I find it crazy walking into a major retail market and their doors are open, the outside air is 85°f and they've managed to cool the inside of the building to 70°f continously. The energy use is terrible. Good luck
Sometimes they use an air curtain to help that. Otherwise it is a waste of energy. Some places so it so you walk by and feel the cool air, and since the door is already open they are more likely to walk in.
Ty when looking btu/Lb of air on psychrometric charts , say it’s 40 btu/Lb of dry air. How is that different than the definition of 1 btu raises one pound of water one degree?? I’m trying to understand the difference here. Thanks
That's a great question. I am in the works for starting a series on air and psychrometrics. A BTU is just a measurement of heat energy and quantity. The definition of BTU gives us a starting point. The same energy I put into 1 pound of water I can put into anything. A furnace will produce ___btu of heat. We are not heating water but we are getting air, and we can heat many things. To cool we remove BTU of heat energy from many things. We can add BTU to air to heat it and remove BTU from air to cool it. Because air has different molecular make up it will change at a different rate than water. This leads to specific heat. The amount of hest energy (BTU) to change 1 pound of A SUBSTANCE one degree F. If we know the specific heat of meat I know how many BTUs I need to remove to cool it or freeze it. Same for ice cream, glycol, blood, wood, soda, beer, and everything else. The typical specific heat for air is .24BTU So if I apply 1 BTU of heat energy to 1 pound of air it will raise the temperature just over 4F from where it was at originally. 1btu if energy divided by .24 specific heat of air = 4.166F temperature change. As you add BTU to air it warm and as you remove BTU it cools
Isn't BTU an heat energy unit? If so shouldn't it be mass times cp of the fluid times temperature difference, being cp the mass thermal capacity of some fluid? In this case water
You started well, BUT u should have measured the weight of gas (or electricty units, if an eletric stove was used), that was used up to heat this 1 pound of water from 38 degrees to 138 degrees F, & then also measured the weight of gas (or units of electricity) used up (for the 100 degrees F rise in the warer temperature), to know how much heat was required in terms of gas cost (or units of electricity cost) to add the 100 BTU of heat energy into that 1 pound of water. So, your experiment was incomplete without measuring the actual approx. cost incurred in terms of money (gas cost or eletric units cost) that U had to spend, to pump 100 BTU heat energy into that water.
You are correct. But it's not a specific experiment, it's just to give value and the basic idea. I would also have to be at sea level ant the heating element would have to be completely submerged into the water, and the pan insulated to not have losses to the air and I would have had to do it over an houers time.
@@love2hvac - U r right, its hard to calculate accurately the electrical units spent to make a rise or fall of 100 BTU in a pound of water, ... but I think, to relate it to the present electrical units costs in refrigeration projects, we can say roughly that, 6000 BTU is equal to about 0.5 ton AC compressor (500 watts), or 12000 BTU is equal to 1 ton (1KV) of an aircondition unit power, .. so that way, we can roughly relate the electric cost to the BTU no. Thanks.
