how to find final temperature of a fluid in an heat exchanger?

Question

I have to find final temperature of water in an heat exchanger.
the heat exchanger must cool down air using water, meaning initial temperature of water must be cold, initial temperature of air is 298.15 Kelvin

I can use this equation to find initial temperature of water, $Q_{max} = C_{min}*\delta T$ where $C_{min} = \frac{kg}{s} * \frac{J}{kg*K} = \frac{W}{K}$

and this is not a problem, but here comes my question: in order to find final temperature from this equation $T_{out, water} = T_{in, water} +/- \frac{Q}{C}$

how can I know which sign to plug into this equation?

this is my reasoning, let me know if it's right: if the text says water is cooling down air, that means initial temperature of water is becoming warmer than before by taking heat from air (i.e the hot fluid), and therefore equation becomes: $T_{out, water} = T_{in_water} + \frac{Q}{C}$

if this is true, then minus sign means the contrary.

but sometimes I have an exercise that say the same thing, but it requires me to flip the signs, for example, "air is being extracted from a server room and it has to be cooled down by cold water", in this case, exercise requires to use this equation: $T_{out, water} = T_{in_water} - \frac{Q}{C}$ so I think my reasoning is wrong or at least it cannot be applied all the time for every situation.

Answer 1

I think your confusion comes from unclear definition of $Q_{max}$ and $\delta T$.

Let's say you have 2 streams with possible heat exchange between them. Following stream parameters will be denoted by subscripts $1$ and $2$:

• mass flow $\dot{m}_i$
• specific heat capacity $c_{p,i}$
• inlet temperature $T_{in,i}$
• outlet temperature $T_{out,i}$

Now it is useful to denote the heat flow between the stream in a way that makes the direction obvious, e.g. $\dot{Q}_{1\rightarrow2}$ for heat flow from stream $1$ to stream $2$. For a heat exchanger in steady state with following parameters:

• overall heat transfer coefficient $U$
• logarithmic mean temperature difference $\Delta T_{1-2}$ between streams $1$ and $2$
• heat exchange area $A$,

two things must be true in steady state:

• energy balance (heat leaving one stream end up in the other one): $$\left(T_{in,1}-T_{out,1}\right)\cdot c_{p,1}\cdot \dot{m}_1 = \left(T_{out,2}-T_{in,2}\right)\cdot c_{p,2}\cdot \dot{m}_2$$
• heat rate proportional to negative temperature difference and overal heat transfer coefficient (Fourier law; heat always flows against temperature gradient): $$\dot{Q}_{1\rightarrow2} = \Delta T_{1-2}\cdot U\cdot A$$

I am not sure if this is obvious, but the equations can be connected in this way: $$\dot{Q}_{1\rightarrow2} = \left(T_{in,1}-T_{out,1}\right)\cdot c_{p,1}\cdot \dot{m}_1$$ $$\dot{Q}_{1\rightarrow2} = \left(T_{out,2}-T_{in,2}\right)\cdot c_{p,2}\cdot \dot{m}_2$$

As you can see, the assumption about which stream is cooler or hotter does not change these equations. It's just sometimes $\dot{Q}_{1\rightarrow2}$ will be positive and other times negative.

Answer 2

First Law of Thermodynamics:

$$ \Delta U = Q + W $$

(or $\Delta U = Q - W$ if you are a weirdo...)

A continuous, steady state heat exchanger at constant pressure can be modeled as

$$ \Delta \dot H = \dot Q $$

(flow work $\dot W = -\Delta (p\dot V)$ combines with $\Delta \dot U$ on the other side of the equation and becomes enthalpy, $H = U+pV$. The dots indicate that's the energy transfer rate in a flow stream).

If there is no phase transition,

$$ \dot m c_p \Delta T = \dot Q $$

$$ T_{out} = T_{in} + \frac{\dot Q}{\dot m c_p} $$

The sign in that equation is always a plus sign, because that derives from the First Law of Thermodynamics, and although there are two competing conventions about the work term $W$, the heat term $Q$ is always written with a plus sign.

However, $\dot Q>0$ when heat flows towards the control volume (the surroundings are hotter than the system) and $\dot Q<0$ when heat flows away from the control volume (the system is hotter than the surroundings). For example, if you're using cold water to cool hot gases, $\dot Q_{water}>0$ and $\dot Q_{gas}<0$. Moreover, $\dot Q_{water} = -\dot Q_{gas}$, because energy is conserved.