All of these are true about using adhesive EXCEPT:

A. A compound gear train is composed of four gears: E,F,G,H. Gear E has 15 teeth and is meshed with gear F. Gear F has 25 teeth

cricket20 [7]

Answer:

What subject is this in?? That would help a lot knowing

Explanation:

5 0

3 years ago

A 75-hp motor that has an efficiency of 91.0% is worn-out and is replaced by a motor that has a high efficiency 75-hp motor that

damaskus [11]

Answer:

the reduction in the heat gain is 2.8358 kW

Given that;

Shaft outpower of a motor = 75 hp = ( 75 × 746 ) = 55950 W

Efficiency of motor = 91.0% = 0.91

High Efficiency of the motor = 95.4% = 0.954

now, we know that, efficiency of motor is defined as; = /

where is the electric input given to the motor

so

= /

we substitute

= 55950 W / 0.91

= 61483.5 W

= 61.4835 kW

now, the electric input given to the motor due to increased efficiency will be;

= /

we substitute

= 55950 W / 0.954

= 58647.79 W

= 58.6477 kW

so the reduction of the heat gain of the room due to higher efficiency will be;

Q = -

we substitute

Q = 61.4835 kW - 58.6477 kW

Q = 2.8358 kW

Therefore, the reduction in the heat gain is 2.8358 kW

Explanation: i hope this answer your question if this is wrong or correct please let me know.also no trying to be rude but can you sent me like a thanks?

8 0

3 years ago

A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 2

Furkat [3]

Answer:

$P_{1} = 403,708\,kPa\,(58.553\,psi)$

Explanation:

Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:

$\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}$

The initial pressure is:

$P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}$

The speed at outlet is:

$v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}$

$v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }$

$v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )$

The initial pressure is:

$P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}$

$P_{1} = 403,708\,kPa\,(58.553\,psi)$

7 0

4 years ago

Read 2 more answers

If two motors are to be started at the same time, the standby power system must have the capacity to provide the _____ of the st

Gwar [14]

Answer:

Sum.

Explanation:

An electric motor can be defined as a machine which is typically used for the conversion or transformation of electrical energy into mechanical energy.

Basically, the mode of operation of an electric motor is simply to supply the motor with an alternating current (AC) voltage to one end, which is then used to power the axle (metal rod) at the other end of the motor.

Two motors can be made to run simultaneously in a synchronous manner through the use of a standby power system that provides the required level of power (alternating current and voltage).

Thus, if two motors are to be started at the same time, the standby power system must have the capacity to provide the sum of the starting kilovolt-ampere (KVA) values for the TWO motors.

7 0

3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

shusha [124]

Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

6 0

3 years ago