2006 Final Exam Solution

ChBE 3200A Final Exam Summer 2006

NAME: ____SOLUTION__________

(1) A tubular reactor is used to convert A to B with a rate constant k [m3/(kg s)] and where cAf is the concentration at the reactor outlet in units of [kg / m3]. The other relevant parameters are the volumetric flow rate Q [m3/s], overall density [kg/m3], viscosity [kg / m s], the tube length L [m], and the tube diameter D [m].

(a) Among the parameters {k, cAf, Q, , , D, L}, how many are independent?

3 are independent – see below

Core variablesr = 3n = 3

k Q DM -1 0 0L 3 3 1T -1 -1 0

Dependent variables 4

cAf L Mu rhoM 1 0 1 1L -3 1 -1 -3T 0 0 -1 0

(b) State one set of independent parameters from the list in (a), but not the

independent combination (Q, k, D) given in part (d).

Possible independent combinations, by taking determinant of 3x3 array of dimensions, are:

kLQ kDQkLmu kDmukLrho kDrhokLcAf kDcAfkmucaf kmurhocaf mu Q rho mu Qrho mu L rho mu Dcaf mu L caf mu Dcaf Q D rho Q D

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(c) State one set of non-independent parameters from the list in (a).

Possible ones:

(d) Using (Q, k, D) as independent variables, find a dimensionless parameter that could be used to scale up this process while maintaining a desired value of cAf.

Find the that balances units of cAf with k, D, Q:

cAf ka Db Qc

M 1 -1 0 0L -3 3 1 3t 0 -1 0 -1

1 -a = 0 a = 1-3 +3a +b +3c = 0 b = 3 -3c – 3a = 3 + 3 -3 = 30 -a +0 -c = 0 c = -a = -1

= cAf k D3 / Q

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k rho cafrho mu cafrho caf Qrho caf Drho caf L


(2) A cylindrical nuclear reactor produces heat at a rate of 10 MW (Megawatts = 1x106 J/s), which can be considered to flow in the radial direction. To manage this heat the cylindrical reactor is immersed in a large pool that is constantly circulated with river water so as to maintain the reactor temperature at 330 K. The reactor is composed of fuel rods and control rods encased in stainless steel, but for this calculation you can assume the reactor is solid stainless steel. At some point in time, the valves in the cooling water supply line suddenly shut and cannot be opened. However, the water in the tank can still be recirculated in order to carry heat away from the reactor. How much time do the operators have to open the valves before the reactor reaches the melting temperature of stainless steel (1528 °C)? The heat transfer coefficient of water is h = 0.1 W/m2K when the flow stops.

Calculate Biot number:V =2RL = 3.926991 m3A =R2L = 15.70796 m2Biot = hV/kSSA Biot = 0.00147

Very low Biot number means a lumped parameter analysis can be used. Convection from surface is limiting and T of cylinder can be considered uniform.

The presence of heat generation, q, requires use of an energy balance that includes heat generation, in the manner of homework problem 18.2.

Energy balance: q – hA = VCp d/dt , where = T - T∞

d/dt = q/VCp – hAVCp

solution (see HW 18.2 solution)

t = (1/b) ln (a / (a – bq)), where a= q/VCp and b = hA/VCp

plugging everything in time to meltdown is 35 minutes. Yikes!

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R

L

ReactorCenter line

q_dot = 10,000,000 J/sL = 5 mR = 0.5 m

kSS = 17 W/mKcpSS = 461 J/kgKSS = 7920 kg/m3

kwater = 0.65 W/mKcpwater = 4200 J/kgK


(3) Heat from a flat plate is to be enhanced by adding straight fins of constant thickness made of stainless steel. The following specifications apply:

h = 60 W/m2 Kk = 8.9 W/ m KTs (surface) = 120 °CT∞ (air) = 20 °CFin thickness = 6 mmFin length = 20 mm (extending normal from the wall)Wall dimensions = 0.5 m x 0.5 m Fin spacing = 5 cm, and fins span entire length of wall.

Determine the number of fins, the fin efficiency and the heat loss from the finned plate.

nf = # fins = 0.5 m / 0.05 m/fin = 10 fins for 5 cm center-to-center distance

or if you included 0.006 m thickness + 0.05m spacing, nf = 0.5 / (0.056) = 9

L(h/kt)1/2 = 0.02m((60 W / m2 K )/ ((8.9 W/mK)(0.006/2 m))1/2 =

A_base = plate width*height – nf*fin thickness*plate widthA_1_fin = 2*fin length*plate width q = h(A_base + nf*A_1_fin*)*(Ts - T∞)

either of the following are acceptable:

nf 10 9.00A_base 0.22 m2 0.22 m2A_fin 0.02 m2 0.02 m2Q 2460 W 2364.00 W

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(4) An oil flows into the tube-side of a heat exchanger. The heat exchanger has 100 tubes each with diameter of 2 cm and length of 5 m. The pressure drop through the heat exchanger is measured to be 2.4 bar. What is the velocity of the oil in the tubes?

Oil viscosity = 0.1 Pa s Oil density = 0.85 g/cm3

P = 2fLv2/ D = 2.4 bar *(100,000 Pa / bar) = 240,000 Pa

Assume laminar flow, f = 16 / Re = 16 / (Dv

240,000 Pa = 2fLv2/ D =32Lv / (D2Solve for v:

v = 240,000 Pa * (D232Lv = 240,000 Pa * (0.022m232Pa s * 5 mv = 6 m/s

check laminar flow assumptionRe = Dv/ = (0.02 m)(6 m/s)(850 kg/m3)/(0.1 Pa s)Re = 1020

Therefore flow is laminar and solution is valid.

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(5) The device shown below is a viscosity pump. It consists of a rotating drum inside of a stationary cylinder, which are concentric with one another. Fluid enters at A and flows through the annulus and exits at B. The pressure at B is higher than that at A due to the action of the pump. The length the fluid travels from A to B in the annulus is L and the width of the annulus, h, is very small compared to the drum radius, R. Hence, the annulus can be treated as flow between two flat plates. Assuming laminar flow and a Newtonian fluid of viscosity , find a relation for the pressure rise from A to B as a function of the flow rate, Q.

This is problem 8.15 in Text, 4th edition

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A B

Or you could start with rectangular Navier Stokes equation:0= - dP/dx + d2vx/dy2


(6) Methane gas at 300 K flows over a packed bed of adsorbent material which removes trace amounts of impurities. The bed is packed with a ceramic adsorbent that is spherical with diameter 1 mm packed to a density that fills 65 % of the bed

volume. The bed has a 1 m diameter and is packed to a height of 3 m. Gas flows at a rage of 0.1 kg/s and the gas viscosity is 2x10-5 Pa s. What is the pressure drop of the

gas across the bed?

Calc using either inlet pressure only, P1, or average pressure based on P1 & P2 = P1 + DP, where DP is a guess for pressure drop.

ideal gas) = PMw / RT = P1 Mw / R / T or = (P1 + P2)*Mw / 2 / R / T

ideal gas) = P(atm)Mw(g /mol)/ 0.08205 (L atm/ mol /K)/ T(K) * 1 L /1000cm3*(1003 cm3)/1m3*1kg/1000g = P Mw / 0.08205 / T kg / m3

A = bed cross-sectional area = D2/4G’ = G/A

Using just rho_1 Using average densityMw 16 g/mol Mw 16 g/molT 300 K T 300 K 0.35 0.35d_ particle 0.001 m d_particle 0.001 mH 3 m H 3 mD 1 m D 1 mG 0.1 kg/s G 0.1 kg/s 2.00E-05 Pa s 2.00E-05 Pa sP_1 1 atm P_1 1 atm

rho_1 0.65001 kg/m3 rho_1 0.65001 kg/m3A 0.785398 rho_2 0.764217 kg/m3G' 0.127324 rho_av 0.707114 kg/m3Nre 9.79E+00 A 0.785398 m2DP 0.19143 atm 19396.65 Pa G' 0.127324 kg/m2/sErgunlh 17.1 Nre 9.79E+00Ergunrh 17.07 DP 0.1757 atm 17802.8 Pa

Ergunlh 17.07367Ergunrh 17.07

Nre = d_particle * G’ / (1 – ) /

Ergunlh = left hand side of Ergun eq.Ergunrh = right hand side of Ergun eq. DP varied until Ergun lh = rh

Note there is a conversion from 101,325 Pa = 1 atm used as well.

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2006 Final Exam Solution