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svetoff [14.1K]
3 years ago
13

A pump operating at steady state receives liquid water at 20°C, 100 kPa with a mass flow rate of 53 kg/min. The pressure of the

water at the pump exit is 5 MPa. The isentropic pump efficiency is 70%. Assume the pump is adiabatic and ignore changes in potential and kinetic energy.
(a) Determine the required power input to the pump.
Engineering
1 answer:
VARVARA [1.3K]3 years ago
5 0

Answer:

Input Power = 6.341 KW

Explanation:

First, we need to calculate enthalpy of the water at inlet and exit state.

At inlet, water is at 20° C and 100 KPa. Under these conditions from saturated water table:

Since the water is in compresses liquid state and the data is not available in compressed liquid chart. Therefore, we use approximation:

h₁ = hf at 20° C = 83.915 KJ/kg

s₁ = sf at 20° C = 0.2965 KJ/kg.k

At the exit state,

P₂ = 5 M Pa

s₂ = s₁ = 0.2965 K J / kg.k    (Isentropic Process)

Since Sg at 5 M Pa is greater than s₂. Therefore, water is in compresses liquid state. Therefore, from compressed liquid property table:

h₂ = 88.94 KJ/kg

Now, the total work done by the pump can be calculated as:

Pump Work = W = (Mass Flow Rate)(h₂ - h₁)

W = (53 kg/min)(1 min/60 sec)(88.94 KJ/kg - 83.915 KJ/kg)

W = 4.438 KW

The efficiency of pump is given as:

efficiency = η = Pump Work/Input Power

Input Power = W/η

Input Power = 4.438 KW/0.7

<u>Input Power = 6.341 KW</u>

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The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8
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Answer:

a.)

US Sieve no.                         % finer (C₅ )

4                                                  100

10                                                95.61

20                                               82.98

40                                               61.50

60                                               42.08

100                                              20.19

200                                              6.3

Pan                                               0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Explanation:

As given ,

US Sieve no.             Mass of soil retained (C₂ )

4                                            0

10                                          18.5

20                                         53.2

40                                         90.5

60                                         81.8

100                                        92.2

200                                       58.5

Pan                                        26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×\frac{100}{w}

∴ we get

US Sieve no.               % retained (C₃ )               Cummulative % retained (C₄)

4                                            0                                           0

10                                          4.39                                      4.39

20                                         12.63                                     17.02

40                                         21.48                                     38.50

60                                         19.42                                     57.92

100                                        21.89                                     79.81

200                                       13.89                                     93.70

Pan                                        6.30                                      100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no.               Cummulative % retained (C₄)          % finer (C₅ )

4                                                     0                                          100

10                                                  4.39                                      95.61

20                                                 17.02                                     82.98

40                                                 38.50                                    61.50

60                                                 57.92                                    42.08

100                                                79.81                                     20.19

200                                                93.70                                   6.3

Pan                                                 100                                        0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu = \frac{D60}{D10}

⇒ Cu = \frac{0.4}{0.12} = 3.33

d.)

Coefficient of Graduation = Cc = \frac{D30^{2}}{D10 . D60}

⇒ Cc = \frac{0.22^{2}}{(0.4) . (0.12)} = 1

3 0
3 years ago
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