Chapter 1 Unit Operations Problems

1. Heat transfer equation

Js-1 = Jm-2 s-1 oC-1 x m2 x oC

On RHS

m= -2 + 2 = 0

s= -1

oC= -1 + 1 = 0

SoJm-2 s-1 oC-1 x m2 x oC = Js-1

2. Specific heat of apples0.86 Btu lb-1 oF-1From Appendix 2

1 Btu= 1055 J

1lb= 0.4536 kg

oF= 5/9 oC

0.86 Btu lb-1 oF-1= 0.86 Btu lb-1 oF-1 x 1055 Jx 1 lb x oF_

1 Btu 0.4536 kg (5/9 oC)

= 3600 J kg-1 oC-1

= 3.6 kJ kg-1 oC-1

3. Viscosity of olive oil

From Appendix 2

1lb= 0.4536 kg

1 ft = 0.3048 m

5.6 x 10-2 lbft-1s-1= 5.6 x 10-2 lb x 0.4536kg x 1 ft x 1 s

ft s 1 lb 0.3048m1 s

= 8.3 x 102 kgm-1s-1

= 83 x 103 kgm-1s-1

= 83 x 103 N s m-2

Compare with data in Appendix 4.

4. Reynolds number for a fluidDv(= [L] x [L] x [t]-1 x [M][L]-3 _

[M][L]-1 [t]-1

L= 2 3 +1= 0

t= -1 + 1= 0

M= 1 1

= 0

Dv is dimensionless

5. Protein content of a food mixture

Ingredient

Weight

% Proteinkg Protein

Maize Starch

100 kg

0.3

0.3

Wheat flour 22.5 kg 12.0

2.7

Skim milk powder 4.31 kg 30.0

1.293

Total

126.81

4.293

Accuracy126.8

4.3

Protein = 3.4%

6. Rate of heating of a sugar syrupUsing Appendix 2

T= 60oC

t= 30 min x 60= 1800s

V= 50 ft3 = 50 x 0.0283 m3 = 1.415 m3= 66.9 lb/ ft3 = 66.9 x 16.01 kg/m3 = 1071 kg/m3W= 1.415 x 1071 kg = 1515 kg

1 Btu = 1055 J

1 lb= 0.4536

1 oF= 5/9 oC = 0.56 oC

(a) Specific Heat = 0.9 Btu x 1055 J x l lb x 1 oF

lb oF 1 Btu 0.4536 kg 0.56oC

= 3737.9 Jkg-1 oC-1

= 3.7 kJkg-1 oC-1

Compare with Appendix 4

(b) Rate of heat input= heat energy/time

= specific heat x kg x oC x 1/t

= 3.7kJ x 1515 kg x 60oC x 1 _

kg oC

1800 s

= 187kJs-1

7. Gas equation

PV = n R T

n= PV

= 2.0 atm x 6 m3 ______

RT

0.08206 m3 atm mole-1K-1 x 300 K

= 0.49 moles

8. Gas law constantR = 0.08206 m3 atm mole-1K-1 (note mole is kg mole, i.e. the molecular weight in kg)

1 m3= 1/0.0283 ft3 = 35.3 ft31 atm= 1.013 x 105 Pa

1 lb = 453.6 g , 1 kg = 2.2lb

1 Pa = 1/6894 lbm-2= 1.45 x 10 4 lbm-2

1atm= 14.7 lbm-2 = 1.013 x 105Nm-2

1 atm = 0.76 mHg = 760mm Hg

1 joule= 1Nm therefore 1 atm =1.013 x 105 Joules m-3 = 1.013 Jm-3

(a) R units= ft3 mm Hg lb-mole-1K-1

R= 0.08206 m3 atm mole-1K-1 x 35.3 ft3 x 760mm Hg x 0.454 kg mole

1 m3 1 atm 1 lb mole

= 999 ft3 mm Hg lb-mole-1K-1(b)R units= m3 Pa mole-1K-1R= 0.08206 m3 atm mole-1K-1 x 1.013 x 105 Pa

1 atm

= 8313 m3 Pa mole-1K-1(c) R units= J g.mole-1K-1 R= 0.08206 m3 atm mole-1K-1 x 1.013 x 105N x 1mole

1 atm m2 1000 g mole

= 8.313 N m g-mole-1K-1

= 8.313 J g-mole-1K-1

9. Liquid pressure in a tank

z= P/(g

[L] = [F] [L] 2____

[M] [L]-3 [L] [t]-2

= [F]__

[M] [t]-2But [F]= [M] [L]_

[t]2

so [L]= [M] [L]_

[t]2 [M] [t]-2

= [L]

10. Grashof number

D3(2(g(T/

= [L]3 ( [M] [L]-3 )2 [L][t-2] [T]

([M][L]-1[t]-1) 2

L= 3-6+1+2= 0

t= -2 + 2= 0

M= 2 2

= 0

T= 1

Therefore [(] = [1/T] and the Grashof number is then dimensionless