**A fluid is confined in a cylinder by a spring-loaded, frictionless piston so that the pressure in the fluid is a linear function of the volume (p = a + bV). The internal energy of the fluid is given by the following equation**

**U = 34 + p V**

**Where U is in kJ, p in kPa, and V in m3. If the fluid changes from an initial state of 170 kPa, \(0.03{m^3}\) to a final state of 400 kPa, \(0.06{m^3}\), with no work other than that done on the piston, find the direction and magnitude of the work and heat transfer.**

**(AMIE, Summer 2023) **

\(\begin{array}{l}{U_2} - {U_1} = 5({p_2}{V_2} - {p_1}{V_1})\\ = 5(400x0.06 - 170x0.03)\\ = 94.5\,kJ\end{array}\)

Now,

p = a + bV

170 = a + b x 0.03

400 = a + b x 0.06

Solving the above two equations, we get

a = -60

b = 7666.67

Therefore

\(\begin{array}{l}W = \int_1^2 {pdV = \int_1^2 {(a + bV)dV} } \\ = a({V_2} - {V_1}) + b\frac{{{V_2}^2 - {V_1}^2}}{2}\\ = \left( {{V_2} - {V_1}} \right)\left[ {a + \frac{b}{2}({V_1} + {V_2})} \right]\\ = (0.06 - 0.03)\left[ { - 60 + \frac{{7666.67}}{2}(0.06 + 0.03)} \right]\end{array}\)

= 8.55 J

\(\begin{array}{l}Q = W + {U_2} - {U_1}\\ = 8.55 + 94.5 = 103.05\,kJ\end{array}\)

**A
turbine is supplied with steam at a gauge pressure of 1.4 MPa The
steam, after expansion in the turbine, flows into a condenser maintained
at a vacuum of 710 mm of Hg. The barometric pressure is 772 mm of Hg.
Express the inlet and exhaust steam pressure in pascal (absolute). Take
density of mercury as \(13600\,kg/{m^3}\) and acceleration due to
gravity as \(9.81\,m/{s^2}\).(AMIE Summer 2023, 10 marks)**

Given data

Gauge pressure = 1.4 MPa = \(1.4x{10^6}\,Pa\)

Vacuum pressure = 710 mm of Hg ;

Barometric pressure = 772 mm of Hg;

p = \(13600\,kg/{m^3}\)

g = \(9.81\,m/{s^2}\)

Atmospheric pressure

\(\rho gh = 13600x9.81x0.772 = 103x{10^3}\,N/{m^2}\)

Inlet steam pressure = Gauge pressure + Atmospheric pressure

\(1.4x{10^6} + 103x{10^3} = 1.503x{10^6}\,N/{m^2}\,or\,Pa\)

= 1.503 MPa

Condenser pressure = Barometric pressure - Vacuum pressure

= 772 - 710 = 62 mm of Hg = 62 x 133.3 = 8265 \(N/{m^2}\) or Pa

= 8.265 kPa

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