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Strike441 [17]
4 years ago
5

How does the modern Diesel engine achieve higher power output without the use of higher compression ratio?

Engineering
1 answer:
krok68 [10]4 years ago
3 0

Explanation:

We know that the efficiency of diesel engine given as follows

\eta=1-\dfrac{\rho^{\gamma}-1}{r^{\gamma-1}\gamma(\rho-1)}

Where r is the compression ratio ,ρ is the cut off ratio and γ is the heat capacity ratio .

Here given that we want to increase the efficiency with out changing the compression ratio.

So from the efficiency formula we can say that when cut off ratio is decreases then the efficiency of diesel cycle will increases.When efficiency is high it means that the power output of engine is high.

The second method to increase the efficiency of diesel engine is that ,use the anti knocking diesel.Use premium diesel instead of use the normal diesel fuel.

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You are traveling along an interstate highway at 32.0 m/s (about 72 mph) when a truck stops suddenly in front of you. You immedi
PolarNik [594]

Answer:

a= - 2.6 m/s².

Explanation:

u = 32 m/s

The speed after 6 s is half of u

v= \dfrac{32}{2}=16\ m/s

t= 6 s

The average acceleration = a

We know v = u +at

v=final velocity

u=initial velocity

Now by putting the values in the above equation

16= 32 + a x 6

a=\dfrac{16-32}{6}\ m/s^2

a=-2.6\ m/s^2

Therefore the acceleration will be - 2.6 m/s².

a= - 2.6 m/s².

Negative indicates that velocity and acceleration is is opposite direction.

8 0
3 years ago
A blown fuse or tripped circuit breaker shows
aleksklad [387]

Answer:

Your house is in need of a service upgrade, or it may indicate that your house has too few circuits.

Explanation:

It's a sign that you are making excessive demands on the circuit and need to move some appliances and devices to other circuits.

6 0
3 years ago
Routers cannot be used to cut through material.<br><br> True<br> False
mario62 [17]

Answer:

Yes a router can be used to cut right through wood and sometimes it makes sense to do so. It leaves nice clean edges, can cut sharp curves and can follow a template

Explanation:

hope thats right

7 0
3 years ago
Read 2 more answers
A small grinding wheel is attached to the shaft of an electric motor which has a rated speed of 3600 rpm. When the power is turn
Ostrovityanka [42]

Answer: a) 150 rev. b) 2105 rev.

Explanation:

a) Assuming a uniformly accelerated motion, we can use the equivalent kinematic equations, replacing linear variables by angular ones.

In order to get the number of revolutions executed, we can use this:

ωf² - ω₀² = 2 γ Δθ (1)

For the first part, we know that ω₀ = 0 (as it starts from rest).

We can find out the value of angular acceleration γ, just applying the definition of angular acceleration, as the change in angular velocity, regarding time, as follows:

γ = (ωf - ω₀) / Δt (2)

As we would want to use SI units, it is advisable to convert the value of ωf, from rpm to rad/sec.

3600 rev/min . (1min/60 sec) . (2π rad/rev) = 120π rad/sec

Replacing in (2), we get γ:

γ = 120 π / 5 rad/sec² = 24 π rad/sec²

Replacing in (1) and solving for Δθ:

Δθ = 120² π² / 2. 24 π = 300 π rad

As 1 rev = 2π rad, Δθ = 150 rev

b) For the second part, we can use exactly the same equations, taking into account that ω₀ = 120 π rad/sec, and that ωf = 0.

The new value for γ is as follows:

γ = -120π  / 70 rad/sec² = -1.71 rad/sec²

Replacing in (1) and solving for Δθ, we get:

Δθ = -120² π² / 2. (-1.71) π = 4210 π rad

As 1 rev = 2π rad, Δθ = 2105 rev

7 0
4 years ago
Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−
sergeinik [125]

Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

v_{i} = 3.986 m/s

Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

v_{e} = 19.33 m/s

Now, the power output can be calculated from the energy balance eqn:

P_{out} = -\dot{m}W_{s}

P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}

P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW

4 0
3 years ago
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