the following two states: i) T-420K, p-l.5bar, 11) T=430K, p-1.5bar. d) A vertical 6in-diameter piston-cylinder assembly containing 1 lb of water, initially a saturated liquid at 14.5psi (IbO), is placed on a hot plate and undergoes the following two processes in series: During the first process (Process A) heating occurs slowly and at constant heat rate of 500W for 10mins and the water expands at Hox plate During the second process (Process B) heating continues at the same i) Calculate the final height of the piston and compare to its initial height. Comment! [10] H,O constant pressure due to the piston\'s weight. power input but the piston is held stationary until the pressure reaches 29psi. ii) Sketch both processes on a p-u diagram (now that you know what it looks like from part iii) Calculate the total internal energy change of the water after the two processes and the a)) identifying process direction and the three relevant states. (This is just as sketch!) duration of both processes in minutes. Organization, clarity, legibility: [5 [5] [10]

Solution

The given pressures are 0.04, 0.5, 1, 10, 50, 100, 200, 220.64 bar.

We know that 1 bar = 100kPa (10⁵ Pa) = 0.1MPa

Hence the pressures are

You can find out the rest of the values by consultingthe tables and find out the X and Y co-ordinates to plot the graphs

(b) To solve this problem, your temperature is given as 420 K which is 147^oC

Checking the values for it in the appendix D.1, we get that the specific volume is v_g= 0.42520 m³/kg

Intermal energy u_g = 2556.2kJ/kg

Specific enthalpy h_g= 2742.1 kJ/kg

(c) You did not provide the steam tables for superheated steam. Hence, assuming standard tables in my mind,

for T=420 K and p = 1.5bar, the specific volume v= 0.753866 m3/kg

Specific enthalpy = 2758.09 kJ/kg

for T = 430 K and p=1.5 bar,

specific volume v= 0.773630 m3/kg

Specific enthalpy h = 2779 kJ/kg

(d) Given initial conditions, mass of water = 1lb = 0.45 kg

Initial pressure = 14.5 psi = 100kPa

I am converting them into SI units because the rest of your content is in SI units

Given it is saturated liquid. Hence, the specific enthalpy of water is 417.50 kJ/kg

Saturation temperature is given as 372.76 K

Hence the totatl enthalpy of water = 417.5 x 0.45 = 187.875 kJ

Specific volume of water = 0.0010432 m3/kg

Hence, initial volume = 0.0010432 x 0.45 = 0.00046944 m3

Given it is contained in a container with diameter of piston = 6 inch = 0.1524 m

Area of the piston = 0.0729 m2

Hence, height of the piston = 0.0064 m = 6.4 mm

Given it is heated at 500 W for 10 min

Total heat addition = 500 J/sec x 10 x 60 = 300 kJ

Hence, final enthalpy of the substance = 487.875 kJ

Hence, specific enthalpy is = 1084.16 kJ/kg

Assuming that x mole fraction has converted into steam, we know that the specific enthalpy of the steam is given by

h = hf + hg

hg = 2674.9 kJ/kg

Hence 1084.6 = 417.5 (1-x) + 2674.9 x

Solving for x, we get x = 0.29

Hence, new specific volume is v = (1-x)vf + x vg = (1-0.29)0.0010432 + 0.29*1.6939 = 0.491974

Hence, total volume = 0.22138 m3/kg

The new height of piston = 3.036 m

This is a drastic new height and very severe expansion. This is mainly due to the fact that water vapour occupies a lot more space than water.

Now, it is given that heat addition continues with the same volume until a pressure of 29 psi is reached = 0.2MPa

We know that the volume remains the same and hence the specific volume remains the same.

But due to the heat addition, we know that more water could become vapour. Let us assume the new mole fraction x

The specific volume of liquid corresponding to 29 psi vf= 0.0010605

vg = 0.88568

Hence 0.491974 = (1-x)0.0010605 + x*0.88568

Hence, new mole fraction x= 0.55

Hence, the specific enthalpy now is given by h = (1-0.55)504.70 + 0.55*2706.2 = 1715.525 kJ/kg

Hence, the total enthalpy of the fluid is = 771.98 kJ

Assuming that there is no heat loss, the heat input from the previous stage is added up as enthalpy.

Hence, heat addition is 771.98 - 487.875 = 284.11 kJ = 284110 J

The rate of heat addition is 500W. Hence time taken to add that much heat = 568.2 seconds = 9.47 minutes.

I am sure looking up the tables, you can draw the pertaining graphs for all the required charts.