Question

Steam at a pressure of 0.08 bar and a quality of 93.2% enters a shell-and-tube heat exchanger where it condenses on the outside of tubes through which cooling water flows, exiting as saturated liquid at 0.08 bar. The mass flow rate of the condensing steam is 3.43 x 10^5 kg/h. Cooling water enters the tubes at 15.8°C and exits at 35.8°C with negligible change in pressure.

1. Neglecting stray heat transfer and ignoring kinetic and potential energy effects, determine the mass flow rate of the cooling water, in kg/h, for steady-state operation.

Answer 1

The answer is "
`\bold{9.09* 10^6 (kg)/(hour)}`".

Step-by-step explanation:

For the reference table we get:

`h1 = 2410 (kJ)/(kg) \ , at \ \ \\\\ \ P = 0.08 \ bar \ and \ \ quality = 0.932`

Through steam tables, they get:

`\ h2 = 173.9 (kJ)/(kg) \ on\\\\ \ P = 0.08 \ bars \ \ \ but \ quality = 0 (sat.liquid),`

Water power transfer =
`\ 3.4 * 10 ^ 5 * (2410-173.9)\\`

It should be comparable to the water enthalpy:

`m_(water)* Cp* (T2-T1)\\\\For \ eg:\\\\ = 3.4 * 10 ^ 5* (2410-173.9) \\\\ = m_(water)*4.18*(35-15)\\`

`m_(water)=9.09* 10^6 (kg)/(hour)`