Question

Water (10 kg/s) at 1 bar pressure and 50 C is pumped isothermally to 10 bar. What is the pump work? (Use the steam tables.) O -7.3J/s O 7.3 kJ/s O -210 kJ/s O 3451 kJ/s

Answer 1

Explanation:

For an isothermal process equation will be as follows.

                W = nRT ln$\frac{P_{1}}{P_{2}}$

It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.

                    No. of moles = $\frac{mass}{\text{molar mass}}$

                                           = $\frac{10000 g/s}{18 g/mol}$

                                           = 555.55 mol/s

                                           = 556 mol/s (approx)

As T = $50^{o}C$ or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.

                  W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]

                      = $556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times ln\frac{1}{10}$    

                     = $556 mol/s \times 8.314 J/ K mol K \times 323.15 K \times -2.303$    

                     = -3440193.809 J/s

Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.

                     $3440193.809 J/s \times \frac{0.001 kJ}{1 J}$

                          = 3440.193 kJ/s

                          = 3451 kJ/s (approx)

Thus, we can conclude that the pump work is 3451 kJ/s.