Thermodynamics question about mass flow rate in an air conditioning unit?

[aero&astro]

1. In an air conditioning unit fresh air at 10°C is mixed with warm air at 30°C. If the mass flowrate ratio of warm to fresh air is 1.6, estimate the temperature at which the mixed air leaves the air conditioner. State any assumptions made.


and also do you always make the air standard assumptions for ALL heat engine and heat pump problems?


my attempt was mdot/mdot = flowrate ratio = change in temperature but then I don't think that makes sense, any help?

 

[kanye20]

Such a correlation would aid in system design and allow the examination of a full range of refrigerant choices.


air conditioner

 

[HyperSniper]

One way to do it would be to may be calculate the energy in and energy out based off the specific heat of the air.

 

[rude man]

Assume constant specific heat vs. temperature and it's a high school-level problem.

 

[DeeNos]

I would approach this question by first considering the fundamental principles of thermodynamics. The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this case, the air conditioning unit is transferring heat energy from the warm air to the fresh air in order to maintain a desired temperature.

To answer the question, we need to use the equation for mass flow rate, which is mass flow rate = density x velocity x cross-sectional area. Assuming the cross-sectional area and velocity remain constant, the mass flow rate is directly proportional to the density of the air.

Given the mass flowrate ratio of 1.6, we can assume that for every 1.6 units of warm air, there is 1 unit of fresh air. This means that the density of the fresh air is lower than the density of the warm air. As the warm and fresh air mix, their densities will also mix, resulting in a final density that is between the two initial densities.

Now, we can use the equation for the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, the work done by the system is negligible, so we can ignore it. The equation can be written as follows:

ΔU = Q - W

Where ΔU is the change in internal energy, Q is the heat added, and W is the work done by the system.

Assuming that the air conditioning unit is operating at steady state, the change in internal energy is zero. This means that the heat added to the system is equal to the heat removed from the warm air. We can express this as follows:

Q = m(warm) x Cp(warm) x (Tf - Tw)

Where m(warm) is the mass of the warm air, Cp(warm) is the specific heat capacity of the warm air, Tf is the final temperature of the mixed air, and Tw is the initial temperature of the warm air.

Similarly, we can express the heat removed from the warm air as follows:

Q = m(fresh) x Cp(fresh) x (Tf - Tf)

Where m(fresh) is the mass of the fresh air, Cp(fresh) is the specific heat capacity of the fresh air, and Tf is the final temperature of the