Question

1 CPD4701 Assignment 2/2019 Question 2 shell and-tube heat exchanger was designed for the following service: Cold stream Hot stream Crude Oil Fluid Cooling water Tube side Stream allocation Shell side Mass flow rate (kg/s) 110 30 Inlet temperature (C) 90 Outlet temperature (C) Heat capacity (J/kg K) Density (kg/m2) Viscosity (Pa-s) Thermal conductivity (W/m-K) Fouling factor (m2 CW) 40 50 2177 4187 787 995 0.72-10 1.89-103 0.122 0.59 0.0002 0.0004 The shell and tube heat exchanger has the following specifications: One (1) shell pass Shell Inside diameter: 1.372 m Tubes: Four (4) tube passes 2 344 tubes of 19.05 mm OD,1.25 tube pitch ratio, square pitch Tube wall thickness = 1.65 mm Tube length= 4.877 m Tube material kw = 50 W/m K Baffle spacing of 620 mm with a 25 % cut Baffles: Tube side: 1-3 m/s Shell side: 0.5-1 m/s Linear velocity range: f the hot fluid increased by 20%. Nonetheless , the inlet After a debottlenecking exercise, the flowrate and outlet temperatures must be maintained at their current values. The pumps feeding the exchanger were designed to allow for an increase in the pressure drop of both fluids must not exceed 90 kPa both fluids. However, the pressure drops 2.1 Calculate the original and the new heat transfer duties. (4) 2.3 Rate e area aVala r for the new service (20% increase in the flow rate of hot fluid). Determine if the existing exchanger will meet the area requirements and the pressure drop requirements. 2.4 If the existing exchanger does not meet the requirements, how can the existing exchanger be modified to meet the requirements. Show (35) calculations (14) [55 TOTAL: 95 UNISA 2

Answer 1

2.1 Heat duty = mCp (T_in - T_out)

= 110 * 2177 *(90-50)

= 9578800 W

In second case, the flow rate is increased by 20% hence the new flow rate is 132 kg/s

Hence heat duty is = 132* 2177 *(90 -50) = 11494560 W

2.2 We know that

A = $\Pi$ d₀ L N_t

Hence the available area is, $\Pi$* (19.05 x 10^-3 ) * 2344 * 4.877 = 684 m²

2.3 Required formulas are given below,

tuber whu N no Np parses Ne exax di 4 NP 2 O-14 Nu O8 O023 x Re O-33 Nu h di Pressure diop eu2 Tube +2.5 NP di fquivalent dia de 8J Ds Shel: xL eu2 - 4 Baffe ipating 2 de 1.2F CP do 2 O785 d de Chdo )x Dexle ALua of Shell do endo t do +£ Da di hi + di cs 2N Scanned w CamS canne

From energy balance we can find the mass flow rate of cooling water

11494560 = m * 4187 * (40-30)

Hence m = 274.53 kg/s

Tube side:

Velocity v =

274.4 4 4 995 II (15.75 10-3)2 2344

= 2.41553 m/s

Reynolds number Re =

995 2.41553 (15.75 * 10) 1.89 10-3

= 52575.52016

Prandtl Number Pr = 5.10955

Nusselt number Nu =

0.02352575.520.85.109550.33

= 235.5706

Heat transfer coefficient = 8824.55 W/m² K

J_f from standard charts = 3.2 x 10^-3

Pressure Drop =

995 2.415332 4.877 4[8(3.2 10* 15.75 10- +2.5] * 2

= 121.05086.8676 Pa

Shell side:  

P_t =1.25 * 19.05 = 23.8125 mm

d_e =  

1.27 (23.8125 0.785 19.052) 19.05

= 18.8104 mm

A_s =

23.8125 19.05 1372 620 23.8125

= 0.170128 m²  

v =

132 787 0.170128

= 0.9587 m/s

Re =

787 0.9587 18.8104 1.89

= 7509.200

Pr = 33.725655

Nu = 485.2256

h = 3146.46 W/m² K

J_f from standard charts is 5 x 10^-2

Pressure Drop =

4877 787 0.95872 1372 85 102 18,814 620 2

= 82958.88 N/m²  

U_overall = 850.433 W/m² K

In
= $\frac{(90-40)-(50-30)}{ln\frac{90-40}{50-30}}$

= 32.7407

R = 4 S =0.16667

LMTD correction factor from charts = 0.931

= 0.931 * 32.7404 = 30.501 C

ATIMTD LMTL

Area required = Q/U

ATIMTD LMTL

= 443.136 m²

Clearly the HX does not meet the pressure drop requirements. But sufficient area is available for heat transfer

2.4

With the given dimensions,

if No. of passes is made to be 2,

Then v = 1.207665 m/s  ( Use the velocity equation, take N_p =2)

Re = 26285.5835

J_f = 4 x 10^-3

Pressure drop = 18.00722 kPa

Nu= 128.83

h = 4826.009 W/m² K

U _overall = 606.075 W/m² K

Area = 621.801 m²

Hence all criteria are met