Recent content by Achievementhunt

Thanks for your help, appreciate it

I can redefine COP as COP= heat to be extracted (Qc)/ Electrical work to operate compressor. We have found the additional heat under faulty conditions, therefor the additional work is Additional heat/COP. This makes sense to me, is this the correct approach?

So an additional 20J/s needs to be extracted from the fridge to keep it cool, reducing the COP?

When the door is closed, the fridge is a closed system, light cannot escape so the entire electrical power consumed by the bulb eventually becomes the additional thermal energy of the fridge, 20 J/s

Thanks for the reply but I am unsure how to calculate the part thermal energy given to the fridge from the light of the bulb

1. Homework Statement Coefficient of performance of a refrigerator is 1.4 A faulty 20w bulb remains on when the fridge is closed. Under normal circumstances, the bulb turns on when the fridge is open. The door is opened 20 times a day for 30 seconds. Calculate the increase in energy...

Thanks for the hint Chet. I will give Pi theorem a try

Homework Statement This is an optional question given to Fluid mechanic students to work on for leisure. P = power ρ = fluid density, rho ω= angular velocity omega μ= dynamic viscosity, mu D= diameter Homework Equations Show the that the power required to rotate the disc is given by...