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How much cooling duty would I need to bring air with a volumetric flow rate of $\dot{V}$ with 50% relative humidity from $T_{\text{high}}$ to $T_{\text{low}}$?

For example, what would the figure be when:

  • $\dot{V}=1.3 \cdot {10}^{5} \frac{\mathrm{m}^3}{\mathrm{hr}}$;

  • $T_{\text{low}}=35\sideset{^\circ }{}{\mathrm{C}}$; and

  • $T_{\text{high}}=45\sideset{^\circ }{}{\mathrm{C}}$?

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  • $\begingroup$ This is probably on-topic here, though for future reference, SE.Engineering might be easier for these sorts of questions. $\endgroup$
    – Nat
    May 10, 2018 at 21:21

2 Answers 2

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You'll need to apply a cooling duty of$$ q ~~=~~\dot{m}\int_{T_{\text{start}}}^{T_{\text{finish}}} {C_{\text{p}}\,\mathrm{d}T} ~~=~~\rho\dot{V}\int_{T_{\text{start}}}^{T_{\text{finish}}} {C_{\text{p}}\,\mathrm{d}T} \,,$$where:

  • $q$ is the heat duty;

  • $C_{\text{p}}$ is the constant-pressure heat capacity of the air;

  • $\dot{m}$ is the mass flow rate of the air;

  • $\dot{V}$ is the volumetric flow rate of the air; and

  • $\rho$ is the density of the air.

For rough approximations over small temperature changes like the $\Delta T= 10\sideset{^\circ}{}{\mathrm{C}}$ change you're asking about, the heat capacity, $C_{\text{p}}$, is often about constant, so you can often get away with$$ q ~~{\approx}~~\dot{m} C_{\text{p}}\left(T_{\text{finish}}-T_{\text{start}}\right) ~~{\approx}~~\rho\dot{V} C_{\text{p}}\left(T_{\text{finish}}-T_{\text{start}}\right) \,.$$

Caution: More work is needed if a phase change may occur. When cooling moist air, the biggest phase change to watch out for tends to be water condensing, since the cooler air can't hold as much water as warmer air.

To be lazy about it, I'll grab values quickly from WolframAlpha:

while noting that those values weren't taken from a proper table for the correct temperatures/pressures/saturations/etc..

So, this looks like$$ \begin{array}{rl} q & {\approx}~~\rho\dot{V} C_{\text{p}}\left(T_{\text{finish}}-T_{\text{start}}\right) \\ & {\approx}~~{\left(1.275\frac{\mathrm{kg}}{\mathrm{m}^3}\right)} \left(1.3 \cdot {10}^{5} \frac{\mathrm{m}^3}{\mathrm{hr}}\right) {\left(1.008 \frac{\mathrm{kJ}}{\mathrm{kg}{\cdot}\mathrm{K}}\right)} \left({\left(35\sideset{^\circ }{}{\mathrm{C}}\right)}-{\left(45\sideset{^\circ }{}{\mathrm{C}}\right)}\right) \\ & {\approx}~~-1.67076 {\cdot} {10}^{6} \frac{\mathrm{kJ}}{\mathrm{hr}} \\ & {\approx}~~-464.1 \, \mathrm{kW} \,, \end{array} $$so it looks like you're in the ballpark of $464.1 \, \mathrm{kW}$ of cooling duty.

Though to be clear, the above was meant to show the general process; the figures used were approximate and I didn't check for condensate, so you'll probably want to find more reliable reference values and get the numbers run more carefully if you're actually building something.

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  • $\begingroup$ ahh cool +1 for teaching me about 'heat capacity'... I couldn't think of what else to type in to WolframAlpha. $\endgroup$ May 11, 2018 at 17:20
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This is no answer, but fun to try to get WolframAlpha to answer... this is the best I could get it to do:

$35000\, \mathrm{m}^3$ volume of moist air at 45 celsius contains $1154\, \mathrm{kg}$ of water (WolframAlpha).

Edit Nat's answer is probably much better than mine... I'm stealing his heat capacity reference, and continuing along my line of thought.

for "heat capacity of air at 50% humidity", wolframalpha tells you humidity ratio:

$$0.00739 \frac{\mathrm{kg}_\text{water}}{\mathrm{kg}_\text{air}}$$

We want to lower it by $10\, \sideset{^\circ}{}{\mathrm{C}}$.

Hooking it all together: $$ 10 ~ {\times} ~ 1\, \mathrm{calorie} ~ {\times} ~ 1000 ~ {\times} ~ \frac{1154}{0.00739} \mathrm{per~hour}~=~1815\,\mathrm{kW} $$

$$ \frac {45 °\mathrm{C} - 35°\mathrm{C}}{\mathrm{hour}}~=~\frac {10 °\mathrm{C}~\text{temperature delta}}{\mathrm{hour}} $$

$$ 35000 \mathrm{m}^3_\text{moist air} ~~ \thickapprox ~~\frac{1154\,\mathrm{kg}_\text{water}} {\frac {0.00739 \text { of water}} {\mathrm{kg}_\text{air}}} \thickapprox 156,000\,\mathrm{kg}_\text{moist air } $$

$$ 1 \,\mathrm{gram}~{\times}~1\, °\mathrm{C}~~=~~1\, \mathrm{calorie} $$

$$ 156,000 \mathrm{kg}_\text{moist air} * \frac {10 °\mathrm{C} \text { temperature delta} }{\mathrm{hour}} = \frac {1,560,000\,\mathrm{KiloCalories}} {\mathrm{hour}} $$

$$ 1 \frac {\mathrm{KiloCalorie}} {\mathrm{hour}}~~=~~1.16\,\mathrm{W} $$

$$ \frac {1,560,000\,\mathrm{KiloCalories}} {\mathrm{hour}} ~ {\times} ~ \frac {1.16\, \mathrm{W}} {\frac {\mathrm{KiloCalorie}} {\mathrm{hour}}}~~=~~1,815,000\,\mathrm{W}~~=~~1815\, \mathrm{kW}\,. $$

Hmm... Nat's answer was ~400-500 kilowatts. Not sure why my estimate is so much higher. Why does WolframAlpha get confused when you add too many units?

So converting to $$ \frac {kiloWatts} {\frac{m^3} {hour}}$$, I get $$ \frac {0.0518kiloWatts} {\frac{m^3} {hour}}$$

But a real world answer would require something like this.

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  • $\begingroup$ It looks like most of your logic's good, though two mistakes. The first is that the specific heat capacity of air, $0.00739$, should be multiplied rather than divided. (The higher the specific heat capacity, the more energy it should take to heat/cool something.) The second is that you want the mass of all of the air, not just the water in it. I'd guess that that may've been confusing since the specific heat capacity was given in units of per-kg-of-water, though that unit's already canceled out by the use of the 1-calorie tactic, which is the heat capacity for water. $\endgroup$
    – Nat
    May 12, 2018 at 11:39
  • $\begingroup$ Beyond that, it looks like we largely used the same approach, with the big difference being that you used the specific heat capacity, $\frac{C_\text{p}^\text{air}}{C_\text{p}^\text{water}}$ to enable the use of the definition of $1\,\mathrm{calorie}$ being the amount of energy needed to raise $1\,\mathrm{kg}$ of water by $1\,\mathrm{K}$, i.e. ${C_\text{p}^\text{water}=\frac{1\,\mathrm{cal}}{\mathrm{kg}{\times}\mathrm{K}}}$. $\endgroup$
    – Nat
    May 12, 2018 at 11:47
  • $\begingroup$ Oh, and the question was asking about $130,000\,\mathrm{m}^3$ of moist air rather than $35,000\,\mathrm{m}^3$, so that may've been part of the issue. $\endgroup$
    – Nat
    May 12, 2018 at 11:54
  • $\begingroup$ Ok, my method is a little convoluted, but I've edited my answer with dimensional analysis to show why i divided by 0.00739 $\endgroup$ May 14, 2018 at 20:37
  • $\begingroup$ I think the OP must have edited the values 130,000 vs. 35,000... now hist post has been edited, and doesn't specify. $\endgroup$ May 14, 2018 at 20:40

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