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Zina [86]
3 years ago
7

One cylinder in the diesel engine of a truck has an initial volume of 650 cm3 . Air is admitted to the cylinder at 35 ∘C and a p

ressure of 1.0 atm. The piston rod then does 450 J of work to rapidly compress the air.
A. What is its final temperature?

Express your answer to two significant figures and include the appropriate units

B.

What is its final volume?

Express your answer to two significant figures and include the appropriate units.
Engineering
1 answer:
kupik [55]3 years ago
3 0

Answer:

1) the final temperature is T2 = 876.76°C

2) the final volume is V2 = 24.14 cm³

Explanation:

We can model the gas behaviour as an ideal gas, then

P*V=n*R*T

since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:

P*V^k = constant = C, k= adiabatic coefficient for air = 1.4

then the work will be

W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)

W = (P1*V1/T1)*(T2-T1)/(1-k)  

T2 = (1-k)W* T1/(P1*V1) +T1

replacing values (W=-450 J since it is the work done by the gas to the piston)

T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C

the final volume is

TV^(k-1)= constant

therefore

T2/T1= (V2/V1)^(1-k)

V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³

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Answer:

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Explanation:

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Replacing in the equation for entropy  

delta S_{12}=\int\limits^2_1 \frac{m*Cp*dT}{T}

m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:  

delta S_{12}= m*Cp*\int\limits^2_1 \frac{ dT }{T}  

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delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is 0.85\frac{kJ}{Kg*K}

We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K

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Q= m*Cp*dT

With dt=T_{2}-T_{1} clearing for T2 we get:

T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K

Now we can estimate the entropy change in the system

delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}

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A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k
goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

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solution

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Q = \frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}

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