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Shkiper50 [21]
3 years ago
14

A two-stroke CI. engine delivers 5000 kWwhile using 1000 kW to overcome friction losses. It consumes 2300 kg of fuel per hour at

an air-fuel ratio of 20 to 1. The heating value of fuel is 42000 kJkg Find the (a) indicated power (b) mechanical efïiciency, (c) air consumption per hr, (d) indicated themal efficiency, and (e) brake thermal efficiency.
Engineering
1 answer:
Lady_Fox [76]3 years ago
5 0

Answer:

(a) Indicating power(IP)=6000 KW

(b)   \eta_{mech}=0.833

(c) Consumption of air per hour =46000 kg/hr

(d) \eta_{BPth}=0.1865

Explanation:

Break power(BP) =5000 KW

Friction power(FP)=1000 KW

Consumption of fuel per hour=2300 kg/hr

CV=42000 KJ/kg

We know that

Indicating power(IP)=Break power(BP)+Friction power(FP)

⇒IP=5000+1000 KW

  IP=6000 KW

(a)

Indicating power(IP)=6000 KW

(b)

Mechanical efficiency  \eta_{mech}=\dfrac{BP}{IP}

 \eta_{mech}=\dfrac{5000}{6000}      

  \eta_{mech}=0.833

(c)

Air fuel ratio=\dfrac{mass \ of \ air}{mass \ of \ fuel}

consumption of air per hour=20\times2300 kg/hr

So consumption of air per hour =46000 kg/hr

(d)

Break thermal efficiency  \eta_{BPth}=\dfrac{IP}{\dot{m_f}\times CV}

\dot{m_f}=\dfrac{2300}{3600}  

                      =0.638 kg/s

\eta_{BPth}=\dfrac{5000}{{0.638}\times 42000}

\eta_{BPth}=0.1865

 

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4

Explanation:

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A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I
Liula [17]

Answer:

  • fire brick / common brick : 1218 °C
  • common brick / magnesia : 1019 °C
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Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

  R = d/k

so the thermal resistances of the layers of furnace wall are ...

  R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

  R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

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The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

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The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

  fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

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For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

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<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0
3 years ago
A labor-intensive process to manufacture a product has a fixed cost of $338,000 and a variable cost of $143 per unit. An automat
ozzi

Answer:

no of unit is 17941

Explanation:

given data

fixed cost = $338,000

variable cost = $143 per unit

fixed cost = $1,244,000  

variable cost = $92.50 per unit

solution

we consider here no of unit is = n

so here total cost of labor will be sum of fix and variable cost i.e

total cost of labor = $33800 + $143 n  ..........1

and

total cost of capital intensive  = $1,244,000 + $92.5 n   ..........2

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solve we get

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Answer:

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Therefore, the gas gained heat by an amount of 142 kJ.

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