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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>

You might be interested in
Calculate the availability of a system where the mean time between failures is 900 hours and the mean time to repair is 100 hour
Debora [2.8K]

Answer:

The availability of system will be 0.9

Explanation:

We have given mean time of failure = 900 hours

Mean time [to repair = 100 hour

We have to find availability of system

Availability of system is given by  \frac{mean\time\ of\ failure}{mean\ time\ of\ failure+mean\ time\ to\ repair}

So availability of system =\frac{900}{900+100}=\frac{900}{1000}=0.9

So the availability of system will be 0.9

7 0
3 years ago
). A 50 mm diameter cylinder is subjected to an axial compressive load of 80 kN. The cylinder is partially
Delicious77 [7]

Answer:

\frac{e'_z}{e_z} = 0.87142

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

                          εx = - [ σx - ν( σy + σz ) ] / E

                          εy = - [ σy - ν( σx + σz ) ] / E

                          εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

                          εx =  [ ν*σz ] / E

                          εy = [ ν*σz ] / E

                          εz = - [ σz ] / E   .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

                         εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

                         εx' = - [ σx' - ν( σy' + σz ) ] / E =  [ ν*σz ] / 2E

                         εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

                         εx' = - [ σy' - ν( σx' + σz ) ] / E =  [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note:  σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

                        - [ σx' - ν( σx'+ σz ) ] =  [ ν*σz ] / 2

                          σx' ( 1 - v ) = [ ν*σz ] / 2

                          σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

                           εz' = - [ σz - ν( σy' + σx' ) ] / E

                           εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

                          εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E  ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

                            \frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ]  }{-\frac{s_z}{E}}  \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142... Answer

5 0
3 years ago
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz
Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s       and ρ= 850 kg/m³  

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ =  ρv

   =850 x 0.00062

   = 0.527 kg/m·s  

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

Q = \dfrac{\Delta P \pi D^4}{128 \mu L}

Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}

Q = 5.06 x 10⁻⁸ m³/s

4 0
2 years ago
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capaci
Serggg [28]

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

8 0
3 years ago
Please help me fast, I don’t have time
Anna71 [15]

Answer: precision

Explanation: Because accuracy is where you keep on getting it right but precision is where you get closer and closer

5 0
3 years ago
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