The elemental unit of an air heater consists of a long circular rod of diameter D, which is encapsulated by a finned sleeve and in which thermal energy is generated by ohmic heating. The N fins of thickness t and length L are integrally fabricated with the square sleeve of width w. Under steady-state operating conditions, the rate of thermal energy generation corresponds to the rate of heat transfer to airflow over the sleeve. (a) Under conditions for which a uniform surface temperature Ts is maintained around the circumference of the heater and the temperature T[infinity] and convection coefficient h of the airflow are known, obtain an expression for the rate of heat transfer per unit length to the air. Evaluate the heat rate for Ts = 320°C, D = 15 mm, an aluminum sleeve (ks = 240 W/m·K), w = 40 mm, N = 16, t = 4 mm, L = 20 mm, T[infinity] = 50°C, and h = 500 W/m2·K. (b) For the foregoing heat rate and a copper heater of thermal conductivity kh = 400 W/m·K, what is the required volumetric heat generation within the heater and its corresponding

a) q´ = 81546.36 W/m

T = 334.8 K

b) q = 4.21x10^8 W/m^3

Explanation:

We will take the following assumptions:

-Stationary state

-constant properties,

- negligible contact resistance

-uniform convection coefficient

-uniform heat generation

a)

Rcond = 1/(r*ks) = ln(1.08w/D)/(2*pi*ks) = ln(2.16)/(2*pi*240) = 5.1x10^-4 mK/W

Lc = L + t/2 = 0.022 m

m = (2h/ks*t) = 32.3 m^-1

mLc = 0.71

nf = tanh(mLc)/mLc = 0.61/0.71 = 0.86

At = NAf + Ab = N(2L+t)+(uw-Nt) = 0.704 + 0.096 = 0.8 m

no = 1-(nAf/At)*(1-nf) = 1-(0.704/0.8)*(0.14) = 0.88

Rto = (no*h*At)^-1 = (0.88*500*0.8)^-1 = 2.8x10^-3 mK/W

q´ = (320-50)/(5.11x10^-4+2.8x10^-3) = 81546.36 W/m

b)

Us*As = Us*pi*D = 298/pi*0.015 = 6323.76 W/m^2K

q = 4Us(Ts-Tinfinity)/D = 4*(6323.76)*250/0.015 = 421584000 W/m^3 = 4.21x10^8 W/m^3

The temperature is equal to:

T = (q*(D/2)^2)/(4ks) + Ts = 4.21x10^8* (0.0075^2)/(4*400) + 320 = 334.8 K

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

object 2 because all the other ones are wrong