Mathcad - Quiz 7 solutions

Mathcad - Quiz 7 solutions - W m K ⋅ := α 1.75 10 7 −...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Bi 2 hr k := Bi 2 4 = C 1 1.4698 := λ 1 1.9081 := For the centerline temperature, use equation 5.49c θ mid C 1 exp λ 1 2 Fo = θ mid T mid T air T i T air := θ mid 0.536 = solve for Fo Fo ln θ mid () ln C 1 () λ 1 2 := Fo 0.277 = Fo is > 0.2, so 1-term solution is appropriate Solve for t t Fo r 2 α := t 10137.984 s = t 168.966 min = t 2.816 hr = 1. A long 16 cm diameter cylindrical piece of hardwood (k = 0.16 W/m•K; α = 1.75 x 10 -7 m 2 /s) is treated by exposure to air at 30?C with a heat transfer coefficient of 8 W/m 2 •K. If the wood is initially at 2?C, how long will it take for the center temperature to rise to 15?C? r8 c m := h8 W m 2 K := k 0.16
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: W m K ⋅ := α 1.75 10 7 − ⋅ m 2 s := T air 303K := T i 275K := T mid 288K := Assumptions 1. Long cylinder, L/R > 10 2. 1-d conduction in r direction (if LC not valid) 3. neglect radiation 4. check to see if Lumped Capacitance is valid 5. If LC is not valid, check to see if 1-term approximate solution is valid Is lumped capacitance valid? L c r 2 := for cylinder L c 0.04 m = Bi h L c ⋅ k := Bi 2 = Bi is > 0.1, so LC method is NOT valid, must consider spatial effects. Recalculate Bi for Table 5.1...
View Full Document

This note was uploaded on 03/11/2011 for the course CHEME 333 taught by Professor Anthony during the Spring '11 term at UNL.

Ask a homework question - tutors are online