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emmainna [20.7K]
3 years ago
11

How do you convert 3.9mL to hL

Chemistry
1 answer:
Reptile [31]3 years ago
5 0

3.9 ml * 1 hl
100000 ml = 0.000039 hl
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A stream of hot wat at 85 deg C at a rate of 1 kg/s is needed for the pasteurizing unit in a milk bottling plant. Such a stream
charle [14.2K]

Answer:

the flow rate for steam from the boiler plant = 0.099kg/s

the flow rate from the city water = 0.901 kg/s

the rate of entropy production in the mixing tank = 0.2044 kJ/k

Explanation:

In a well-insulated mixing tank where:

Q_{cv} = 0  & W_{cv}=0

The mass flow rates can be calculated using the formula:

Q_cv+m_1h_1+m_2h_2=m_3h_3+W_{cv      ------ equation (1)

so;

0+m_1h_1+m_2h_2=m_3h_3+0               ------- equation (2)

Given that:

From the steam in the boiler plant;

The temperature (T₁) = 200°C

Pressure (P₁) = 10 bar

The following data from compressed water and super-heated steam tables were also obtained at: T₁ = 200°C

h₁ = 2828.27 kJ/kg

s₁ = 6.95 kJ/kg K

m₁ (flow rate for steam in the boiler plant) = ????

Also, for city water

The temperature (T₂) = 20°C

Pressure (P₂) = 1 bar

Data obtained from compressed water and super-heated steam tables are as follows:

h₂ = 84.01 kJ/kg

s₂ = 0.2965 kJ/kg K

m₂ (flow rate for city water) = ???

For  stream of hot wat at 85 deg C

Temperature (T₃) = 85°C

h₃(h_f) = 355.95 kJ/kg

s₃(s_f) = 1.1344 kJ/kg K

m₃ = 1 kg/s

so since:

m₁ + m₂ = m₃       (since m₃  = 1)

m₂ = 1 -  m₁

From equation (2);

0+m_1h_1+m_2h_2=m_3h_3+0    

= m_1(2828.27)+(1-m_1)(84.01)=1(355.95)

= 2828.27m_1+(84.01-84.01m_1)=(355.95)

= 2828.27m_1-84.01m_1=355.95-84.01

= m_1(2828.27-84.01)=355.95-84.01

m_1 =  \frac{355.95-84.01}{2828.27-84.01}

m_1 =  \frac{271.94}{2744.26}m_1 =  0.099 kg/s

∴ the flow rate for steam from the boiler plant = 0.099kg/s

since; m₂ = 1 -  m₁

m₂ = 1 -  0.099 kg/s

m₂ = 0.901 kg/s

∴ the flow rate from the city water = 0.901 kg/s

b)

rate of entropy production in the mixing tank can be determined using the formula:

ΔS_{production} = m_3}s_3-(m_1s_1+m_2s_2)

ΔS_{production} = (1)(1.1344)-(0.099)(6.6955)-0.901(0.2965)

ΔS_{production} = 1.1344-0.6628545-0.2671465

ΔS_{production} = 1.1344 - 0.930001

ΔS_{production} = 0.204399

ΔS_{production} ≅ 0.2044 kJ/k

∴ the rate of entropy production in the mixing tank = 0.2044 kJ/k

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