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sammy [17]
3 years ago
12

An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1200 K and a heat rejection at 400 K. During t

he heat addition the volume triples. Find the two specific heat transfers (q) in the cycle and the overall cycle efficiency. Solution:

Physics
1 answer:
lana [24]3 years ago
4 0

Answer:

The specific heat capacity is q_{L}=126.12kJ/kg

The efficiency of the temperature is n_{TH}=0.67

Explanation:

The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

q_{h} = R* T_{h} in(V_{2}/V_{1})

   =0.287 * 1200 ln(3)

   =0.287*1318.33

   =378.36kJ/kg

The specific heat capacity:

q_{L}=q_{h}*(T_{L}/T_{H})

q_{L}=378.36 * (400/1200)

q_{L}=378.36 * 0.333

q_{L}=126.12kJ/kg

The efficiency of the temperature will be:

n_{TH}=1 - (T_{L}/T_{H})

n_{TH}=1-(400/1200)

n_{TH}=1-0.333

n_{TH}=0.67

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Consult Multiple-Concept Example 15 to review the concepts on which this problem depends. Water flowing out of a horizontal pipe
kiruha [24]

Answer:

The pressure of the water in the pipe is 129554 Pa.

Explanation:

<em>There are wrongly written values on the proposal, the atmospheric pressure must be 101105 Pa, and the density of water 1001.03 kg/m3, those values are the ones that make sense with the known ones.</em>

We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:

A_1v_1=A_2v_2

We want v_2, and take into account that the areas are circular:

v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}

Substituting values we have (we don't need to convert the cm because they cancel out between them anyway):

v_2=\frac{r_1^2 v_1}{r_2^2}=\frac{(1.8cm)^2 (0.56m/s)}{(0.49cm)^2}=7.56m/s

For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:

P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2

Since the tube is horizontal h_1=h_2 and those terms cancel out, so the pressure of the water in the pipe will be:

P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}

And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:

P_1=101105 Pa+\frac{1001.03Kg/m^3 ((7.56m/s)^2-(0.56m/2)^2)}{2}=129554 Pa

3 0
3 years ago
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