I'm working on a thermodynamics problem which is as follows:

"Saturated humid air at 200 kPa and 15C is heated to 30C as it flows through a 4-cm diameter pipe with a velocity of 20 m/s. Disregarding the pressure losses, calculate the relative humidity at the pipe outlet and the rate of heat transfer, in kW, to the air."

Here's a diagram of the setup:

enter image description here

Specifically, I'm stuck on the part of the problem where I calculate the rate of heat transfer to the air. Without going into too much detail, one of the things I need in order to calculate the rate of heat transfer is the mass flow rate of dry air, $\dot m_a$, in kg dry air / s.

I took a look at the solution guide to get some help on this. Here's what it said to do:

(1) Solve for $\dot V_1$, the volumetric flow rate, using the velocity $v_1$ and the diameter of the pipe $d$:

$\dot V_1 = v_1 * \frac{\pi d^2}{4} = (20 m/s) * \frac{\pi (0.04 m)^2}{4} = 0.02513 m^3/s$

(2) Find $\nu_1$, the specific volume of the dry air, by assuming ideal gas behavior:

$P_{a1}\nu_{1} = R_aT_1,$

where $P_{a1}$ is the partial pressure of the dry air at the inlet, $R_a$ is the gas constant for dry air, and $T_1$ is the temperature at the inlet. Again, without going into too much detail, it's possible to solve for $P_{a1}$ and get $\nu_{a1} = 0.4168$ m^3 / kg dry air.

(3) Divide $\dot V_1$ by $\nu_1$ to get $\dot m_a$:

$\dot m_a = \frac{\dot V_1}{\nu_1} = \frac{0.02513 m^3/s}{0.4168 m^3 / kg dry air} = $ 0.06029 kg dry air / s.

Here's why I'm confused: as far as I can tell, $\dot V_1$ is the volumetric flow rate of the mixture of air and water, i.e., $m^3$ mixture / $s$. By contrast, $\nu_1$ seems to be the specific volume of the dry air, i.e., $m^3$ dry air / $kg$ dry air. Thus, the "$m^3$" terms in $\frac{0.02513 m^3/s}{0.4168 m^3 / kg dry air}$ should not cancel out!

In my opinion, the solution guide should have solved for $\dot V_{a1}$, the volumetric flow rate of dry air only, and then used this to solve for $\dot m_a$. This would have been relatively easy to do, since, for ideal gases, $\frac{P_{partial}}{P_{total}} = \frac{V_{partial}}{V_{total}}$, and we know the partial pressure $P_{a1}$ at the inlet.

Using this information, I solved the problem my way and got $\dot m_a$ = 0.05978 kg dry air / s.

I know the difference between my answer and the solution guide is small, but I wanted to confirm that my rationale is correct. I assume that the solution guide decided that the dry air made up such a large portion of the flow stream by volume, that they could simply assume $\dot V_{a1}$ = $\dot V_1$.

Please let me know if I'm missing something. Thank you!


1 Answer 1


I can't figure out what you did, but the vapor pressure of water at 15 C is 1.71 kPa. So the partial pressure of the dry part of the air is 198.29 kPa at the inlet. That means that the mole fraction of the water vapor in the inlet flow is 0.0087 and the mole fraction of dry air is 0.9913. So the molecular weight of the moist air is (0.9913)(29)+(0.0087)(18)=28.9. So, does it really matter?


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