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FromTheMoon [43]
3 years ago
11

A single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of

5m. Determine the V.R, M.A and efficiency of the machine.​
Engineering
1 answer:
Allushta [10]3 years ago
4 0

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%

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Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum driver output
Vera_Pavlovna [14]

Answer:

112.5 watts

Explanation:

The output voltage (V) = 15 volts

The equivalent resistance for the speakers connected in parallel (R_T) is gotten by using the formula:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\\\But\ R_1=R_2=R_3=R_4=8\ ohm\\\\\frac{1}{R_T} =\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \\\\\frac{1}{R_T} =\frac{1+1+1+1}{8} \\\\\frac{1}{R_T} =\frac{4}{8} \\\\R_T=\frac{8}{4} \\\\R_T=2\ ohm

The current flowing through the amplifier (I) is:

I = V / R_T = 15 / 2 = 7.5 A

The audio power (P) when outputting this maximum voltage is given by:

P = V² / R_T = I²R_T. Therefore:

P = V² / R_T = 15² / R_T = 112.5 watts

6 0
3 years ago
Air enters the compressor of an ideal Brayton refrigeration cycle at 100 kPa, 270 K. The compressor pressure ratio is 3, and the
trapecia [35]

Answer:

its 56 78 92

Explanation:

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6 0
3 years ago
Water (cp = 4180 J/kg·°C) enters the 2.5 cm internal diameter tube of a double-pipe counter-flow heat exchanger at 17°C at a rat
aksik [14]

Answer:

Length = 129.55m, 129.55m

Explanation:

Given:

cp of water = 4180 J/kg·°C

Diameter, D = 2.5 cm

Temperature of water in =  17°C

Temperature of water out = 80°C

mass rate of water =1.8 kg/s.

Steam condensing at 120°C

Temperature at saturation = 120°C

hfg of steam at 120°C = 2203 kJ/kg

overall heat transfer coefficient of the heat exchanger = 700 W/m2 ·°C

U = 700 W/m2 ·°C

Since Temperature of steam is at saturation,

temperature of steam going in = temperature of steam out = 120°C

Energy balance:

Heat gained by water = Heat loss by steam

Let specific capacity of steam = 2010kJ/Kg .°C

Find attached the full solution to the question.

3 0
3 years ago
Two mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surro
loris [4]

Answer:

(a)The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) The final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) The volume flow rate at exit is V₃ =V₁ + V₂

Explanation:

Solution

Now

The system comprises of two inlets and on exit.

Mass flow rate enthalpy of fluid from inlet -1 be m₁ and h₁

Mass flow rate enthalpy of fluid from inlet -2 be m₂ and h₂

Mass flow rate enthalpy of fluid from  exit be m₃ and h₃

Mixing chambers do not include any kind of work (w = 0)

So, both  the kinetic and potential energies of the fluid streams are usually negligible (ke =0, pe =0)

(a) Applying the mass balance of mixing chamber, min = mout

Applying the energy balance of mixing chamber,

Ein = Eout

min hin =mout hout

miCpT₁ + m₂CpT₂ +Qin =m₃CpT₃

T₃ = miCpT₁/m₃CpT₃ + m₂CpT₂/m₃CpT₃ + Qin/m₃CpT₃ +

T₃ =m₁T₁/m₃+ m₂T/m₃ + Qin/m₃Cp

The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) From the ideal gas equation,

v =RT/PT

v₃ = RT₃/P₃

The volume flow rate at the exit, V₃ =m₃v₃

V₃ = m₃ RT₃/P₃

Substituting the value of T₃, we have

V₃=m₃ R/P₃ (=m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp)

V₃ =  R/P₃ (m₁T₁+ m₂T₂ + Qin/Cp)

Now

The mixing process occurs at constant pressure P₃=P₂=P₁.

Hence V₃ becomes:

V₃=m₁RT₁/P₁ +m₂RT₂/P₂ + RQin/P₃Cp

V₃ =V₁ + V₂ + RQin/P₃Cp

Therefore, the final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) Now for an adiabatic mixing, Qin =0

Hence V₃ becomes:

V₃ =V₁ + V₂ + r * 0/P₃Cp

V₃ =V₁ + V₂ + 0

V₃ =V₁ + V₂

Therefore the volume flow rate at exit is V₃ =V₁ + V₂

8 0
3 years ago
A very large plate is placed equidistant between two vertical walls. The 10-mm spacing between the plate and each wall is filled
Vikentia [17]

Answer:

Force per unit plate area is 0.1344 N/m^{2}

Solution:

As per the question:

The spacing between each wall and the plate, d = 10 mm = 0.01 m

Absolute viscosity of the liquid, \mu =1.92\times 10^{- 3} Pa-s

Speed, v = 35 mm/s = 0.035 m/s

Now,

Suppose the drag force that exist between each wall and plate is F and F' respectively:

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F = \tau A

where

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A = Cross - sectional Area

Therefore,

Net Drag Force, F = (\tau ' +\tau '')A

\frac{F}{A} = \tau ' +\tau ''

Also

F = \frac{\mu v}{d}

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\mu = dynamic coefficient of viscosity

Pressure, P = \frac{F}{A}

Therefore,

\frac{F}{A} = \frac{\mu v}{d} + \frac{\mu v}{d} = 2\frac{\mu v}{d}

\frac{F}{A} = 2\frac{1.92\times 10^{- 3}\times 0.035}{0.010} = 0.01344 N/m^{2}

8 0
3 years ago
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