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andrey2020 [161]
3 years ago
5

The intake and exhaust processes are not considered in the p-V diagram of Otto cycle. a) true b) false

Engineering
1 answer:
vovangra [49]3 years ago
6 0

Answer:

b) false

Explanation:

We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.

It consist four processes

1-2:Reversible adiabatic compression

2-3:Constant volume heat addition

3-4:Reversible adiabatic expansion

3-4:Constant volume heat rejection

Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.

But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.

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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
White raven [17]

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

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