Thermodynamics: Calculating Entropy Change for Mixing Streams

[roam]

Homework Statement

http://img401.imageshack.us/img401/6641/problem2t.jpg

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.

 

[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.

 

[roam]

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.

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.

 

[rock.freak667]

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.

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?)

 

[roam]

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?)

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?

 

[rock.freak667]

Thanks, so the mass flow rates are

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

This is correct.

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

$Q=mc\Delta T$

Also correct.

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?

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 T_f AND one for the heat lost by the 20°C stream to the final temperature T_f?

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

 

[roam]

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 T_f AND one for the heat lost by the 20°C stream to the final temperature T_f?

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

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?

 

[rock.freak667]

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

 

[roam]

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

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

 

[rock.freak667]

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

Yes, try it.