Exam (elaborations)

Exam MTX311 2015 memo and practices questions

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Course
MTX 311 (MTX311)

Institution
University Of Pretoria (UP)

Book
Fundamentals of Thermodynamics

Practice material exam 2015+memo worked out solutions. The memorandum has solutions are 100% correct including alternative working methods. The solutions are done in a step by step matter

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A smart inventor has developed a piston-based continuous (control volume) “turbine” that can handle
liquid-vapour mixtures of any gas quality (x) Saturated ammonia vapour at 120°C is fed to this
machine where it reversibly and adiabatically generates work, until the pressure is 119.5kPa. How
much work is generated (in kJ/kg)?

Saturated sulphur dioxide vapour, 400K, is heated to 600K in a constant pressure heating process
(without any shaft work). How much heat is required?

100 g/s of air (mass flow rate of the dry air) at 1 bar, 40°C and 60% relative humidity has to be dried
to 25%relative humidity, 40°C
We do this by first cooling down the air (1bar) and condensing excess water vapour. In a second step,
the air is reheated to 40°C. How much heat is added/extracted in step 1 and 2, and how much water
vapour is removed?

We have a mixture of 20% (m/m) water vapour, 6% (m/m) hydrogen and 74% (m/m) oxygen at 0.85
bar, 120°C. We want to cool down this mixture at constant pressure, but to avoid corrosion, no
condensation is allowed to take place. What is the lowest acceptable temperature?

, A smart inventor has developed a piston-based continuous (control volume) “turbine” that can handle
liquid-vapour mixtures of any gas quality (x) Saturated ammonia vapour at 120°C is fed to this
machine where it reversibly and adiabatically generates work, until the pressure is 119.5kPa. How
much work is generated (in kJ/kg)?

Solution
This looks like a difficult question, but actually it is quite simple.
First of all, it is a CONTROL MASS system, so we have to use control mass equations:

Δu-q-w

Secondly, the system is reversible and adiabatic:
Δq=TΔs=0 (since the system is adiabatic)
Δs=0  Send=Sbegin

Strategy:
 Look up u and s of saturated ammonia vapour at 120°C
 Interpolate at 119.5kPa to find the vapour fraction. Now calculate the internal energy of the
two phase mixture (with the same entropy as the sat. vapour at 120°C)
 Calculate work: w=Δu

At 120°C: u=1240.3kJ/kg, s=3.8861 kJ/kgK
At 119.5 kPa (-30°C), ul=44.08 kJ/kg, uv=1288.9kJ/kg, sl=0.1856kJ/kgK, sv=5.778kJ/kgK

Vapour fractions: x=(3.8861-0.1856)/(5.778-0.1856)=66.17%
umix=44.08+0.6617*(1288.9-44.08)=867.78 kJ/kg
Δu-q-w=-w
w=-Δu=-(867.78-1288.9)=372.52 kJ/kg

Saturated sulphur dioxide vapour, 400K, is heated to 600K in a constant pressure heating process
(without any shaft work). How much heat is required?

Solution

A difficult question, but one thing is simple: for a constant pressure process without shaft work, Δh=q-
w (w=0)q=Δh
So what we have to find to solve this question is the change in enthalpy of sulphur dioxide between
400 and 600K.

From here, this is quite a difficult question because there is so little information. First of all, there are
no thermodynamic tables of sulphur dioxide available. However, we can get the specific heat from
table A5 (at room temperature) or table A6.
For an ideal gas, Δh=CpΔT. For a real gas, we will have to correct for enthalpy deviations. However,
enthalpy deviation depends on pressure and we do not know the vapour pressure (yet).

The solution strategy therefore is:

Calculate Δhig from CpΔT
Check for non-ideality

When non-ideal:
Read off Δhd of saturated vapour at 400K from enthalpy deviation chart. Also read off Pr of saturated
vapour at 400K from enthalpy deviation chart. This is a constant pressure heating process; Pr,2=Pr,1
Calculate Tr,2 and read off enthalpy deviation of state 2.
Subtract Δhd,1 from Δhd,2 to get the change in enthalpy deviation.
Subtract this value from the ideal gas change in enthalpy to get the total change in enthalpy.

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