Two streams of water are mixed in a steady-state process: one stream of 15 kg/s

Question

Two streams of water are mixed in a steady-state process: one stream of 15 kg/s at 950 C and a second stream of undetermined flow at 250 C. It is desired to produce a product stream at a temperature of 750 C. The mixing vessel is not well-insulated and heat is lost at a rate of 1.0 kW. Do not assume that the heat capacity value is constant.

(a) What is the flowrate of the 250 C stream and the 750 C stream?

(b) If the mixing vessel was perfectly insulated, what would be the temperature of the product stream if the flows are the same as in part (a)? Ignore any mixing effects. (c) Repeat parts (a) and (b) for another liquid of your own choosing. Specify the liquid and the source for your heat capacity values.

Explanation / Answer

Ans. Given,

            Stream A = Flow rate of 15 kg/s at temperature 9500C

            Stream B = Flow rate of X kg/s at temperature 2500C       , where, X = unknown.

            Loss of heat during mixing of stream A & B = 1.0 kW

Now,

            Final mass (per second) of stream after A & B are mixed = 15 kg + X kg = (15+X) kg

            Given, temperature of mixed stream (A & B mixed) = 7500C

Using, Q = m s T         -- equation 1   

Where, q = heat content                                                , m= mass in kilogram,

s= specific heat (in terms of kJ/kg0C)                  

T = Temperature

Before proceeding any further, note the following points-

I. Water is in vapor phase. Water vapor has specific heat value of 1.855 kJ/kg0C at 2500C, 2.113 kJ/kg0C at 7500C, and 2.252 kJ/kg0C at 9500C.

II. Considering the calculation for 1 second (better say, per second)- when stream A and B are mixed, the final (sum) mass of stream becomes (15+X) kg and reaches an equilibrium temperature of 7500C.

III. While reaching thermal equilibrium, the mixing vessel loses 1.0 kW energy.

IV. So, total heat content of stream (A + B mixed per second) is equal to that of final stream at 7500C minus 1.0 kW (heat dissipation).

Therefore,

            Q1 (stream A) + Q1 (stream B) = Q (final stream mixture) – 1.0 kW

[15 kg x (2.252 kJ/kg0C) x 9500C ] + [X kg x (1.855 kJ/kg0C) x 2500C ] =

[(15+X) kg x (2.113 kJ/kg0C) x 7500C ] – 1.0 kW

Or, 32091 kJ + 463.75X kJ = 23771.25 kJ + 1584.75X kJ – 1 kJ                   ; [note: 1 W = 1 J/s]

Or, 32091 kJ - 23771.25 kJ + 1 kJ = 1584.75X kJ - 463.75X kJ

Or, 8320.75 kJ = 1121X kJ

Or, X = (8320.75 kJ) / (1121 kJ) = 7.42

So, mass of stream B (second stream) flowing in 1 second = 7.42 kg

Thus, flowrate of stream B = 7.42 kg /sec        

Part B: In case the mixing vessel is perfectly insulated, there is no heat dissipation of 1.0 kW.

Let the equilibrium temperature of mixed stream = T0C

So,

[15 kg x (2.252 kJ/kg0C) x 9500C ] + [7.42 kg x (1.855 kJ/kg0C) x 2500C ] =

[(22.42) kg x (2.113 kJ/kg0C) x T0C ]

Or, 32091 kJ + 3441.025 kJ = 47.37346 kJ T

Or, T = (35532.025 kJ) / (47.37346 kJ) = 750.04

Thus, temperature of insulated vessel at thermal equilibrium = 750.040C

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