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lys-0071 [83]
3 years ago
6

The density of a fluid is given by the empirical equation rho 70:5 exp 8:27 107 P where rho is density (lbm/ft3 ) and P is press

ure (lbf/in2 ). (a) What are the units of 70:5 and 8:27 107?
Engineering
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

The unit of 70.5 is lbm/ft^3

The unit of 8.27×10^7 is in^2/lbf

Explanation:

The unit of 70.5 has the same unit as density which is lbm/ft^3 because exponential is found of constant values (unitless values)

The unit of 8.27×10^7 (in^2/lbf) is the inverse of the unit of pressure P (lbf/in^2) because the units have to cancel out so a unitless value can be obtained. Exponential is found of figures with no unit

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Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.
Eva8 [605]

Correct answer is option E. No dimensions

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since all terms get cancelled

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2 years ago
Takt time is the rate at which a factory must produce to satisfy the customer's demand. a)- True b)- False
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A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
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:

mass flow rate of the air

rates of heat addition and rejection

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

                                           

                     

5 0
3 years ago
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