Recent content by timmayy

ahh ok, that's great thanks :) Tim

Hi, whether it was counter or co current wasnt given in the question, we also don't have inlet or outlet temperatures to work it out with (im not entirely sure what the temperatures given are), all we could do was work out cross sectional areas and we tried taking log mean values for k...

Hi, I've recently been given a series of questions on heat transfer to do and have done most of them with general ease, but this one question I've been stuck on for ages and i can't seem to figure out: "A double pipe heat exchanger is made up from a length of 25mm i.d. steel pipe of 2.5mm...

Ah ok, thanks a lot :)

i also tried changing the T1 and T2 in the conduction equation: Qconduction = (0.65 x (T1 - 425)) / 0.05 = 13T1 - 5525 and so the final answer will be 439.5 celsius - which seems to high to me

my working is as follows: Qconduction - Qconvection - Qradiation = 0 Qconduction = (0.65 x (425 - T2)) / 0.05 = 5525 - 13T2 (the 425 being 152 + 273 to convert into kelvin) Qconvection = 20 x (152 - 22) = 2600 Qradiation = 0.8 x 5.67x10-8 x (4254 - 2954) = 1136.36 so, (5525 - 13T2) - 2600...

Hi, I've been working through this same question and found an answer of 439 centigrade - which seems far too high for me, i assumed the back wall would be a lower temperature than the front wall, and so must be less than 152 (the front wall) and larger than 0 obviously