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Consider the cooling requirements for a cooling system that cools a classroom. Assume that there is no external supply or return of air to the room and that it is occupied by 100 students, each who has heat transfer at a rate of 100 W (360 kJ/h). You can also assume typical lighting of 15 W per m2 of floor area and floor dimensions 30 m by 15 m. The classroom is an internal space that is not connected to outside walls. Thus, only internal gains need to be considered. Assuming that the temperature is held constant, determine the rate of cooling required (kW). Be sure to show an appropriate system with all energy flows, list assumptions, and start with basic equations.

Respuesta :

Answer:

16.75 KW

Explanation:

Given:

- The rate of heat rejected to the air by each student Q_in = 100 W / student

- The amount of electrical wok done due to lighting W_in = 15 W / m^2

- The area of floor A = 30 m x 15 m

- Number of students in room N = 100

Find:

Assuming that the temperature is held constant, determine the rate of cooling required (kW)

Solution:

- In this case we will consider the air in room to be a move-able system. The interaction of air with the surroundings i.e students and the lighting in the room can be modeled by an energy balance:

                                    E_in - E_out = Δ E_sys

- The above balance can be re-written for steady state system as:

                                    Q_net,in - W_net,out = flow(m) * ( u_2 - u_1)

- We know that the temperature of the room remains constant. Hence, the change in internal energy Δ U = 0. Hence the heat balance becomes:

                                    (Q_in - Q_L ) - (- W_in) = 0

- Where Q_L is the rate of cooling, now rearrange the above expression:

                                     Q_L = W_in + Q_in

- Plug in the values:

                                     Q_L = 100*100 W + 15*(30*15) W

                                     Q_L = 16,750 W = 16.75 KW

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