Heat exchanger calculation

katchum

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

What happens when in counter flow heat exchangers, you diminish the mass flow of the low temperature stream?

2. Relevant equations

UA = cte
m2 = cte
Tin = cte
tin = cte
Q=cm1(tout-tin)
Q=cm2(Tout-Tin)
Q=UA(Tout-tout)+(Tin-tin)
----------------------
ln(Tout-tout/Tin-tin)

T = warm stream
t = cold stream

3. The attempt at a solution

I've inputted these equations in mathcad, but I am unable to find the mass flow m1 in function of Q. I get weird numbers like W(...) and root(...)

I want to know if m1 decreases, Q will increase or decrease.
So basically:

When I want to heat up 1 kg of cold water as high as possible.
Should I use a low flow or a high flow of cold stream?

It looks like I always get warm water in bath when I use low inlet flows.

Hmm, I also think, that the hot stream will decrease a little bit in temperature when the cold stream has a low flow, so the temperature difference will be very high, so basically the cold stream should be higher in temperature at the outlet.

So is it true that if m1 decreases, the Q will increase?

Edit:

This is weird:

Apparently when you decrease m1, then Q will decrease too. So less heat will be transferred.
But, the temperature of the outlet streams will be higher.

Is there any explanation for this?

My conclusion is, to have an efficient heat transfer you should open the inlet valve fully open to your bath. But when you want warm water you should close the inlet valve a little bit.

Funny to know is, when you want to save energy you should use a low inlet flow to your bath, you'll get warm water and there is less heat transfer so you'll use up less energy. Or am I wrong?

Abninfamy

Well, if you look at it from a first law perspective for one of the streams.
Q=m(h(out)-h(in)) If you decrease m, regardless of the factors of h, Q will decrease.

katchum

How do you know that h(out)-h(in) doesn't increase more than m decreases?

h(in) = cte and h(out) will increase.