An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power produced is 2000 W, what is the corresponding rate of heat flow? What is the rate of lost work?

Rate of flow of ideal gas , n = 4 kmol/hr = [tex]\frac{4\times 1000mol}{3600s}=1.11mol/s[/tex] (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

[tex]\Delta U=0[/tex] (For isothermal process)

So, by applying first law of thermodynamics:

[tex]\Delta U=\Delta q-\Delta W[/tex]

[tex]\Delta q=\Delta W[/tex] .......(1)

Now, calculating the work done for isothermal process, we use the equation:

[tex]\Delta W=nRT\ln (\frac{P_1}{P_2})[/tex]

where,

[tex]\Delta W[/tex] = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

[tex]P_1[/tex] = initial pressure = 100 kPa

[tex]P_2[/tex] = final pressure = 50 kPa

Putting values in above equation, we get:

[tex]\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW[/tex]

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

[tex]\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW[/tex]

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.