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d1i1m1o1n [39]
3 years ago
11

In an air standard diesel cycle compression starts at 100kpa and 300k. the compression ratio is 16 to 1. The maximum cycle tempe

rature is 2031K. Determine the thermal efficiency.
Engineering
1 answer:
KIM [24]3 years ago
7 0

Answer:

\eta=0.60

Explanation:

Given :Take \gamma=1.4 for air

      P_1=100 KPa  ,T_1=300K

  \frac{V_1}{V_2}=r ⇒ r=16

As we know that  

   T_2=T_1(r^{\gamma-1})

So T_2=300\times 16^{\gamma-1}

  T_2=909.42K

Now find the cut off ration \rho

      \rho=\frac{V_3}{V_2}  

         \frac{V_3}{V_2}=\frac{T_3}{T_2}

\rho=\frac{2031}{909.42}

 \rho=2.23

So efficiency of diesel engine

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

Now by putting the all values

\eta =1-\dfrac{2.23^{1.4}-1}{1.4\times 16^{1.4-1}(2.23-1)}

So \eta=0.60

So the efficiency of diesel engine=0.60

     

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Answer:

A benchmark

Explanation:

Most times a benchmark serves as the better measure when assessing a computer's performance, this is because CPU speeds can only evaluate an aspect of a computer's performance whereas a benchmark offers the advantage of measuring all the aspects of a computer's performance for a specific type of computing problem.

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3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
max2010maxim [7]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

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3 years ago
A 6V battery is connected in series with two bulbs. What would the voltage drop be across each bulb?
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3V

Explanation:

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When adding two 8 bit binary numbers, which of the following statements is true?
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2 years ago
The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any frictio
Dmitrij [34]

Complete Question

The cars of a roller-coaster ride have a speed of 19.0 km/h as they pass over the top of the circular track. Neglect any friction and calculate their speed v when they reach the horizontal bottom position. At the top position, the radius of the circular path of their mass centers is 21 m, and all six cars have the same mass.V = -18 m What is v?X km/h

Answer:

v=23.6m/s

Explanation:

Velocity v_c=18.0km/h

Radius r=21m

initial velocity uu=19=>5.27778

Generally the equation for Angle is mathematically given by

\theta=\frac{v_c}{2r}

\theta=\frac{18}{2*21}

\theta=0.45

\theta=25.7831 \textdegree

Generally

Height of mass

h=\frac{rsin\theta}{\theta}

h=\frac{21sin25.78}{0.45}

h=20.3m

Generally the equation for Work Energy is mathematically given by

0.5mv_0^2+mgh=0.5mv^2

Therefore

v=\sqrt{u^2+2gh}

v=\sqrt{=5.27778^2+2*9.81*20.3}

v=23.6m/s

3 0
3 years ago
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