# Thermodynamics: Waters Mixing

1. Mar 16, 2012

### roam

1. The problem statement, all variables and given/known data

http://img401.imageshack.us/img401/6641/problem2t.jpg [Broken]

3. The attempt at a solution

I'm not sure how to approach this problem. I know of this equations for when they reach thermal equilibrium:

$T_f = \frac{C_1 T_{1,i} + C_2T_{2,i}}{C_1+C_2}$

And this for entropy change:

$\Delta S = C \ln \left( \frac{T_f}{T_i} \right)$

But how do we relate the speed of the two stream flows to all this? I appreciate if anyone could please help me to get started on this problem.

Last edited by a moderator: May 5, 2017
2. Mar 16, 2012

### rock.freak667

Since you have the density of water as 1000 kg/m3 you can convert the volumetric flow rate given to a mass flow rate. Which then becomes a simple use of equating the rate of heat loss by one to the rate of heat gain by one.

3. Mar 17, 2012

### roam

Thank you. So the mass flow rates would be

$\dot{m} = 1000 \times v \times A$ (I do not have the cross-sectional area)

There is another equation $Q = \frac{\dot{m}}{\rho}$, but how do I relate this to the tempratures? Because I need to find the final temprature of the mixture.

4. Mar 17, 2012

### rock.freak667

You have your volumetric flow rate in m3/s, you want mass flow rate in kg/s.

So you know that ρ= m/V so m = ρV, so that will give yu the mass flow rate m.

What would be the heat required to raise the temperature of water from T1 to T2 i.e. what is the formula? (Hint: What equation relates, mass, specific heat capacity and a temperature difference?)

5. Mar 17, 2012

### roam

Thanks, so the mass flow rates are

$\left\{\begin{matrix}m_1 = 1000 \times 5 = 5000 \ kg /s\\ m_2 = 1000 \times 2 = 2000 \ kg/s \end{matrix}\right.$

An equation for heat that relates mass, specific heat capacity and temprature difference is

$Q=mc\Delta T$

So, do I need to find the heat required to raise the water at 0°C to the 20°C temprature of the second stream?

$Q= m \times (4.18 \times 10^3) \times (293.15 \ K - 273.15 \ K)$

Which vlaue of m do I need use? And how does finding Q help us to determine the final temprature of the mixture?

6. Mar 18, 2012

### rock.freak667

This is correct.

Also correct.

Here's where your thinking is wrong. The final temperature will not be 20°C. What you should do is consider conservation of energy with no heat loss.

The stream at 0°C will gain the heat lost by the stream at 20°C.

Give your formula of Q=mcΔT can you make two expressions, one for the heat gained by the 0°C stream to the final temperature Tf AND one for the heat lost by the 20°C stream to the final temperature Tf?

When you do that, just put them equal to one another!

7. Mar 18, 2012

### roam

Thank you very much, that's a great idea! It worked out as follows

$5000 \times (4.18 \times 10^3) \times (T_f - 293.15) = 2000 \times (4.18 \times 10^3) \times (T_f - 273.15)$

$5000T_f-(5000 \times293.15) = 2000T_f - (2000 \times 273.15)$

$3000 T_f = 919450 \ \therefore T_f = 306.483 \ K$

But now for the second part of the problem I need to calculate "the total rate of change of entropy". So I know of the following equation which gives the total change of entropy for the object+environment:

$\Delta S_{o+e} = C_o \ln \left( \frac{T_f}{T_o} \right) +C_e \ln \left( \frac{T_f}{T_e} \right)$

Is this the correct equation in this case for the total rate of change of entropy?

Last edited: Mar 18, 2012
8. Mar 18, 2012

### rock.freak667

I think you need to add the entropy change of stream 1 to the entropy change of stream 2.

9. Mar 18, 2012

### roam

Using the equation $\Delta S = C \ln \left( \frac{T_f}{T_i} \right)$?

10. Mar 18, 2012

Yes, try it.