Problems

Problem 1:-

Evaluate the slopes of a constant-pressure process and a constant- volume process lines on an enthalpy versus pressure diagram and on an enthalpy versus entropy diagram, respectively, for an ideal gas. Redraw the cycle shown below on h-P and h-s diagrams?

Solution 1:-

For an ideal gas: and

For an ideal gas:

Problem 2:-

The decrease in the melting point of ice with increase in pressure is largely responsible for our ability to skate on ice. At the line of contact between the skate and the ice a high pressure is exerted by the person who is skating because of the small area of contact. Therefore, ice melts and provides a thin layer of liquid water which acts as a lubricant and help in skating. Suppose a man of mass 75 kg desires to skate on ice. The area of contact between the skates and ice is 20 mm². Will he able to skate on ice, which is at –2⁰ C?

Solution 2:-

If DT=-2K, pressure required = 27MPa

Pressure increase due to the person on skates

Since the increase in pressure is greater than the required pressure, it is possible to skate.

Problem 3:-

Determine the sublimation pressure of water vapour at –60⁰C using data available in the steam tables.

Control mass: water

From steam tables,

For saturated solid- saturated vapour water,

Temp.,T C	Pressure, P kPa	Evap.Enthalpy, hig KJ/kg
-30	0.03810	2839.0
-40	0.01286	2838.9

Solution 3:-

For sublimation, the change from solid ‘i’ directly to vapour ‘g’, the Clapeyron equation reduces to the form,

h_ig is relatively constant in these temperature ranges. Hence, integrate between the limits –40⁰C and –60⁰C.

Let

P₂ = 0.0129kPa T₂=233.2K T₁=213.2K

Then,

P₁ = 0.00109kPa

Problem 4:-

The pressure on a block of copper having a mass of 1kg is increased in a reversible process from 0.1 to 100 MPa while the temperature is held constant at 15⁰C. Determine the work done on the copper during this process, the change in entropy per kilogram of copper, the heat transfer, and the change of internal energy per kilogram.

Over the range of pressure and temperature in this problem, the following data can be used:

Volume expansivity = a_P =5.0*10^-5K^-1

Isothermal compressibility=b_T=8.6*10^-12m²/N

Specific volume= 0.000114 m³/kg

Analysis:-

Control mass: copper block

States: initial and final states known

Process: Constant temperature, reversible

The work done during the isothermal compression is

The isothermal compressibility has been defined as

Therefore, for this isothermal process,

Since v and b_T remain essentially constant, this is readily integrated:

The change of entropy can be found by considering the Maxwell relation, and the definition of volume expansivity.

This equation can be readily integrated, if we assume that v and a_P remain constant:

The heat transfer for this reversible isothermal process is

q = T(s₂-s₁)

The change in internal energy follows directly from the first law,

(u₂-u₁)= q-w

Solution 4:-