HELP PLEASE

Patient people vote more often than non-patient people.<br> True<br> O<br> False

xxMikexx [17]

Answer:

True

Explanation:

8 0

3 years ago

Which contribution of Science is the most important discovery of man? Explain why?

Vsevolod [243]

Answer:

<h2>The Invention of the Internal Combustion Engine (ICE)</h2>

Explanation:

The internal combustion engine is an engine in which ignition and combustion take place in the engine(in one place), the invention of the ICE was an integral part of the industrial revolution, as there was increasing demand for power, and manual labor could not suffice, especially during the mid 19 century.

The ICE made it possible for tasks that demand intensive power consumption to come through to reality, it was as a result of the invention of the ICE that road transportation was made easier for mankind, as the means of transport then was the use of beast of burden, now we have cars, airplanes ship, etc, essentially the invention of ICE reduced the tedious task man would have to engage in for his daily needs

5 0

3 years ago

The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the

soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

(dP - y*L*sin(Q))/ L = 2*t / r

dP / L - y*sin(Q) = 2*t / r

Where dP / L = - dP/dx,

dP / dx = -2*t / r - y*sin(Q)

Where r = D /2 ,

dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)

dP / dx = -8 + 0

dP/dx = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

dP / dx = 8 - 62.4

dP/dx = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

dP / dx = -8 + 62.4

dP/dx = 54.4 lb / ft^3

7 0

3 years ago

An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the

Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe $T_p$ = -15° C = (-15 +273)K =258 K

Temperature of water $T_w$ = 3° C = (3 + 273)K = 276 K

Temperature of ice $T_i$ = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K ; R = 0.05

convection coefficient $Lh_l$ =2000 W/m².K

The energy balance can be expressed as:

$q_{conduction} =q_{convention}$

where;

$q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}$ ------------- equation (1)

$q_{convention} = \pi DLh_l(T_w-T_i)$ ------------ equation(2)

Equating both equation (1) and (2); we have;

$\frac{2\pi LK(T_i-T_p)}{In(R/r)}$ $= \pi DLh_l(T_w-T_i)$

Replacing the given data; we have:

$\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)}$ $= \pi (0.1)*2000(276-273)$

$\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96$

$In(\frac{0.05}{r})*1884.96 = 182.84$

$In(\frac{0.05}{r}) = \frac{182.84}{1884.96}$

$In(\frac{0.05}{r}) =0.0970$

$\frac {0.05}{r} =e^{0.0970}$

$\frac {0.05}{r} =1.102$

$r=\frac{0.05}{1.102}$

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

6 0

4 years ago

Air is compressed by an adiabatic compressor from 100 kPa and 20°C to 1.8 MPa and 400°C. Air enters the compressor through a 0.1

Tamiku [17]

Answer:

(a) the mass flow rate of air is 5.351 kg/s

(b) the input power required is 2090.786 kW

Explanation:

Given;

initial pressure, P₁ = 100 kPa

initial temperature, T₁ = 20 °C

Final pressure, P₂ = 1.8 MPa

Final temperature, T₂ = 400 °C

Inlet area of the compressor = 0.15 m²

outlet area of compressor = 0.078 m²

velocity of air = 30 m/s

Part (a) mass flow rate of air through the inlet

Mass flow rate = Area x velocity = density x volumetric rate

m = Av = ρV

from ideal gas law, PV = nRT and ρ = m/V

substitute these values in the above equations, we will have;

$m = \frac{PAv}{RT}$

$m = \frac{100000*0.15*30}{287*(20+273)}= 5.351 \ kg/s$

Part (b) the required power input

$W + m(h_1+\frac{v_1{^2}}{2}) = mh_2$

where;

W is the input power

m is the mass flow rate

h₁ is the initial enthalpy

h₂ is the final enthalpy

initial and final enthalpy are obtained from steam table using interpolation;

h₁ = 293.166 kJ

h₂ = 684.344 kJ

$W + m(h_1+\frac{v_1{^2}}{2}) = mh_2\\\\W = mh_2 - m(h_1+\frac{v_1{^2}}{2})\\\\W = 5.351 ( 684.344) - 5.351 (293.166 + \frac{30^2}{2000}) \\\\W = 3661.925 \ kW -1571.139 \ kW\\\\W = 2090.786 \ kW$

7 0

3 years ago

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