Suggested Response - Q4

** Using the steam table below, answer the following questions.

a) Steam is available in your brewery at 1.5 MPa (1.4 MPa g). What steam flow rate is required to increase the temperature of 1,000 kg of wort from 77°C to 100°C in 25 min? The specific heat capacity of wort is 4,050 J kg-1 K-1.

b) How much steam would be required if the pressure of steam was increased to 2.5 MPa (2.4 MPa g)?

Table: Steam table

Temperature

°C

Pressure

MPa

Specific Volume

m3 kg-1

Specific Enthalpy

Liquid

kJ kg-1

Specific Enthalpy of Vaporisation

kJ kg-1

Specific Enthalpy Vapour

kJ kg-1

0.01

0.000612

205.9912

0

2,500.92

2,500.92

5

0.000873

147.0113

21.02

2,489.04

2,510.06

10

0.001228

106.3

42.02

2,477.19

2,519.21

30

0.004247

32.8783

125.73

2,429.82

2,555.55

50

0.012400

12.0269

209.34

2,381.95

2,591.29

70

0.031200

5.0395

293.07

2,333.03

2,626.10

90

0.070200

2.3591

377.04

2,282.49

2,659.53

100

0.101400

1.6718

419.17

2,256.40

2,675.57

125

0.232200

0.7700

525.07

2,188.03

2,713.10

150

0.476200

0.3925

632.18

2,113.75

2,745.93

179.88

1

0.1944

762.52

2,014.59

2,777.11

200

1.55490

0.1272

852.27

1,939.74

2,792.01

225

2.54970

0.0784

966.80

1,835.35

2,802.15

263.94

5

0.0394

1,154.64

1,639.57

2,794.21

280

6.41660

0.0302

1,236.88

1,542.99

2,779.87

285.83

7

0.0274

1,267.66

1,504.97

2,772.63

295.01

8

0.0235

1,317.31

1,441.37

2,758.68

311.00

10

0.0180

1,408.06

1,317.43

2,725.49

 

Response:

a) We first need to calculate the energy required to raise the wort from 77°C to 100°C in 25 min. We can calculate this sensible heat in the following way:

Q equals M C increment T

Q equals 1 comma 000 space cross times space 4 comma 050 space cross times space left parenthesis 100 minus 77 right parenthesis

Q equals 93 comma 150 comma 000 space J equals 93 comma 150 space k J

This heating occurs in 25 min, so, to calculate the heat transfer rate, we can divide the total energy by the time.

q equals Q over t equals fraction numerator 93 comma 150 over denominator 25 end fraction equals 3 comma 726 space k J space m i n to the power of negative 1 end exponent equals 62.1 space k W space left parenthesis space J space s to the power of negative 1 end exponent right parenthesis

We can now set this equal to the latent heat energy associated with condensing saturated steam at 1.5 MPa. Refer to the table above, and you will find that the specific enthalpy of vaporisation of steam at 1.5 MPa is 1,939.74 kJ kg-1. So, the rate of latent energy transfer is calculated using:

q equals m increment H subscript v a p end subscript

Setting this equal to the desired heating rate and solving for, the steam flow rate can be calculated as:

q equals m space cross times space 1 comma 939.74 equals 3 comma 726 space k J space m i n to the power of negative 1 to the power of blank end exponent

m equals 1.92 space k g space m i n to the power of negative 1 end exponent equals 115 k g space h r to the power of negative 1 end exponent

b) If the pressure is changed then the specific enthalpy of vaporisation changes. By looking at the steam table, we will find that the specific enthalpy of vaporisation for steam at 2.5 MPa is 1,835.35 kJ kg-1. We can then follow the same process as before.

q equals m space cross times space 1 comma 835.35 equals 3 comma 726 space k J space m i n to the power of negative 1 to the power of blank end exponent

m equals 2.03 space k g space m i n to the power of negative 1 end exponent equals 122 space k g space h r to the power of negative 1 end exponent

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