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AleksandrR [38]
3 years ago
8

Briefly discuss if it would be better to operate with pumps in parallel or series and how your answer would change as the steepn

ess of the system curve changes?
Engineering
1 answer:
Aleksandr [31]3 years ago
5 0

Answer:

1) In series, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.

2) In parallel, the combined head and volume flow will move along the system curve from point 1 to point 3.

Explanation:

1) Pump in series:

When two or more pumps are connected in series, their resulting pump performance curve will be obtained by adding their respective heads at the same flow rate as shown in the first diagram attached.

In the first diagram, we have 3 curves namely:

- system curve

- single pump curve

- 2 pump in series curve

Also, we have points labeled 1, 2 and 3.

- Point 1 represents the point that the system operates with one pump running.

- Point 2 represents the point where the head of two identical pumps connected in series is twice the head of a single pump flowing at the same rate.

- Point 3 is the point where the system is operating when both pumps are running.

Now, since the flowrate is constant, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.

2) Pump in parallel:

When two or more pumps are connected in parallel, their resulting pump performance curve will be obtained by adding their respective flow rates at same head as shown in the second diagram attached.

In the second diagram, we have 3 curves namely:

- system curve

- single pump curve

- 2 pump in series curve

Also, we have points labeled 1, 2 and 3

- Point 1 represents the point that the system operates with one pump running.

- Point 2 represents the point where the flow rate of two identical pumps connected in series is twice the flow rate of a single pump.

- Point 3 is the point where the system is operating when both pumps are running.

In this case, the combined head and volume flow will move along the system curve from point 1 to point 3.

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Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
MariettaO [177]

Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

Heat,e= 3×10^6 W/m^3

Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0
3 years ago
A piston-cylinder device contains 0.58 kg of steam at 300°C and 0.5 MPa. Steam is cooled at constant pressure until one-half of
Mumz [18]

Answer:

a) Tբ = 151.8°C

b) ΔV = - 0.194 m³

c) The T-V diagram is sketched in the image attached.

Explanation:

Using steam tables,

At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.

Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C

b) The volume change

Using data from A-5 and A-6 of the steam tables,

The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume

(αբ) (which is calculated from the final quality and the consituents of the specific volumes).

ΔV = m(αբ - αᵢ)

αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)

q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg

αբ = 0.00109 + 0.5(0.3748 - 0.00109)

αբ = 0.187945 m³/kg

αᵢ = 0.5226 m³/kg

ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³

c) The T-V diagram is sketched in the image attached

3 0
3 years ago
A(n) ____ is an object setting used to control the visible display of objects.
KatRina [158]
Remote?? maybe I’m not really sure
3 0
3 years ago
) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p
andrey2020 [161]

Answer:

number of pulses produced =  162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = \frac{linear\ displacement}{circumference} × No of pulses per revolution .................2

number of pulses produced = \frac{200}{314.16} × 256

number of pulses produced =  162 pulses

5 0
3 years ago
For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea
miskamm [114]

Answer:

k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

where r_1 and r_2 be the inner radius, outer radius of the annalus.

Explanation:

Let r_1, r_2 and L be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, T_1=30^{\circ}C\\ and at the outer surface, T_2=40^{\circ}C.

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

40 W/m^2.

\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40

\Rightarrow q=2.4\pi L \;W

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

q=-kA\frac{dT}{dr}\;\cdots(i)

where,

K=Thermal conductivity of the material.

T= temperature at any radial distance r.

A=Area through which heat transfer is taking place.

Here, A=2\pi rL\;\cdots(ii)

Variation of temperature w.r.t the radius of the annalus is

\frac {T-T_1}{T_2-T_1}=\frac{\ln(r/r_1)}{\ln(r_2/r_1)}

\Rightarrow \frac{dT}{dr}=\frac{T_2-T_1}{\ln(r_2/r_1)}\times \frac{1}{r}\;\cdots(iii)

Putting the values from the equations (ii) and (iii) in the equation (i), we have

q=\frac{2\pi kL(T_1-T_2)}{\LN(R_2/2_1)}

\Rightarrow k= \frac{q\ln(r_2/r_1)}{2\pi L(T_2-T_1)}

\Rightarrow k=\frac{(2.4\pi L)\ln(r_2/r_1)}{2\pi L(10)} [as q=2.4\pi L, and T_2-T_1=10 ^{\circ}C]

\Rightarrow k=0.12\ln(r_2/r_1)\frac {W}{ m^{\circ} C}

This is the required expression of k. By putting the value of inner and outer radii, the thermal conductivity of the material can be determined.

7 0
3 years ago
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