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3 years ago

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1 A power transmission includes a belt drive, a chain drive and a gear drive. Which of the following is the best arrangement bet

ween a motor and a working machine? (a) belt drive gear drive chain drive (b) chain drive gear drive belt drive (c) belt drive chain drive gear drive (d) chain drive belt drive gear drive

1 answer:

Setler79 [48]3 years ago

3 0

Answer:d

Explanation:

(d) chain drive belt drive gear drive

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A town is designing a rectangular, 4m deep settling tank for treating surface water intake. The tank will have a flow velocity o

Oksana_A [137]

Answer:

333335790089

Explanation:

blh

7 0

3 years ago

Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.

vekshin1

Solution :

<u>Sieve Size</u> (in) <u>Weight retain</u><u>(g)</u>

3 1.62

2 2.17

$1\frac{1}{2}$ 3.62

$\frac{3}{4}$ 2.27

$\frac{3}{8}$ 1.38

PAN 0.21

Given :

Sieve weight % wt. retain % cumulative % finer

size retained wt. retain

No. 4 59.5 10.225% 10.225% 89.775%

No. 8 86.5 14.865% 25.090% 74.91%

No. 16 138 23.7154% 48.8054% 51.2%

No. 30 127.8 21.91% 70.7154% 29.2850%

No. 50 97 16.6695% 87.3849% 12.62%

No. 100 66.8 11.4796% 98.92% 1.08%

Pan <u> 6.3 </u> 1.08% 100% 0%

581.9 gram

Effective size = percentage finer 10% ( $$$D_{20}$ )

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x , 10%

$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$

$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$

x = 0.2634 mm

Effective size, $D_{10} = 0.2643 \ mm$

Now, N 16 (1.19 mm) , 51.2%

N 8 (2.38 mm) , 74.91%

x, 60%

$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$

x = 1.6317 mm

$\therefore D_{60} = 1.6317 \ mm$

Uniformity co-efficient = $\frac{D_{60}}{D_{10}}$

$Cu= \frac{1.6317}{0.2643}$

Cu = 6.17

Now, fineness modulus = $\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$

$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$

= 4.41

which lies between No. 4 and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

7 0

2 years ago

A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a

Dafna1 [17]

Answer:

a) $\dot W = 1.062\,kW$

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}$

$COP_{HP} = 15.692$

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

$\dot Q_{H} = 60000\,\frac{kJ}{h}$

The minimum power supplied to the heat pump is:

$\dot W = \frac{\dot Q_{H}}{COP}$

$\dot W = \frac{\left(60000\,\frac{kJ}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)}{15.692}$

$\dot W = 1.062\,kW$

5 0

3 years ago

The following is a list of metals and alloys:

viktelen [127]

Answer:

A) Gray cast iron

B) Aluminum

C) Titanium alloy

D) Tool steel

E) Titanium alloy

F) magnesium

G) Tungsten

5 0

3 years ago

Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°

cricket20 [7]

Answer:

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

$m\cdot (h_{in} - h_{out}) = 0$

$h_{in} = h_{out}$

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

$P = 1500\,kPa$

$T = 198.29\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 6.3068\,\frac{kJ}{kg\cdot K}$

$x = 0.967$

State 2 - Outlet (Superheated Vapor)

$P = 200\,kPa$

$T = 130\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 7.1776\,\frac{kJ}{kg\cdot K}$

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

$\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}$

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

6 0

3 years ago