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guapka [62]
3 years ago
12

A cold air-standard Diesel cycle has a compression ratio of 18 and a cutoff ratio of 1.5. Determine the maximum temperature of t

he air and the rate of heat addition to this cycle when it produces 150 kW of power and the state of the air at the beginning of the compression is 90 kPa and 57 °C. Use constant specific heats at room temperature.
Engineering
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

\dot Q_{in} = 228.659\,kW, T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)

Explanation:

The ideal efficiency of the Diesel cycle is given by this expression:

\eta_{th} = \left\{1 - \frac{1}{r^{k-1}} \cdot \left[\frac{r_{c}^{k}-1}{k\cdot (r_{c}-1)} \right]\right\}\times 100\%

Where r and r_{c} are the compression and cutoff ratios, respectively.

\eta_{th} = \left\{1-\frac{1}{18^{0.4}}\cdot \left[\frac{1.5^{1.4}-1}{1.4\cdot (1.5-1)} \right] \right\}\times 100\%

\eta_{th} = 65.648\,\%

The heat addition to the cycle is:

\dot Q_{in} = \frac{\dot W}{\eta_{th}}

\dot Q_{in} = \frac{150\,kW}{0.656}

\dot Q_{in} = 228.659\,kW

The temperature at state 2 is:

T_{2} = T_{1} \cdot r^{k-1}

T_{2} = (330.15\,K)\cdot 18^{0.4}

T_{2} = 1049.108\,K\,(775.958\,^{\textdegree}C)

And the temperature at state 3 is:

T_{3} = T_{2}\cdot r_{c}

T_{3} = (1.5)\cdot (1049.108\,K)

T_{3} = 1573.662\,K\,(1300.512\,^{\textdegree}C)

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Answer:

Fuel efficiency for highway = 114.08 miles/gallon

Fuel efficiency for city = 98.79 miles/gallon

Explanation:

1 gallon = 3.7854 litres

1 mile = 1.6093 km

Let's first convert the efficiency to km/gallon:

48.5 km/litre = (48.5 * 3.7854) km/gallon

48.5 km/litre =  183.5919 km/gallon (highway)

42.0 km/litre = (42.0 * 3.7854) km/gallon

42.0 km/litre = 158.9868 km/gallon (city)

Next, we convert these to miles/gallon:

183.5919 km/gallon = (183.5919 / 1.6093) miles/gallon

183.5919 km/gallon = 114.08 miles/gallon (highway)

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3 0
3 years ago
Two very long concentric cylinders of diameters D1 = 0.42 m and D2 = 0.5 m are maintained at uniform temperatures of T1 = 950 K
MrRa [10]

Answer:

Q=33.34 KW/m

Explanation:

Given that

D₁=0.42 m

A₁= π D₁ L

For unit length

A₁= π D₁ = 0.42 π m²

D₂=0.5 m

A₂= 0.5 π m²

ε₁= 1 ,ε₂= 0.55

T₁=950 K  ,T₂ = 500 K

Q=\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{A_1\epsilon _1}+\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+\dfrac{1}{A_1F_{12}}}

F₁₁+F₁₂= 1

F₁₁= 0

So, F₁₂= 1

Q=A_1\dfrac{\sigma (T_1^4-T_2^4)}{\dfrac{1-\varepsilon _1}{\epsilon _1}+A_1\dfrac{1-\varepsilon _2}{A_2\epsilon _2}+1}

Q=0.42\pi \dfrac{5.67\times 10^{-8}(950^4-500^4)}{\dfrac{1-1}{1}+0.42\pi\times \dfrac{1-0.55}{0.5\pi\times 0.55}+1}

Q=33.34 KW/m

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Marie and James are bubbling dry pure nitrogen (N2) through a tank of liquid water (H2O) containing ethane (C2H6). The vapor str
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What is the pitch circle diameter of a 50-tooth spur gear having a circular pitch of 0.375 inch
Dmitrij [34]

Answer:

Pitch circle diameter = 5,66 inch

Explanation:

<u>According to the pitch circle diameter formula</u>

      <em>Dc= Dp - 2b</em> ; where Dp= primitive diameter, b= dedendum of the tooth and also the dedendum formula is b= 1,25 Pc/π.

     The definition of primitive diameter is equal to (Pc*Z)/π ; Pc= circular pitch and Z= number of teeths.

    <u>Replacing in the formula </u>

        ⇒ Dc= (Pc*Z)/π - 2* 1,25 Pc/π,    Pc= 0,375 ; Z=50

  In conclusion the pitch circle diameter is equal to:

            Dc= 5,66 inch.      

   

         

<h3>           </h3>

 

   

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