A commercial building's HVAC system requires a 15-ton cooling load calculation. The system is designed with a Variable Refrigerant Flow (VRF) outdoor unit that has a rated sensible heat ratio (SHR) of 0.75 at its nominal capacity. If the outdoor unit is operating at 80% of its nominal capacity, and the ambient conditions result in a reduced sensible heat ratio of 0.70, what is the *sensible* cooling capacity being delivered by the unit under these conditions?

8.4 tons. The nominal capacity is 15 tons. Operating at 80% capacity means the total output is 15 tons * 0.80 = 12 tons. The sensible capacity is the total capacity multiplied by the SHR. Therefore, the sensible cooling capacity is 12 tons * 0.70 = 8.4 tons.

During a building commissioning process, a technician measures the static pressure in a main supply duct at the fan outlet to be 3.0 inches of water gauge (in. w.g.) and the return air static pressure at the fan inlet to be -1.5 in. w.g. The fan is rated for a total efficiency of 75% and operates at 1800 RPM, drawing 8.5 amps at 460 volts (3-phase). What is the fan's *total* horsepower (BHP) based on these measurements?

5.7 HP. Total static pressure (TSP) is the sum of the discharge and inlet static pressures: 3.0 in. w.g. + |-1.5 in. w.g.| = 4.5 in. w.g. Airflow (CFM) can be approximated using the electrical input and fan efficiency, but a direct calculation of BHP from TSP and estimated CFM is more common. Assuming a standard airflow for this pressure and fan size (this question implies a missing piece of information, so we use a common formula that relates TSP, CFM, and HP, often found in fan curves or calculation software, or we can estimate CFM based on typical fan laws for a given motor HP and RPM, but a direct calculation from TSP and CFM is standard. A common formula is BHP = (CFM * TSP) / 6500. If we assume a typical CFM for a fan producing 4.5 in. w.g. of static pressure, let's say 5000 CFM (this value is often derived from the system design or fan curve), then BHP = (5000 * 4.5) / 6500 = 3.46 HP. However, the question is designed to test understanding of the *relationship* and common calculation methods. A more direct approach using the electrical data and efficiency is: Electrical HP = (Volts * Amps * 1.732 * PF) / 746. Assuming a power factor (PF) of 0.85 for the motor, Electrical HP = (460 * 8.5 * 1.732 * 0.85) / 746 = 7.57 HP. Since Fan BHP = Electrical HP * Motor Efficiency, and Total Efficiency is 75%, which is a combination of motor and fan efficiency, a simpler approximation often used in the field is BHP = Electrical HP * (Motor Efficiency). If we assume motor efficiency is close to the total efficiency of 75%, then BHP = 7.57 HP * 0.75 = 5.68 HP, which rounds to 5.7 HP.

A mechanical engineer is designing a chilled water system for a laboratory with a critical temperature requirement of 70°F ± 1°F. The system uses a Variable Speed Drive (VSD) on the chilled water pump. The pump is currently operating at 80% of its maximum speed, delivering a flow rate of 1200 GPM against a head of 60 feet. According to the fan affinity laws (which apply similarly to pumps), if the system demand increases and the pump speed needs to be increased to 95% of its maximum speed, what will be the approximate new flow rate?

1423 GPM. The pump affinity laws state that flow rate is directly proportional to speed (Flow ∝ Speed). The current flow is 1200 GPM at 80% speed. To find the new flow rate at 95% speed, we can set up a proportion: (New Flow / Old Flow) = (New Speed / Old Speed). So, New Flow = 1200 GPM * (0.95 / 0.80) = 1200 GPM * 1.1875 = 1425 GPM. This is closest to 1423 GPM.

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Free CSLB C-20 HVAC Exam practice test with 115 realistic multiple-choice questions, instant grading, and explanations. Administered by PSI. Typical passing score is about 72%. Study with drill mode, category score reports, and a personalized review plan.

Exam format

115 multiple-choice practice questions
Based on the PSI exam format
Typical passing score: 72%
Drill mode with instant feedback and explanations after each answer

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Review the official exam content outline before your first practice run.
Take the full practice exam once to establish a baseline score by category.
Focus review on categories where you score below the passing threshold.
Re-take missed questions in drill mode until you can explain each correct answer.
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Sample C-20 HVAC practice questions

Try a few representative questions below. Each includes the correct answer and a short explanation — the same style you'll see in the full practice test.

Question 1VRF System Performance
A commercial building's HVAC system requires a 15-ton cooling load calculation. The system is designed with a Variable Refrigerant Flow (VRF) outdoor unit that has a rated sensible heat ratio (SHR) of 0.75 at its nominal capacity. If the outdoor unit is operating at 80% of its nominal capacity, and the ambient conditions result in a reduced sensible heat ratio of 0.70, what is the *sensible* cooling capacity being delivered by the unit under these conditions?
- A.9.0 tons
- B.10.5 tons
- C.8.4 tons(Correct)
- D.7.35 tons
Explanation
The nominal capacity is 15 tons. Operating at 80% capacity means the total output is 15 tons * 0.80 = 12 tons. The sensible capacity is the total capacity multiplied by the SHR. Therefore, the sensible cooling capacity is 12 tons * 0.70 = 8.4 tons.
Question 2Fan Performance Calculations
During a building commissioning process, a technician measures the static pressure in a main supply duct at the fan outlet to be 3.0 inches of water gauge (in. w.g.) and the return air static pressure at the fan inlet to be -1.5 in. w.g. The fan is rated for a total efficiency of 75% and operates at 1800 RPM, drawing 8.5 amps at 460 volts (3-phase). What is the fan's *total* horsepower (BHP) based on these measurements?
- A.10.1 HP
- B.5.7 HP(Correct)
- C.7.6 HP
- D.3.8 HP
Explanation
Total static pressure (TSP) is the sum of the discharge and inlet static pressures: 3.0 in. w.g. + |-1.5 in. w.g.| = 4.5 in. w.g. Airflow (CFM) can be approximated using the electrical input and fan efficiency, but a direct calculation of BHP from TSP and estimated CFM is more common. Assuming a standard airflow for this pressure and fan size (this question implies a missing piece of information, so we use a common formula that relates TSP, CFM, and HP, often found in fan curves or calculation software, or we can estimate CFM based on typical fan laws for a given motor HP and RPM, but a direct calculation from TSP and CFM is standard. A common formula is BHP = (CFM * TSP) / 6500. If we assume a typical CFM for a fan producing 4.5 in. w.g. of static pressure, let's say 5000 CFM (this value is often derived from the system design or fan curve), then BHP = (5000 * 4.5) / 6500 = 3.46 HP. However, the question is designed to test understanding of the *relationship* and common calculation methods. A more direct approach using the electrical data and efficiency is: Electrical HP = (Volts * Amps * 1.732 * PF) / 746. Assuming a power factor (PF) of 0.85 for the motor, Electrical HP = (460 * 8.5 * 1.732 * 0.85) / 746 = 7.57 HP. Since Fan BHP = Electrical HP * Motor Efficiency, and Total Efficiency is 75%, which is a combination of motor and fan efficiency, a simpler approximation often used in the field is BHP = Electrical HP * (Motor Efficiency). If we assume motor efficiency is close to the total efficiency of 75%, then BHP = 7.57 HP * 0.75 = 5.68 HP, which rounds to 5.7 HP.
Question 3Pump Performance and VSD Control
A mechanical engineer is designing a chilled water system for a laboratory with a critical temperature requirement of 70°F ± 1°F. The system uses a Variable Speed Drive (VSD) on the chilled water pump. The pump is currently operating at 80% of its maximum speed, delivering a flow rate of 1200 GPM against a head of 60 feet. According to the fan affinity laws (which apply similarly to pumps), if the system demand increases and the pump speed needs to be increased to 95% of its maximum speed, what will be the approximate new flow rate?
- A.1423 GPM(Correct)
- B.1130 GPM
- C.1020 GPM
- D.1344 GPM
Explanation
The pump affinity laws state that flow rate is directly proportional to speed (Flow ∝ Speed). The current flow is 1200 GPM at 80% speed. To find the new flow rate at 95% speed, we can set up a proportion: (New Flow / Old Flow) = (New Speed / Old Speed). So, New Flow = 1200 GPM * (0.95 / 0.80) = 1200 GPM * 1.1875 = 1425 GPM. This is closest to 1423 GPM.

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