Calculate the number of plates required for a heat exchanger

[sci0x]

Homework Statement

Calc wort flow rate ratio and no. Of plates required to cool 360 hl hr-1 wort, in counter current flow from 98 degrees C to 15 degrees C using plate heat exchanger

Relevant Equations

Q=UAdT

Cooling water inlet temp is 4 degrees C
Cooling water outlet temp is 80 degrees C
Density of water is 1000 kgm-3
Specific heat water is 4.2kJkg-1K-1
Density wort is 1060 kgm-3
Specific heat wort is 4.0Kjkg-1K-1
Overall heat transfer coeff is 3000Wm-2k-1
Area of each plate is 0.75m2

For water: wort flow rate ratio I've said
Heat lost by hot wort = heat gained by cold
Mh x Cph x (Tih-Toh) = Mc x Cpc x (Toc-Tic)
(360)(4000)(98-15) = Mc(4200)(80-4)
Mc = 374.436
374.436:360
1.04:1

This is expected answer.

Im having trouble calculating no. Of plates req which should be 111
I should use log mean temp differemce in heat transfer rate eq

So q=UA(LMTD)
Overall heat trans co is in K so change C to K
LMTD is 83-75/ln(83/76) is 79.44 degrees C or 352.99 Kelvin
Q=3000(A)(352.99)

Q is also Mh x Cph x (Tih-Toh) = (360)(4000)(98C-15C) or 360(4000)(356.15K)
512856000=3000A352.99
A = 484.29
Each plate is 0.75m2 x 484.29 = 363.21 plates
Can someone show me how to get 111 here.
Q from past exam paper

 

[BvU]

Relevant Equations:: Q=UAdT
Overall heat trans co is in K so change C to K

what about changing a temperature difference from $\Delta ^\circ C$ to $\Delta K$ ?

LMTD is 83-75/ln(83/76)

You forgot the brackets
and the numbers are weird -- look up what LMTD means

 

[Chestermiller]

The heat load is $Q=(36)(1060)(4.0)(98-15)=12700000\ kJ/hr=3519\ kj/s=3519000\ W$

The two end temperature differences are 18 C and 11 C. What is the LMTD based on these?

 

[sci0x]

Hi Chester, i messed up there:
dt1 is 98-80 = 18
dt2 is 15-4 is 11

LMTD is 18-11/ln(18/11) is 14.21

The heat load is 3519000W

3519000 = (3000)(A)(14.21)
A= 82.54m2

82.54/0.75 = 110 plates

Just wondering in your calculation Q=(36)(1060)(4.0)(98-15)
What units are the 36 in, to convert from 360hl hr-1

 

[Chestermiller]

Hi Chester, i messed up there:
dt1 is 98-80 = 18
dt2 is 15-4 is 11

LMTD is 18-11/ln(18/11) is 14.21

The heat load is 3519000W

3519000 = (3000)(A)(14.21)
A= 82.54m2

82.54/0.75 = 110 plates

Just wondering in your calculation Q=(36)(1060)(4.0)(98-15)
What units are the 36 in, to convert from 360hl hr-1

How many hl are there in a m^3?

 

[sci0x]

Okay m3
1hl = 0.1m3
360hl = 36m3

Cheers