Thermodynamics Numerical Problems and Answers
1. Pressure Measurement with a Manometer
Q.1.1 A manometer is used to measure the pressure
inside a tank having its specific gravity (Sg) of 0.85 and the
manometer column height of 50cm. If we consider the local atmospheric pressure
is 70 kPa, then what is the value of absolute pressure within a tank.
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Manometric readings at absolute pressure |
#Solution1.1
As per the above given data, we should take an
assumption that the fluid inside the tank of gas having much lower than density
of manometer fluid.
Where, we have Sg =0.85, Patm = 96kPa and Manometric column height(h)= 55cm, here g value is 9.81m/s2
Here, for the
density of the given fluid should be determined by multiplying its specific
gravity by the density(ρ) of
water is specified by 1000kg/m3.
Or we can do ρ=Sg= (0.85)
x (1000) = 850 kg/m3
We have an equation for manometer to calculate the
pressure at a depth of h from the free surface is P= Patm + ρgh.
So, now we can
calculate the absolute pressure(P)
=96kPa+(850kg/m3)
(9.81m/s2) (0.55m) (1N/1kg.m/s2) (1kPa/1000N/m2)
=100.6kPa
=4682.175 Pa
Or we have the value
of gage pressure is 1kPa= 1000Pa
1P= (1/103)
kPa or 4682.175 Pa=4.6kPa
(Solved.)
Q.1.2 Take a
pipe line of the gas pressure is recorded by the mercury manometer having one
limb open to the atmosphere. If the difference between the mercury height in
two limbs is 500mm, then determine the gas pressure in bar.
Give data acceleration due to gravity(g)=9.81 m/s2, density of the mercury(ρ)=13.640kg/m3 and barometer readings 761mm of Hg.
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Manometric pressure readings in a pipe line |
#Solution1.2
Here, we have taken from the above figure, the plane
line 1-2, we can write
P=P0
+ 𝛒gh or we have P0 = 𝛒gh0 , where h0= barometric height, 𝛒=
density of the mercury and P0 = atmospheric pressure.
Again, we can have P =𝛒g(h+h0) =13.640kg/m3 x 9.81m/s2 (0.500+0.761)
m = 168.732 N/m2 = 169kPa = 1.69bar (Solved.)
2.The weight of Piston effect on pressure in a
cylinder
Q.2.1 The piston-cylinder arrangement, a vertical
piston contains a gas having mass of 60kg and its cross-sectional area of 0.04m2.
The local atmospheric pressure is 0.97 bar and the gravitational acceleration (g)
is 9.81m/s2.
(i) Calculate the pressure inside the cylinder
(ii) If we supply some heat to the gas, then its volume is doubled, have any effect of changes in pressure inside the cylinder.
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Free body diagram of the piston |
#Solution 2.1
Here, we should take the friction between the piston
and the cylinder is negligible.
The gas pressure in the piston-cylinder device depends
on the atmospheric pressure and the weight of the piston as per the above
figure shown and balancing the vertical forces yield.
PA=Patm + W
Taking P by substitute in both sides,
Now, we have P= Patm +mg/A
=0.97bar + (60kg) (9.81m/s2) / (0.04m2)
x (1N/1kg.m/s2) x (1bar/105N/m2)
=1.12bar (Solved.)
(ii) Due to change in
volume, there is no effect on the free body and the pressure inside the cylinder
wall remain in same.
N.B. :- If the gas behaves as an ideal gas, then the absolute temperature becomes double when the volume doubles at constant pressure.
Q.2.2 A rigid
tank contains a hot fluid and cooled by paddle wheel. Initially, the internal
energy of fluid is 800 kJ. During cooling process, the fluid extracts the heat
of 500 kJ and the paddle wheel works 100 kJ on the fluid. Calculate the final
internal energy of the fluid. Neglect the energy stored in the paddle wheel.
#Solution 2.2
A fluid in a rigid tank loses heat while being starred
by paddle wheel. The final internal energy of the fluid is to be calculated.
Let’s take some assumptions, for process-1, the tank
is stationary in condition, thus kinetic energy and potential energy changes
are zero 𝛁KE
= ∆PE
= 0, for that 𝛁E = ∆U
And for process-2, the
internal energy is only in the form of system’s energy which may alter. The
storage of energy in the paddle wheel is negligible.
 |
| Cooling of a hot fluid in a tank |
From the above figure, this a closed system, where
there is no mass crossing the boundary during the process. Here, the volume of
a rigid tank is constant. And there is no work in moving boundary. The heat lost
from the system and shaft work is done on the system.
Let’s applying the energy balance on the system
Ein – Eout = ∆Esystem
Wsh.in − Qout = ∆U = U2 −U1
100 kJ − 500 kJ =U2 – 800 kJ
Or we can write U2 = 400 kJ
Thus, the internal energy of the system is 400 kJ.
3.The piston and cylinder arrangement with work
transfer for the system
Q.3.1 A piston-cylinder arrangement
contains some fluid with its stirring device inside the cylinder. Here the
piston is frictionless and it moves down against the fluid due to the
atmospheric pressure of 101.325 kPa. Let suppose, the total no. of revolutions
is 10,000 by the stirring device with its average torque produced by 1.275 mN.
Meanwhile, the piston having its diameter moves out 0.8 m. Then calculate the
total work transfer for the system?
#Solution 3.1
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Piston-Cylinder Arrangement containing fluid |
From the above figure, we work done by the
stirring device on the system.
W1 = 2𝛑TN
Or we can write, 2𝛑 x 1.275 x 10,000Nm = 80 kJ
Here, this is a
negative work done for the system, thus the work done by the system W2 = (pA). L = 101.325 kN/m2 x 𝛑/4(0.6)2
m2 x 0.80 m = 22.9 kJ
But here this is
a positive work done for the system.
The total work transfer
for the system
W = W1 + W2 = -80 + 22.9 = -57.1 kJ (Solved.)
