NCEES HVAC Practice Problem Help

I have been going through the official NCEES practice exam for HVAC. I understand the main concepts and am combing through each problem to make sure I fully understand each step. I am stumped on one portion of the following problem though:

A room in a building has the following characteristics:

Sensible heat gain = 90,000 Btu/hr
Latent heat gain = 40,000 Btu/hr
Supply air to the room = 3,600 cfm @ 55ºF db

The room is kept at 78ºF db/ 45%rh. Outdoor air conditions are 92ºF db/ 76ºF wb (h=39.4 BTU/lb). Outdoor air ventilation for the room is 700 cfm. The end goal of the problem is to find the supply wet-bulb temperature if the supply dry bulb is 55ºF. Our mixed air condition would be 80.7ºF db and 66.2ºF wb (h=30.99 BTU/lb) with the above information.

A portion of problem requires us to calculate the load of the OA entering the room. The solution manual has the load defined as a function of the mixed air temperature and OA. In other words, the test has q_t_OA = 4.5 x Q_OA x delta_h where delta_h is h_OA and h_MixedAir. I am lost as to why the system load by the OA would be impacted by the mixed condition. If we would push the example to the extreme and bring in 3,599 CFM of OA and only recirculate 1 CFM of room air, the q_t_OA load would be smaller than if we brought in less OA (assuming the enthalpy difference is between the OA and mixed air conditions as shown in the NCEES solution. Under these conditions delta_h moves closer to zero as more OA is added). Shouldn't the OA load use the enthalpy differences between the OA and room condition instead? This would mean that as more OA is added the cooling capacity on the system would also increase.

I'm not following your logic here (maybe a diagram would help clarify). If the supply air is cooler because room air is being recirculated, that reduces the load on the heat exchanger, to maintain the 78 degree room temperature.

Perhaps my wording wasn’t the best. Basically what I’m asking is, how do you calculate the OA load? I believe the load is based on the enthalpy difference of OA and RA (q_t_OA = 4.5 x Q_s_OA x (h_OA - h_RA)). The NCEES solutions manual, which I believe is incorrect, says the load is based on the enthalpy difference of OA and mixed air.

If the rate of the OA is known (700cfm from what you posted), you can use equation 43.28(a) in Merm edition 13 gives qtot= q sensible,room + q latent, room + (4.5)(h outside - h inside)Vdot outside

The NCEES equation is unfamiliar to me. Does it give the same answer as the formula above?

But I still want to address something you said earlier, because it's critical to understanding the problem. As the OA flow increases, the load on the heat exchanger INCREASES, thus the cooling capacity DECREASES.

Unintended Max P.E. said:

Ok, I think I understand what you're getting at here. So, the best case scenario for cooling the room is to recirc 100% room air at the same flow rate, but sometimes it's desireable to add some fresh air to the loop (for health reasons, typically). Adding in outside air adds a load to it but some of that effect is offset by adding additional airflow. We don't know what the airflow would be, necessarily, if no OA was supplied (unless that was given and you just didn't post it). So your assumption of the same supply flow is incorrect if no OA is coming in. So to figure out the load of the OA you have to do the delta from the mixed condition. Make sense?

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No, it does not make sense. We do know what the airflow is. The airflow is 700 cfm of OA. The SA is 3,600 cfm as stated. The assumption that the supply flow doesn't remain the same is irrelevant (we could have a constant volume system that always provides 3,600 cfm air). In my original post, I tried to explain a scenario. I'll try to make it clearer but keep in mind, all I am looking for, is how to calculate the q_t_OA load. We know the q_t required to cool the room. Any additional air (OA) brought into the system will need to be cooled before it enters the space. I believe the air needs to be cooled to room conditions. The NCEES solutions manual says for this problem that the OA load is based of mixed air conditions. I will again try to explain why I believe this is incorrect.

OA Load Scenario 1 with low OA intake - This isn't the actual problem statement. I am using this to illustrate a point.
We have a constant volume system that brings in 3,600 cfm air. Let's say we bring in 500 cfm of OA under 92ºF db/ 76ºF wb (h=39.4). This would mean 3,100 cfm of room condition air at 78ºF db/ 45%rh (h=28.8) is recirculated. Under these conditions, we can calculate the mixed air condition to be 79.9ºF db/65.3ºF wb (h=30.1). To calculate the q_t we use the standard q_t equation q_t = 4.5 x Q_s x delta_h.

If we calculate the load using OA and room conditions, the OA load would be:
q_t_OA=4.5*500cfm*(39.4-28.8)=23,850 BTU/hr
If we calculate the load using OA and mixed air conditions as outlined in the NCEES solutions manual, the OA load would be:
q_t_OA=4.5*500cfm*(39.4-30.1)=20,925 BTU/hr

OA Load Scenario 2 with high OA intake:
Let's say we bring in 3,000 cfm of OA under 92ºF db/ 76ºF wb (h=39.4). This would mean 600 cfm of room condition air at 78ºF db/ 45%rh (h=28.8) is recirculated. Under these conditions, we can calculate the mixed air condition to be 89.6ºF db/74ºF wb (h=37.4).

Again, If we calculate the load using OA and room conditions, the OA load would be:
q_t_OA=4.5*3000cfm*(39.4-28.8)=143,100 BTU/hr
If we calculate the load using OA and mixed air conditions as outlined in the NCEES solutions manual, the OA load would be:
q_t_OA=4.5*3000cfm*(39.4-37.4)= 27,000 BTU/hr

In the above scenario, we increase the amount of OA by 600%. This should also increase the cooling required by the system to cool the OA by 600%. If we follow the NCEES solution, we would only increase the cooling by 30%. I am not going to type up all the math here but if the cooling is not enough to cool the OA, the room conditions will increase in temperature and eventually lead to issues. If I am incorrect, please be specific about where I am making a mistake so I can understand.

Sorry, I realized my error and made a significant edit (I really read your initial problem wrong, please forgive). Please see the edit to my post.

Unintended Max P.E. said:

If the rate of the OA is known (700cfm from what you posted), you can use equation 43.28(a) in Merm edition 13 gives qtot= q sensible,room + q latent, room + (4.5)(h outside - h inside)Vdot outside

The NCEES equation is unfamiliar to me. Does it give the same answer as the formula above?

But I still want to address something you said earlier, because it's critical to understanding the problem. As the OA flow increases, the load on the heat exchanger INCREASES, thus the cooling capacity DECREASES.

Click to expand...

Thank you. The MERM equation is correct I was just looking for verification. As I’ve outlined, the NCEES answer gives a different answer which I believe is incorrect.

As for your last point, this is not a heat exchanger problem. I’ve given all the information needed to solve the problem. The problem implies that we’re dealing with an AHU cooling coil. As part of the problem, we determine the cooling capacity of the coil (determined in the MERM equation). As OA increases, more heat is brought into the system. The coil must reject this heat. It is not relevant to the problem what the cooling coil fluid does.

SpudTheDog said:

Thank you. The MERM equation is correct I was just looking for verification. As I’ve outlined, the NCEES answer gives a different answer which I believe is incorrect.

As for your last point, this is not a heat exchanger problem. I’ve given all the information needed to solve the problem. The problem implies that we’re dealing with an AHU cooling coil. As part of the problem, we determine the cooling capacity of the coil (determined in the MERM equation). As OA increases, more heat is brought into the system. The coil must reject this heat. It is not relevant to the problem what the cooling coil fluid does.

Click to expand...

Your assessment is correct. The equation in the NCEES solution is wrong. OA load is calculated with the enthalpy difference between outside and return air.

Regarding a different question, I'm looking at #25 in the practice exam. Although the the numbers are plugged into the correct equation, I believe the equation shown on the 1st and 2nd lines switch the CFM1 and CFM2.

Regarding #38, if water pump inlet pressure is 20 psig, and discharges 160 ft above to atmosphere with 20 ft of friction losses. Isn't the pressure required after the pump 160 ft + 20 ft = 180 ft / 2.307 = 78 psig? I'm not sure why the solution adds 20 psig to that...

Personal opinion, #39 seems to be written poorly.

sheela34 said:

Regarding #38, if water pump inlet pressure is 20 psig, and discharges 160 ft above to atmosphere with 20 ft of friction losses. Isn't the pressure required after the pump 160 ft + 20 ft = 180 ft / 2.307 = 78 psig? I'm not sure why the solution adds 20 psig to that...

Click to expand...

The pressure rating of the piping system will have to be greater than the highest pressure in the system. The greatest pressure in this system is at the pump discharge. The pressure at the pump discharge is 20psi + Δp_pump, where Δp_pump is the pressure rise across the pump. With the given information Δp_pump=78psi hence the pressure at the pump discharge is 98 psig.

The pressure rating of the piping system will have to be greater than the highest pressure in the system. The greatest pressure in this system is at the pump discharge. The pressure at the pump discharge is 20psi + Δp_pump, where Δp_pump is the pressure rise across the pump. With the given information Δp_pump=78psi hence the pressure at the pump discharge is 98 psig.

I agree with everything you say, except for the numbers. 78 psi is the pump discharge pressure and with a pump delta of 58 psi.

Regarding #52, I don't understand why they included heat pump power in the answer when the question is just asking for the value of the supplemental electric heat.

Regarding #54, this one I did not comprehend at all what the situation is that is presented in the problem. oh well

Regarding #78, technically all the answer choices would meet the requirement of having a deflection of less than 0.5 in., but I guess they want the answer closet to the cutoff spring constant. Though the question doesn't make it absolutely clear.