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3 years ago

What is the ideal cooling system for low horsepower motor? For example1hp motor

Engineering

1 answer:

Papessa [141]3 years ago

7 0

Answer:

Air cooling.

Explanation:

Low power motors are supposed to be low cost, and they dissipate little heat. Therefore a low cost solution is ideal.

Air cooling can be achieved with very little cost. Fins can be added to a cast motor casing and a fan can be places on the shaft to use a small amount of the motor power to move air to cool it.

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The best grade of hardwood lumber that is generally available is _____

Vesnalui [34]

Answer:

FAS

Explanation:

first and second grade

5 0

2 years ago

According to the

zysi [14]

Answer:

The part of the system that is considered the resistance force is;

Explanation:

The simple machine is a system of pulley that has two pulleys

The effort, which is the input force at A gives the value of the tension at C and D which are used to lift the load B

Therefore, we have;

A = C = D

B = C + D = C + C = 2·C

∴ C = B/2

We have;

C = B/2 = A

Therefore, with the pulley only a force, A equivalent to half the weight, B, of the load is required to lift the load, B

The resistance force is the constant force in the system that that requires an input force to overcome in order for work to be done

It is the force acting to oppose the sum of the other forces system, such as a force acting in opposition to an input force

Therefore, the resistance force is the load force, B, for which the input force, A, is required in order for the load to be lifted.

3 0

3 years ago

A rigid tank contains 2 kg of N2 and 4 kg of Co2 at temperature of 25 C and 1 MPa. Find the partial pressure of each gas respect

lions [1.4K]

Answer: Partial pressures are 0.6 MPa for nitrogen gas and 0.4 MPa for carbon dioxide.

Explanation: Dalton's Law of Partial Pressure states when there is a mixture of gases the total pressure is the sum of the pressure of each individual gas:

$P_{total} = P_{1}+P_{2}+...$

The proportion of each individual gas in the total pressure is expressed in terms of mole fraction:

$X_{i}$ = moles of a gas / total number moles of gas

The rigid tank has total pressure of 1MPa.

Nitrogen gas:

molar mass = 14g/mol

mass in the tank = 2000g

number of moles in the tank: $n=\frac{2000}{14}$ = 142.85mols

Carbon Dioxide:

molar mass = 44g/mol

mass in the tank = 4000g

number of moles in the tank: $n=\frac{4000}{44}$ = 90.91mols

Total number of moles: 142.85 + 90.91 = 233.76 mols

To calculate partial pressure:

$P_{i}=P_{total}.X_{i}$

For Nitrogen gas:

$P_{N_{2}}=1.\frac{142.85}{233.76}$

$P_{N_{2}}$ = 0.6

For Carbon Dioxide:

$P_{total}=P_{N_{2}}+P_{CO_{2}}$

$P_{CO_{2}} = P_{total}-P_{N_{2}}$

$P_{CO_{2}}=1-0.6$

$P_{CO_{2}}=$ 0.4

Partial pressures for N₂ and CO₂ in a rigid tank are 0.6MPa and 0.4MPa, respectively.

4 0

2 years ago

What does the simplify command do

uranmaximum [27]

Answer:The simplify command is used to apply simplification rules to an expression. The simplify routine searches the expression for function calls, square roots, radicals, and powers and invokes the appropriate simplification procedures. For detailed information on the simplify command, see simplify/details.

Explanation:

4 0

2 years ago

The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable

Fittoniya [83]

Answer:

a) 35%

b) yes it can be improved by moving the tray near the top

Tray should be located ( 1 to 2 meters below surface )

max removal efficiency ≈ 70%

c) The maximum removal will drop as the particle settling velocity = 0.5 m/h

Explanation:

Given data:

flow rate = 8000 m^3/d

Detention time = 1h

depth = 3m

Full length movable horizontal tray : 1m below surface

a) Determine percent removal of particles having a settling velocity of 1m/h

velocity of critical sized particle to be removed = Depth / Detention time

= 3 / 1 = 3m/h

The percent removal of particles having a settling velocity of 1m/h ≈ 35%

b) Determine if the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray and the maximum removal efficiency

The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the

Total Maximum removal efficiency

= percent removal $_{above}$ + percent removal $_{below}$