# Relative humidity problem: definition of specific volume in terms of kg dry air

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:

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!