Question

1. Ammonia is initially at a temperature of -10° C and a specific volume of 0.07 m3/kg. The ammonia undergoes an isobaric expansion to a final specific volume of 0.22 m3/kg. Evaluate the specific work done on the ammonia in kJ/kg and the specific heat transfer to the ammonia in kJ/kg. Neglect changes in kinetic energy and potential energy.

2. Carbon dioxide is contained in a piston-cylinder assembly with an initial pressure and temperature of 8 lbf/in2 and 100° F, respectively. The carbon dioxide has a mass of 0.05 lb. The carbon dioxide is expanded isothermally to a final volume of 1 ft3. Model the carbon dioxide as an ideal gas with constant specific heats. Evaluate the specific heats at 100° F. Determine the amount of work done on the gas in Btu and the heat transfer to the gas in Btu. Neglect changes in kinetic energy and potential energy.

3. An ideal gas with constant specific heats undergoes a process from an initial pressure and specific volume of 80 kPa and 40 m3/kg to a final specific volume of 20 m3/kg. During the process, the hydrogen’s pressure and specific volume are related through the equation given below. In the equation, ?? and ?? are the initial pressure and specific volume of the hydrogen, respectively. Determine the specific work done on the gas in MJ/kg and the specific heat transfer to the gas in MJ/kg. The ideal gas has a molar mass of 2.0 kg/kmol and a specific heat at constant volume of 7.5 kJ/(kg∙K). Neglect changes in kinetic energy and potential energy.

   ? = ?? ⋅ [2 − (?/??)]

Answer 1

Note: On seeing through questions it is clear that basically there are three different questions. So, We will be looking these questions one by one.

Let's start with the very first question itself that of related to ammonia.

Data: Initial temperature of ammonia, Ti = - 10  °C = 273 - 10 = 263 K

Specific volume of ammonia(initially), v₁ =  0.07 m3/kg, and

Specific volume of ammonia(finally), v₂ = 0.22 m3/kg.

Observation: The ammonia undergoes an isobaric expansion. We have to evaluate the followings:

(a) The specific work done on the ammonia, Ws in kJ/kg and

(b) The specific heat transfer to the ammonia, Qs in kJ/kg.

Assunption: Following assumption are necessary:

(i) Assume ammonia is behaving as ideal gas at given conditions.

(ii) Neglect changes in kinetic energy and potential energy.

Analysis: The given isobaric process can be shown on p - v diagram as given the figure below:

Clearly as mentioned in the diagram, p_A(=p_{1) =} p_B(=p₂) and the work done on the ammonia will be equal to the area under p - v diagram on v - axis.

Hence, the work done = p_A(v₂ -v_{1) =} p₁*(v₂ -v_{1) (i)}

Substituting values of specific volume , we get he specific work done on the ammonia,

Ws = p₁*(0.22 - 0.07) = 0.15p₁ (ii)   

Now, for ideal gas we have ideal gas equation

pv = RT (iii)

Here symbols have usual meaning.

R is characteristic gas constant, for ammonia, R = 0.884 kJ/kgK

or, pressure p = RT/v

Therefore, p₁ = RT₁/v₁ = 0.884*263/0.07 = 3321.3 kPa

putting thi value of p₁in equation (ii), we obtained,

Ws = 0.15*3321.3 = 498.2 kJ/kg (iv) Ans

Now for Qs:

From the first law of thermodynamics

Qs = Ws +Us (v)

Where Us is the change in the specific internal energy of the system,

Us = C_V(T₂-T₁) (vi)

At ambient pressure and temperature the isobaric specific heat, C_P, of ammonia is 2.2 [kJ/kg K] or 0.52 [Btu/lb °F] = [cal/g K], while the isochoric specific heat, C_V, is 1.6 [kJ/kg K}

Also for isobaric process:

T₂/T₁ = v₂ /v₁

Thus, T₂ = T_1*v₂ /v₁ = 263*0.22/0.07 = 826.6 K (vii)

So, substituting the values of data in equation (vi), we get;

  Us = 1.6*(826.6- 263) = 901.8 kJ/kg   (viii)

Finally, from equation (v), we get the specific heat transfer to the ammonia;

Qs = 498.2 + 901.8 =  1400 kJ/kg (ix) Ans