Air flows at 45m/s through a right angle pipe bend with a constant diameter of 2cm. What is the overall force required to keep the pipe bend in place? Is it: a) 0.76 N b) 1.08 N c) 1.52 N d) 0.25 N e) 2.56 N

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced  read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

[tex] P = T\times \omega[/tex]

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

[tex]\omega[/tex] angular speed [tex]=\frac{2 \pi N}{60} [/tex]

Therefore Power is

[tex]P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65  watt[/tex]

P = 80.922 KW

Answer:

1.29 lbf

Explanation:

Weight is a force, it is the product of a mass by the acceleration of gravity.

f = m * a

Gallons are a unit of volume. The relationship of volume with mass is:

m = V * δ

δ is the density (mass per unit of volume)

Then the weight of a liquid is:

w = V * δ * g

Rearranging:

δ = w / (V * g)

The acceleration of gravity on Earth is 32.2 ft/s^2

δ = 50 / (20 * 32.2) = 0.078 lbm/gallon

Knowing this and the gravity on the Moon we can calculate how much would 3 gallons of this liquid weight on the Moon.

w = V * δ * g

w = 3 * 0.078 * 5.51 = 1.29 lbf

Explanation:

Step1

Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.

The expression for absolute pressure is given as follows:

[tex]P_{ab}=P_{g}+P_{atm}[/tex]

Here, [tex]P_{ab}[/tex] is absolute pressure, [tex]P_{g}[/tex] is gauge pressure and[tex]P_{atm}[/tex] is atmospheric pressure.

Step2

Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.  

Answer:

Mass of the combined system = 201.8 Kg

Weight of the combined system = 1977.64 N

Explanation:

Given:

Mass of the plastic tank = 2 kg

Volume of the plastic tank = 0.2 m³

Density of water = 999 Kg/m³

Now, the volume of water in the tank =  volume of tank = 0.2 m³

Also,

Mass = Density × Volume

therefore,

Mass of water = 999 × 0.2 = 199.8 Kg

Thus, the combined mass = Mass of tank + Mass of water in the tank

or

the combined mass = 2 + 199.8 = 201.8 Kg

Also,

Weight = Mass × Acceleration due to gravity

or

Weight = 201.8 × 9.8 = 1977.64 N

Answer:

105mg of calcium chloride [tex]CaCl_{2}[/tex]

Explanation:

The molecular formula of calcium chloride is [tex]CaCl_{2}[/tex] and the molecular formula of soda ash is [tex]Na_{2}CO_{3}[/tex]

First of all you should write the balanced reaction between both compounds, so:

[tex]CaCl_{2}+Na_{2}CO_{3}=2NaCl+CaCO_{3}[/tex]

Then you should have the molar mass of the two compounds:

Molar mass of [tex]CaCl_{2}=110.98\frac{g}{mol}[/tex]

Molar mass of [tex]Na_{2}CO_{3}=105.98\frac{g}{mol}[/tex]

Now with stoichiometry you can find the mass of calcium chloride that reacts with 100mg of soda ash, so:

[tex]100mgNa_{2}CO_{3}*\frac{1molNa_{2}CO_{3}}{105.98gNa_{2}CO_{3}}*\frac{1molCaCl_{2}}{1molNa_{2}CO_{3}}*\frac{110.98gCaCl_{2}}{1molCaCl_{2}}=105mgCaCl_{2}[/tex]

Answer with Explanation:

In a tensile test in an tensile machine the following steps followed to obtain the young's modulus:

1) A specimen of material with known cross sectional area and gauge length is loaded axially.

2) The load is increased and the corresponding change in the gauge length of the material is noted by using a deflectometer or a dial gauge.

3) The ratio between Force applied and the nominal cross section known as stress is calculated for each applied value of force until the material fails.

4) The ratio between the change in gauge length and the original gauge length is found this ratio is known as strain.

5) A graph is plotted between the values calculated in step 3 and 4 above.

6) The slope of the line at the origin of the graph gives the young's modulus of the material.

Answer:

Star topology

Explanation:

Topology:

Topology is the connection of networks .These network connect by nodes.

Type of topology:

1.Bus topology

2.Star topology

3.Ring topology

4.Mesh topology

Star topology is the most reliable topology .Because failure of node node does not affect the other nodes.In star topology  network  are connect in star form.

Ring topology is the most undesirable topology.In ring topology network connects in the ring form.In ring topology failure of one node affect the whole network.

Answer:

23.2 m/s

Explanation:

A jetpack appliang a force on a body would result in an acceleration following Newton's equation:

f = m * a

Rearranging:

a = f / m

The problem doesn't state a mass, so I'll use mine, which is

m = 72 kg

Then:

a = 334 / 72 = 4.64 m/s^2

The equation for speed under constant acceleration is:

V(t) = V0 + a * t

Since I was standing before turning the jetpack on

V0 = 0

So:

V(5) = 0 + 4.64 * 5 = 23.2 m/s

Answer:

The correct answer is option 'd':Fluid distribution system

Explanation:

In fluid mechanics 'manifold' is a term used to define a pipe into which many other pipes drain or a pipe which branches into many different smaller pipes.

Manifolds are of many types such as exhaust manifold,hydraulic manifold, inlet manifold ,e.t.c.

Manifolds serve an important purpose as they are used to exhaust the gas from different piston chambers into a single exhaust.

Answer:

The heat is transferred is at the rate of 752.33 kW

Solution:

As per the question:

Temperature at inlet, [tex]T_{i} = 20^{\circ}C[/tex] = 273 + 20 = 293 K

Temperature at the outlet, [tex]T_{o} = 200{\circ}C[/tex] = 273 + 200 = 473 K

Pressure at inlet, [tex]P_{i} = 80 kPa = 80\times 10^{3} Pa[/tex]

Pressure at outlet, [tex]P_{o} = 800 kPa = 800\times 10^{3} Pa[/tex]

Speed at the outlet, [tex]v_{o} = 20 m/s[/tex]

Diameter of the tube, [tex]D = 10 cm = 10\times 10^{- 2} m = 0.1 m[/tex]

Input power, [tex]P_{i} = 400 kW = 400\times 10^{3} W[/tex]

Now,

To calculate the heat transfer, [tex]Q[/tex], we make use of the steady flow eqn:

[tex]h_{i} + \frac{v_{i}^{2}}{2} + gH  + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH' + p_{s}[/tex]

where

[tex]h_{i}[/tex] = specific enthalpy at inlet

[tex]h_{o}[/tex] = specific enthalpy at outlet

[tex]v_{i}[/tex] = air speed at inlet

[tex]p_{s}[/tex] = specific power input

H and H' = Elevation of inlet and outlet

Now, if

[tex]v_{i} = 0[/tex] and H = H'

Then the above eqn reduces to:

[tex]h_{i} + gH + Q = h_{o} + \frac{v_{o}^{2}}{2} + gH + p_{s}[/tex]

[tex]Q = h_{o} - h_{i} + \frac{v_{o}^{2}}{2} + p_{s}[/tex]                (1)

Also,

[tex]p_{s} = \frac{P_{i}}{ mass, m}[/tex]

Area of cross-section, A = [tex]\frac{\pi D^{2}}{4} =\frac{\pi 0.1^{2}}{4} = 7.85\times 10^{- 3} m^{2}[/tex]

Specific Volume at outlet, [tex]V_{o} = A\times v_{o} = 7.85\times 10^{- 3}\times 20 = 0.157 m^{3}/s[/tex]

From the eqn:

[tex]P_{o}V_{o} = mRT_{o}[/tex]

[tex]m = \frac{800\times 10^{3}\times 0.157}{287\times 473} = 0.925 kg/s[/tex]

Now,

[tex]p_{s} = \frac{400\times 10^{3}}{0.925} = 432.432 kJ/kg[/tex]

Also,

[tex]\Delta h = h_{o} - h_{i} = c_{p}\Delta T =c_{p}(T_{o} - T_{i}) = 1.005(200 - 20) = 180.9 kJ/kg[/tex]

Now, using these values in eqn (1):

[tex]Q = 180.9 + \frac{20^{2}}{2} + 432.432 = 813.33 kW[/tex]

Now, rate of heat transfer, q:

q = mQ = [tex]0.925\times 813.33 = 752.33 kW[/tex]

Answer:

First we must know the efficiency of the gearbox in which energy is lost by friction between its components, we divide this efficiency by the nominal power and find the real power.

With this value knowing the speed of rotation of the gearbox we select an engine that has a power just above.

Also keep in mind that the motor torque must be greater than the resistance offered by the gearbox and the 0.5Kw machine

Answer:

The velocity of the pulling plate is 14.98 m/s.

Explanation:

The shear stress caused by the 20 Newton force on the plate is given by

[tex]\tau =\frac{20}{Area}=\frac{20}{0.3\times 1.0}=66.66N/m^2[/tex]

Now by newton's law of viscosity we have

[tex]\tau =\mu\cdot \frac{\Delta u}{\Delta y}[/tex]

where

[tex]\mu [/tex] is the dynamic viscosity of the water at 20 degree

[tex]\frac{\Delta u}{\Delta y}[/tex] is the velocity gradient

Applying values we get

[tex]66.66=8.90\times 10^{-4}\times \frac{U}{0.2\times 10^-3}\\\\\therefore u=14.98m/s[/tex]

To find the weight of a 90 kg person on Earth, use the formula w = mg, where g is 9.80 m/s², resulting in a weight of 882 N to the nearest 1 N.

To calculate the weight of a person on Earth using Newton's law of universal gravitation, you use the formula:

w = mg

Where w is the weight in newtons (N), m is the mass in kilograms (kg), and g is the acceleration due to gravity on Earth, which is 9.80 m/s².

Given a person has a mass of 90 kg, their weight can be calculated as follows:

w = (90 kg) × (9.80 m/s²) = 882 N

Therefore, the weight of a 90 kg person standing on the surface of the Earth is 882 N to the nearest 1 N.

Answer: 0.57 hp,  24.2 Btu/min

Explanation:

Hi!

The power P delivered when rotating a crank at angular speed ω, applying a torque τ is given by:

P = τ*ω

In this case torque is τ = 12 in * 25 lbf = 300 lbf*in

Angular speed is 2 round per seconds, which is 120 RPM. Then power is:

P = 300*120 lbf*in*RPM

1 hp = (lbf*in*RPM) / 63,025

P = 0.57 hp

1 hp = 42,4 Btu/min

Then P = 24.2 Btu/min