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Andre45 [30]
3 years ago
5

When using levers like scissors or hedge clippers, what can be done to increase the cutting force so that you don’t have to sque

eze so hard?
Engineering
1 answer:
enot [183]3 years ago
6 0

An inclined plane has a small force applied over a large distance. Ramps, ladders and stairs are inclined planes. Both wedges and pulleys make work easier by changing the direction of the force.

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The pump of a water distribution system is powered by a 6-kW electric motor whose efficiency is 95 percent. The water flow rate
Sonja [21]

Answer:

a) Mechanical efficiency (\varepsilon)=63.15%  b) Temperature rise= 0.028ºC

Explanation:

For the item a) you have to define the mechanical power introduced (Wmec) to the system and the power transferred to the water (Pw).

The power input (electric motor) is equal to the motor power multiplied by the efficiency. Thus, Wmec=0.95*6kW=5.7 kW.

Then, the power transferred (Pw) to the fluid is equal to the flow rate (Q) multiplied by the pressure jump \Delta P. So P_W = Q*\Delta P=0.018m^3/s * 200x10^3 Pa=3600W.

The efficiency is defined as the ratio between the output energy and the input energy. Then, the mechanical efficiency is \varepsilon=3.6kW/5.7kW=0.6315=63.15\%

For the b) item you have to consider that the inefficiency goes to the fluid as heat. So it is necessary to use the equation of the heat capacity but in a "flux" way. Calling <em>H</em> to the heat transfered to the fluid, the specif heat of the water and \rho the density of the water:

[tex]H=(5.7-3.6) kW=\rho*Q*c*\Delta T=1000kg/m^3*0.018m^3/s*4186J/(kg \ºC)*\Delta T[/tex]

Finally, the temperature rise is:

\Delta T=2100/75348 \ºC=0.028 \ºC

7 0
3 years ago
In a vapor-compression refrigeration cycle, ammonia exits the evaporator as saturated vapor at -10°C. The refrigerant enters the
sergij07 [2.7K]

Answer:  (a) 0,142 (b) 52.99 (c) 2.83 (d) 88.26

Explanation:

If the refrigarating capacity is 150kw

(a) the mass flow rate of refrigerant, in kilograms per second  is 0.142

(b) the power input to the compressor, in kilowatts is 52.99

(c) the coefficient of performance is 2.83

(d) the isentropic compressor efficiency is 68.6 per cent

8 0
3 years ago
Technician A says that after replacing a power steering hose, the system should be flushed, refilled, and bled. Technician B say
dedylja [7]
The. Answer will be D
3 0
4 years ago
1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem
BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁  = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}

P_2= \dfrac{1060}{5^{1.33}}  = 124.65 \ kPa

\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2}   \right

5^{0.33} = \left \dfrac{T_1}{T_2}   \right

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3×c_v×(T₂ - T₁)

c_v = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86  kJ/(kg·K)

We use cv given by  (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3×c_v×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = U_g = h_g -pV_g

h_g = 2777.12 kJ/kg

V_g = 0.194349 m³/kg

p = 1000 kPa

U_{g1} = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, U_{g1} + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0
3 years ago
How is the fuel introduced into the Diesel engine?
Ugo [173]

Answer:

diesel fuel is pumped at high pressure to the injectors which are responsible for entering the fuel into the combustion chamber,

when the piston is at the top the pressure is so high that it explodes the fuel (diesel) that results in a generation of mechanical power

5 0
3 years ago
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