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zmey [24]
3 years ago
14

A petrol engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency, and how much powe

r is rejected to the ambient surroundings?
Engineering
1 answer:
hoa [83]3 years ago
5 0

Answer:

efficiency =42.62%

AMOUNT OF POWER REJECTED IS 20.080 kW

Explanation:

given data:

power 20 hp

heat energy = 35kW

power production = 20 hp = 20* 746 W = 14920 Watt   [1 hp =746 watt]

efficiency = \frac{power}{heat\ required}

efficiency = \frac{14920}{35*10^3}

                = 0.4262*10^100

                 =42.62%

b) heat\ rejected = heat\ required - amount\ of\ power\ generated

                           = 35*10^3 - 14920

                           = 20.080 kW

AMOUNT OF POWER REJECTED IS 20.080 kW

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Given :- the volume of the pyramid is 36 cubic cm , find the volume of the prism on same base and same height as pyramid .

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Volume of pyramid = (1/3) * Base area * height .

Volume of prism = Base area * height .

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→ Volume of pyramid = 36 cm³

→ (1/3) * Base area * height = 36

→ Base area * height = 36 * 3

→ Base area * height = 108 cm³.

then,

→ Volume of prism = Base area * height .

→ Volume of prism = 108 cm³ (Ans.)

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Air at a pressure of 6000 N/m^2 and a temperature of 300C flows with a velocity of 10 m/sec over a flat plate of length 0.5 m. E
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Answer:

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

Explanation:

To solve this problem we use the expression for the temperature film

T_{f}=\frac{T_{\inf}+T_{w}}{2}=\frac{300+27}{2}=163.5

Then, we have to compute the Reynolds number

Re=\frac{uL}{v}=\frac{10\frac{m}{s}*0.5m}{16.96*10^{-6}\rfac{m^{2}}{s}}=2.94*10^{5}

Re<5*10^{5}, hence, this case if about a laminar flow.

Then, we compute the Nusselt number

Nu_{x}=0.332(Re)^{\frac{1}{2}}(Pr)^{\frac{1}{3}}=0.332(2.94*10^{5})^{\frac{1}{2}}(0.699)^{\frac{1}{3}}=159.77

but we also now that

Nu_{x}=\frac{h_{x}L}{k}\\h_{x}=\frac{Nu_{x}k}{L}=\frac{159.77*26.56*10^{-3}}{0.5}=8.48\\

but the average heat transfer coefficient is h=2hx

h=2(8.48)=16.97W/m^{2}K

Finally we have that the heat transfer is

Q=hA(T_{w}-T_{inf})=16.97*0.5(27-300)=-2316.4J

In this solution we took values for water properties of

v=16.96*10^{-6}m^{2}s

Pr=0.699

k=26.56*10^{-3}W/mK

A=1*0.5m^{2}

I hope this is useful for you

regards

8 0
3 years ago
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