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Katarina [22]
3 years ago
11

What is the most common type of pump?

Engineering
1 answer:
Charra [1.4K]3 years ago
7 0

Answer:

The most common type of pumps are Positive displacement and Non positive displacement pumps.

Explanation:

<u> Pumps are two types:</u>

     (A) Positive displacement pump

             (a)Gear pump

                      (1) Ge rotor pumps

                      (2)Internal gear pumps

                      (3)Lobe pumps

                      (4) External gear pumps

               (b)Piston pump

                       (1)Radial piston

                       (2)Axial piston

               (c)vane pump

        (B) Non positive displacement pump

             (a) Centrifugal pump

   

     

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Which bulb has the lowest total cost of operation? (a) Incandescent (b) Fluorescent (c) LED
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Explanation:

If we ignore the initial procurement cost of the items the operational cost of any device consuming electricity is given by

Cost=Energy\times cost/unit

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4 years ago
A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at
DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

dQ = m \times c_p \times (T_2 -T_1)

For ideal gas, c_p = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

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3 0
3 years ago
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astra-53 [7]

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6 0
3 years ago
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Answer:

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System.out.println("Tails"); }

else {

System.out.println("Heads"); }

rand = 0;

}

}

7 0
3 years ago
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