Chemical Engineering Thermodynamics: First and second Law problems

1.   100 mol of an ideal gas is compressed isothermally at 400 K from 100 kPa to 500 kPa. The process is irreversible and requires 20% more work than for reversible compression. Calculate the entropy change of the gas, entropy change of the surrounding and universe. The surrounding is at 300K.   
Sol: W =  nRTln(P2/P1)
ΔSsys   =  nRln(P2/P1)
 Qsurr  =   1.2xW
 ΔSsurr  =  Qsurr/T(300)
ΔSuniv = ΔSsys + ΔSsurr

2.   A copper rod is of length 1 m and diameter 0.01 m. One end of the rod is at 100°C and, the other at 20°C. The rod is perfectly insulated along its length and the thermal conductivity of copper is 380 W/mK. Calculate the rate of entropy production due to irreversibility of this heat transfer.   
Sol: Q= -kAdT/dx (as heat is transferred axially)
       ΔS = Q[1/293-1/373]  as heat is being transferred from high temperature to low temperature) 

3.   A rigid tank of volume 1 m³ contains 200 moles of CO₂ at 25°C. Electric current of 5 A at 440 V passes through a resistor, placed inside the tank, for 15 min. Determine the final state of CO₂ in the tank. CO₂ may be treated as a vander Waal gas with constants a = 363.077 × 10^-3 Pa (m³/mol)² and b = 0.043 × 10^-3 m³/mol. The change in internal energy of CO₂ can be expressed as 

     dU = c_v dT + dV, with c_v = 32.34 J/molK. 
Sol: dU = dQ + dW
      nc_v dT = V.I.t  
      Find T2 and  calculate P2 using vander Waal equation of state.
 

4.  0.3 m³ of water at 1 bar and 20% quality is enclosed between a cylinder and a piston resting on stops 1 (initial state) as shown in figure. The atmospheric pressure and the weight of piston are such that a pressure of 3 bar is required to lift the piston. The system is heated until the piston reaches the upper stops 2 (final state) where the volume is 0.45 m³. Heating is continued further until water exists as saturated vapour. Show the processes on T-V diagram and determine the final pressure of water, overall heat transfer, work done and entropy change for the process.           
     Sol: intially v = 0.2m(vg) + 0.8m(vl)
           calculate m and get final specific volume as v/m
           find saturation pressure from steam table.
           W = -PextdV
            Find ΔU from initial and final conditions and then calculate q as ΔU-W
            similarly find ΔS 

Chemical Engineering: Great Opportunities Ahead

Chemical Engineering is the skill to scale up the operation from a laboratory to the plant. It is about converting material from its raw form to useful products in an economic and environmental friendly way. The job of a chemical engineer is to optimize and control the various parameters in the process like temperature, pressure, flow rate, composition; that affect the process, and design various equipments used in chemical industries.
In this scenario, there are lots of opportunities for a chemical engineer, like to fulfill the demand of energy (Petroleum industries, Nuclear Energy, Renewable Energy), environment challenges (air and water pollution, waste management), Pharmaceutical industries, Fertilizer Industries, Food Industries, Plastic Industries etc. Chemical branch seems to have a great scope in future.

A chemical engineer studies courses like  Chemical process calculation, Fluid flow, Heat and Mass transfer, Thermodynamics, Chemical reaction engineering, chemical technology, process and plant design, Process modeling and simulation etc.

There are various public sector units that hire chemical engineers like
IOCL, HPCL, BPCL, ONGC, BARC, NPCIL, NFL, NAlCo, PDIL, EIL.
In addition to engineering jobs there are opportunities for research in this stream like R&D units of refineries, BARC and various labs like CSIR, NCL.

Mass Transfer: Molecular Diffusion flux (JA) and Total Diffusion Flux (NA)

Diffusion is the process of transfer of molecules when there is a concentration difference. Diffusion can occur in two ways; one is molecular diffusion and other is convective diffusion. In molecular diffusion the driving force is concentration difference while in convective diffusion there is bulk velocity of particles and medium.

For example (assume unidirectional diffusion) if a person has put perfume on his body, when he and observer both are either stationary or moving with same velocity, then the flux of perfume molecules with reference to observer is due to concentration difference i.e. molecular diffusion flux (J_A) and if the person is moving and the observer is stationary then diffusion of molecules is observed having both the effects: molecular as well as convective (due to motion of person) i.e. total flux (N_A).

Total Flux = molecular flux + convective flux

N_A          =          J_A + c_A.v

c_A.v_{A     =}              c_A(v_{A –} v) + c_A.v

Where v_A is the avg. velocity of all A like species and v is the avg. velocity of the carrier relative to observer means average of carrier velocity and velocity of A molecules.  In the above example, v_A is the velocity of all perfume particles and v is the avg. velocity of the person and perfume particles. So, the difference c_A(v_{A –} v) corresponds to molecular diffusion and c_A.v corresponds to convective diffusion. If a person is moving with perfume put on body and other person is stationary with perfume put on his body, then the difference between the two fluxes corresponds to convective diffusion. The other example can be taken of the fragrance particles of air freshener, the particles are diffusing through molecular diffusion and if air blows, then there will be combined effect of molecular diffusion and convective diffusion.

In distillation column, the flow of liquid and vapor streams leads to molecular as well as convective diffusion.

Heat Transfer: Heat flux and Temperature relation in Insulation and Convection

Insulation is added to a system to decrease the rate of heat loss. In a system operating at steady state, if we increase the thickness of insulation, will it decrease the rate of heat loss?

For a steady state system, the heat in to the system should be equal to heat out of the system. So, by increasing the thickness of insulation will not decrease the heat loss. For example, if we are heating a copper disc with a 50 W heater, under steady state 50 W heat will be lost to the surrounding. If we add insulation on the sphere, the heat lost to the surrounding must be 50 W to make it steady. Then, heat loss has not been reduced. But by adding insulation we can increase the temperature of the system without losing more heat, i.e. if the temperature of the sphere was 40 °C with 50 W heat input earlier, by adding more insulation we can raise the temperature of sphere with same amount of heat input.
Natural and Forced Convection:
In steady state, the heat transferred by natural convection and forced convection will be same. But in forced convection, the temperature of the surface will be lowered as compared to the natural convection. If we maintain the same temperature in natural and forced convection, the heat transferred by the forced convection will be more than natural convection.  

Chemical Engineering Thermodynamics: Questions

    3. The change in internal energy for an isothermal expansion of a real gas is greater than, less than or  equal  to 0

4. Is it possible in a cycle a heat interaction of 50 kJ from one reservoir and 50 kJ of work transfer?

5. Work done in a clockwise cycle is work done by system or work done on the system i.e. +ve/-ve.

6. Under what condition ΔV=0 and W≠0 and  ΔV≠0 and W=0.

7. How would you calculate the work done for an irreversible process?

8. Effect of temperature on melting point of water

9. Entropy change for irreversible adiabatic expansion process greater than, less than or equal to 0.