Question 4 A thermal wheel is used to recover energy from the extract air leaving a building at 5 kg/s mass flow rate an

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899604
Joined: Mon Aug 02, 2021 8:13 am

Question 4 A thermal wheel is used to recover energy from the extract air leaving a building at 5 kg/s mass flow rate an

Post by answerhappygod »

Question 4 A Thermal Wheel Is Used To Recover Energy From The Extract Air Leaving A Building At 5 Kg S Mass Flow Rate An 1
Question 4 A Thermal Wheel Is Used To Recover Energy From The Extract Air Leaving A Building At 5 Kg S Mass Flow Rate An 1 (179.99 KiB) Viewed 34 times
Question 4 A thermal wheel is used to recover energy from the extract air leaving a building at 5 kg/s mass flow rate and 23 °C dry bulb temperature. The heat is used to preheat the fresh air supply coming in at 5.5 kg/s and temperature of 12 °C. The thermal wheel speed is 20 rev/ min. (a) Draw a schematic of a typical thermal wheel with the temperature distribution of the air across the wheel - assume counter flow. (20%) (b) Using the data below calculate; i. The effectiveness of the matrix wheel. (40%) ii. The rate of actual heat recovery. (10%) iii. The temperatures of the air leaving the wheel. (10%) c) Calculate the temperature of the air leaving the wheel if the speed is increased to 25 revs/ min. (20%) Data Assume mass of the thermal wheel is 25 kg Specific heat capacity of Air, Cp air = 1.005 kJ/kg-K Specific heat capacity of Matrix, Cp matrix = 0.4 kJ/kg-K Heat transfer coefficient for extract air stream = 20 W/m2K Heat transfer coefficient for supply air stream = 25 W/m2K Face area for the wheel = 1.2 m2 Width of the wheel = 0.2 m Assume the matrix has a surface area to volume ratio of 3200 m2/m3 1 UA + NTU = hiAi hodo Assume that the effectiveness of a recuperative heat exchanger & is: 1 1 = UA (mcp) min eft 1.93 1 E = Ep 1 [(mcy) matrix/(mcp) min) Where &c is the equivalent effectiveness for a counter flow heat exchanger. 1 - exp[-NTU(1–R)] Ec 1-Rx exp[-NTU(1-R)] R is the ratio of thermal capacities and NTU = Number of Transfer Unit (mcp) min R = (mcp) max Total marks for Question 4 = 100%
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply