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3 years ago

What is the process of a Diesel engine uses to convert fuel to mechanical energy

Engineering

1 answer:

faust18 [17]3 years ago

6 0

Answer:

A diesel engine is a type of heat engine that uses the internal combustion process to convert the energy stored in the chemical bonds of the fuel into useful mechanical energy. ... First, the fuel reacts chemically (burns) and releases energy in the form of heat.

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Trava [24]

Answer:

Explanation:

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3 years ago

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_______________ is an effective way to manage waste in a shop.

Murrr4er [49]

Answer:

To reduce waste.

Explanation:

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3 years ago

By increasing the cross-sectional area of the restriction, one can significantly increase the flow velocity. a) True b) False

natka813 [3]

Answer:

b)false

Explanation:

As we know that

Volume flow rate Q

Q = A x V

For constant volume flow rate,if velocity will increase then automatically area will decrease and vice versa.

Generally nozzle are used to increase the velocity and diffuser are used to decrease the exit velocity of flow.

So by increasing the cross sectional area of the restriction ,the velocity of the flow will decrease.

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3 years ago

1. (15) A truck scale is made of a platform and four compression force sensors, one at each corner of the platform. The sensor i

Elanso [62]

Answer:

a). 139498.24 kg

b). 281.85 ohm

c). 10.2 ohm

Explanation:

Given :

Diameter, d = 22 m

Linear strain, $\epsilon$ = 3%

= 0.03

Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = $0.03 \times 30 \times 10^9$

$\frac{P}{A} =0.03 \times 30 \times 10^9$

$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$

= 342119.44 N

For the four sensors,

Maximum weight = 4 x P

= 4 x 342119.44

= 1368477.76 N

Therefore, weight in kg is $m=\frac{W}{g}=\frac{1368477.76}{9.81}$

m = 139498.24 kg

b). Change in resistance

$k=\frac{\Delta R/R}{\Delta L/L}$

$\Delta R = k. \epsilon R$ , since $\epsilon= \Delta L/ L$

$\Delta R = 6.9 \times 0.03 \times 340$

$\Delta R = 70.38$ Ω

For 4 resistance of the sensors,

$\Delta R = 70.38 \times 4 = 281.52$ Ω

c). $k=\frac{\Delta R/R}{\epsilon}$

If linear strain,

$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$ , where k = 1

$\Delta R = \frac{\Delta L}{L} \times R$

$\Delta R = 0.03 \times 340$

$\Delta R = 10.2$ Ω

4 0

3 years ago

Three single-phase, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a three-phase 460/208 V transformer bank. The eq

evablogger [386]

Answer:

A) attached below

B) I₁ = 18.1 A , I₂ = 69.39 A

C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%

Explanation:

A) Schematic diagram attached below

attached below

B) magnitude of primary and secondary winding currents

I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )

VL = Line voltage = 208

cos∅ ( power factor ) = 0.8

P = 20 * 10^3 watts

insert values into equation 1

I₂ = 69.39 A

I₁ ( primary current ) = I₂V2 / V1

I₁ = ( 69.39 * 120 ) / 460 = 18.1 A

C ) Calculate the Primary voltage magnitude and the Voltage regulation

V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )

= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )

= 460 + 40.544 ∠ 100.3°

∴ V( magnitude) = 454.5 ∠ 5.04° V

Voltage regulation

= ((Vmag - V1) / V1 )) * 100

= (( 454.5 - 460 / 460 )) * 100

= -1.195 % ≈ -1.2%

8 0

3 years ago