Engineering, 12.03.2020 05:39 emthebest123
A thermocouple is made by welding two dissimilar metals into a small spherical bead. The potential difference between the two legs of the termocouple is a measure of the average temperature in the bead. Suppose that you have a chromel-alumel thermocouple bead, with the following material properties: \[Rho]=2700 kg m^-3, k=100 W/(m K), Subscript[C, p]=400 J/(kg K). The thermocouple bead has a radius of 1 mm. It is to be plunged suddenly from room temperature into boiling water (100 \[Degree]C), and the heat transfer coefficient is 10^4 W/(m^2 K). Determine the response time of the thermocouple, by calculating how long it takes for the thermocouple bead temperature to be within 0.1% of 100 \[Degree]C, i. e. how long does it take for the thermocouple temperature to rise to a temperature of 99.9 \[Degree]C? Hint: This is a small bead so we might want to test to see if we can simplify the handling of this problem as we could do for a thin sheet of Aluminum, i. e. check to see if Newtonian cooling applies.
Answers: 2
Engineering, 04.07.2019 19:10, gabrielaperezcz
Air inially occupying a volume of 1 m2 at 100 kpa, 27 c undergoes three internally reversible processes in series. process 1-2 compression to 500 kpa during which pv constant process 2-3 adiabatic expanslon to 100 kpa process 3-1: constant-pressure expansion to 100 kpa (a) calculate the change of entropy for each of the three processes. (b) calculate the heat and work involved in each process. (c) is this cycle a power cycle or refrigeration cycle?
Answers: 3
Engineering, 04.07.2019 19:20, Katmcfee7681
Acommercial grade cubical freezer, 4 m on a side, has a composite wall consisting of an exterior sheet of 5.0-mm thick plain carbon steel (kst= 60.5 w/m k), an intermediate layer of 100-mm thick polyurethane insulation (kins 0.02 w/m k), and an inner sheet of 5.0- mm thick aluminium alloy (kal polyurethane insulation and both metallic sheets are each characterized by a thermal contact resistance of r 2.5 x 104 m2 k/w. (a) what is the steady-state cooling load that must be maintained by the refrigerator under conditions for which the outer and inner surface temperatures are 25°c and -5°c, respectively? (b) for power saving purpose, which wall material should be increased/reduced in. thickness in order to reduce 50% of the cooling load found in part (a)? redesign the thickness of the proposed material. 177 w/m-k). adhesive interfaces between the q=575.93 w
Answers: 2
A thermocouple is made by welding two dissimilar metals into a small spherical bead. The potential d...
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