Question

A hot-water stream at 60oC enters a mixing chamber with a mass flow rate of 0.5 kg/s where it is mixed with a stream of cold water at 10oC. If it is desired that the mixture leave the chamber at 41oC, determine the mass flow rate of the cold-water stream. Assume all the streams are at a pressure of 250 kPa.

Answer 1

Answer: 0.306 kg/s

Explanation:

If we assume that the specific heat for all stream is constant, we can then say, the mass flow rate of cold water can be calculated by using the energy balance equation.

The energy balance equation basically states that, the heat gained is equal to the heat lost

m'1h1 + m'2h2 = m'3h3

m'1T1 + m'2T2 = (m'1 + m'2) T3

0.5 (60) + m'2(10) = (0.5 + m'2)41

30 + 10m'2 = 20.5 + 41m'2

30 - 20.5 = 41m'2 - 10m'2

9.5 = 31m'2

m'2 = 9.5 / 31

m'2 = 0.306 kg/s

Answer 2

Answer:

the mass flow rate of the cold-water stream = 0.3065 kg/s

Explanation:

To solve this, we will work under the assumption that all the streams have constant specific heats. Thus, using energy balance equation, we can calculate the mass flow rate of the cold water from ;

m'1h1 + m'2h2 = m'3h3

This transforms to;

m'1T1 + m'2T2 = (m'1 + m'2) T3

Where;

m'1 is the flow rate of hot stream water = 0.5 kg/s

T1 is the temperature of the hot stream water = 60°C

T2 is the temperature of the cold water = 10°C

T3 = Temperature at which the mixture leaves the chamber = 41°C

m'2 is mass flow rate of cold stream water.

Let's make m'2 the subject;

m'1T1 + m'2T2 = (m'1 + m'2) T3

m'1T1 + m'2T2 = m'1•T3 + m'2•T3

m'1T1 - m'1•T3 = m'2•T3 - m'2T2

m'2(T3 - T2) = m'1(T1 - T3)

m'2 = (m'1)[(T1 - T3)/(T3 - T2)

Thus, plugging in relevant values to get ;

m'2 = 0.5[(60 - 41)/(41 - 10)

m'2 = 0.5(19/31)

m'2 = 0.3065 kg/s