Acartridge electrical heater is shaped as a cylinder of length l = 200 mm and outer diameter d = 20 mm. under normal operating conditions the heater dissipates 2 kw while submerged in a water flow that is at 20°c and provides a convection heat transfer coefficient of h = 5000 w/m2 · k. neglecting heat transfer from the ends of the heater, determine its surface temperature ts. if the water flow is inadvertently terminated while the heater continues to operate, the

Ts=51.83C

Explanation:

First we calculate the surface area of the cylinder, neglecting the top and bottom covers as indicated by the question

Cilinder Area= A=πDL

L=200mm=0.2m

D=20mm=0.02m

A=π(0.02m)(0.2m)=0.012566m^2

we use the equation for heat transfer by convection

q=ha(Ts-T)

q= heat=2Kw=2000W

A=Area=0.012566m^2

Ts=surface temperature

T=water temperature=20C

Solving for ts

Ts=q/(ha)+T

Ts=2000/(5000*0.012566m^2)+20=51.83C

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2