Which of the following is a advantage of a chain and sprocket over a pulley and belt system?

A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he

sergejj [24]

Answer:

T₂ =93.77 °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure ,P₂= 367+1 = 368 kPa

Lets take temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

$\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}$

${T_2}=T_1\times \dfrac{P_2}{P_1}$

Now by putting the values in the above equation we get

${T_2}=300\times \dfrac{368}{301}\ K$

The temperature in °C

T₂ = 366.77 - 273 °C

T₂ =93.77 °C

8 0

3 years ago

How does heat conduction differ from convection?

Helga [31]

Explanation:

Conduction:

Heat transfer in the conduction occurs due to movement of molecule or we can say that due to movement of electrons in the two end of same the body. Generally, phenomenon of conduction happens in the case of solid . In conduction heat transfer takes places due to direct contact of two bodies.

Convection:

In convection heat transfer of fluid takes place due to density difference .In simple words we can say that heat transfer occur due to motion of fluid.

7 0

3 years ago

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At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t

nlexa [21]

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

$PGP = \rho *\frac{\pi r^2}{2} *V*e \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}$

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

5 0

3 years ago

Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotation

lions [1.4K]

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

h2 - h1 = 0.09 m

The outer part will be 9 cm higher than the inner part.

3 0

3 years ago

Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate

Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - $t_{A2}$ ) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - $t_{A2}$ = 1100/3

$t_{A2}$ = 733.33 K

$\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}$

Where

$\Delta \bar{t}_{a}$ = Arithmetic mean temperature difference

$t_{A_{1}$ = Inlet temperature of the gas = 1100 K

$t_{A_{2}$ = Outlet temperature of the gas = 733.33 K

$t_{B_{1}$ = Inlet temperature of the air = 300 K

$t_{B_{2}$ = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

$\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K$

Hence, from;

$\dot{Q} = UA\Delta \bar{t}_{a}$ , we have

5912500 = 90 × A × 341.67

$A = \frac{5912500 }{90 \times 341.67} = 192.3 \, m^2$

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0

3 years ago