Mathcad - 2010 Exam 2 solutions

Mathcad - 2010 Exam 2 solutions - Springs in suspension...

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T inf 1.272 10 3 × K = T inf T f T i exp X () 1 exp X := X 0.578 = Xh A s t ρ Vc p := solve algebraically for Tinf T f T inf T i T inf exp h A s t ρ V c p = t 600 s = V 623.41 cm 3 = V 6.234 10 4 × m 3 = V π D 2 4 L := A s 997.456 cm 2 = A s 0.1m 2 = A s π D L := Bi is < 0.1, so LC method is valid; temperature is ~uniform in the rod Bi 1.956 10 3 × = Bi hL c k := L c 6.25 10 3 × m = L c D 4 := Geometry is a cylindrical rod. Lc is R/2 Assume 1. Unsteady state 2. Constant properties 3. Check Bi for LC method validity 4. Neglect radiation effects 5. Neglect end areas of cylinder - consider lateral surface area t 10min := T f 723 K = T f 450 273 + K := T i 293K := h2 0 W m 2 K := L 1.27m := D 2.5cm := k 63.9 W mK := c p 424 J kg K := ρ 7832 kg m 3 := Springs in suspension systems of cars and trucks are made of steel rods that are heated and wound into coils when they are ductile. Consider steel rods ( ρ = 7832 kg/m 3 , c p = 424 J/kg•K, k = 63.9 W/m•K) with a diameter of 2.5 cm and a length of 1.27 m. The rods are heated in an oven with a uniform convection heat transfer coefficient of 20 W/m 2 •K. The rods need to be heated from an initial temperature of 20 o C to the desired temperature of 450 o C. Determine the ambient temperature the oven has to be if the heating process is to take 10 minutes.
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For surface temperature x 1.0 := θ T f T inf T i T inf := θ 0.361 = Using eqn 5.40a and solve for Fo cos λ 1 x () 0.908 = Fo ln θ C 1 cos λ 1 x λ 1 2 := Fo 5.085 = note - one term solution is valid!!!! t Fo L 2 α := t 5.588 s = Takes minimum of 5.6 seconds to chill steak Solve for centerline temp θ o C 1 exp λ 1 2 Fo := θ o 0.398 = T fo θ o T i T inf T inf + := T fo 276.32 K = In a meat processing plant, 2 cm thick steaks (k = 0.45 W/m•K, α = 0.91 x 10 -4 m 2 /s) that are initially at 25 o C are to be cooled by passing them through a refrigeration compartment (via conveyor belt). The air in the refrigeration compartment is -11 o C. The heat transfer coefficient on both sides of the steaks is 9 W/m 2 •K. If both surfaces of the steaks are to be cooled to 2 o C, what is the minimum residence time for the steaks in the refrigeration compartment? What is the corresponding centerline temperature of the steaks at this time? L1 c m := k 0.45 W mK := α 0.91 10 4 m 2 s := T i 298K := T inf 262K := h9 W m 2 K := T f 275K := Assume 1. Unsteady state 2. Constant properties 3. Symmetry of the steak and convection conditions 4. Neglect radiation 5. Check Bi to see if LC method is valid 6. If LC method is not valid - check to see if 1-term approximate solution is valid For a slab geometry - L = t/2 Bi hL k := Bi 0.2 = Bi is > 0.1, so LC method is NOT valid For surface temperature - assume the 1-term approximate solution is valid (check this later) From Table 5.1 Bi 0.2 = λ 1 0.4328 := C 1 1.0311 :=
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C r C min C max := C r 1 = q max C min T hin T cin () := q max 1.9171 10 5 × W = q max 191.71 kW = Calculate NTU NTU UA C min := NTU 11.893 = Calculate effectiveness Use Eqn 11.32 - crossflow with both fluids unmixed (note that this eqn is exact for Cr = 1) ε 1 exp 1 C r NTU 0.22 exp C r NTU 0.78 1
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This note was uploaded on 03/11/2011 for the course CHEME 333 taught by Professor Anthony during the Spring '11 term at UNL.

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Mathcad - 2010 Exam 2 solutions - Springs in suspension...

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