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anzhelika [568]
4 years ago
5

An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the g

ravitational potential energy of the object, in kJ, relative to the surface of the earth.
Engineering
1 answer:
Harlamova29_29 [7]4 years ago
4 0

Answer:

Gravitational Potential =58.914 KJ

Explanation:

We know that

Gravitational Potential Energy = mass\times g\times Height

Given mass = 251 kg

Height= 24 m

g is acceleration due to gravity = 9.78m/s^{2}

Applying values in the equation we get

Gravitational Potential Energy=251X9.78X24 Joules

Gravitational Potential Energy=58914.72 Joules

Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ

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Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

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Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

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Answer:

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