- #1

sci0x

- 83

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- 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

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