Adhil M.

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Mechanical Engineering

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Question:

(a) Calculate the power required to compress air in a steady state, steady flow process with no change in elevation at a rate of 2 kg/s from 0.101 MPa, 40 degree C, 10 m/s to 0.300 MPa,50 degree C, at 125 m/s. During this process, the enthalpy of the air increases by 40.15 kJ/kg, while 8 kJ/s of heat is lost to the environment.

Adhil M.

Answer:

(a) Initial condition : $dm/dt = 2 kg/s$ P = $0.101 $ MPa T =

Physics

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Question:

(a) A feather is dropped on the moon from a height of 1.40 meters. The acceleration of gravity on the moon is 1.67 m/s2. Determine the time for the feather to fall to the surface of the moon. (b) A car travels up a hill at a constant speed of 37 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the average speed for the whole trip.

Adhil M.

Answer:

(a) $$ S = 1.40 $$ m $$ a = 1.67 $$ $$m/s^{2}$$ We have the relation of motion, $$ S = u \times t + 1/2 \times a \times t^{2} $$ $$ 1.40 = 0 + 1/2 \times 1.67 \times t^{2} $$ $$ t^{2} = \frac{1.40}{1/2 \times 1.67} $$ $$ t = 1.294 $$ s (b) By definition the average speed is the ration of the total traveled distance and the total traveled time. Let us introduce the total traveled distance of the car as L. Then the time of the travel up the hill is $$ t_{1} = \frac{L/2}{v_{1}} $$ $$ t_{1} = \frac{L}{2 \times 37} $$ $$ = L/74 $$ The time of the travel down the hill is $$ t_{2} = \frac{L/2}{v_{2}} $$ $$ t_{1} = \frac{L}{2 \times 66} $$ $$ = L/132 $$ The total traveled time is $$ t = t_{1} + t_{2} $$ $$ t = \frac{L/74}{L/132} $$ $$ = 0.0211 \times L $$ Then the average velocity is, $$ v_{av} = \frac{L}{t} $$ $$ = \frac{L}{0.0211 L} $$ $$ = 47.4 $$ km/h

Aerospace Engineering

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Question:

An aircraft is powered by a turbojet engine and has a maximum speed of 790 kmph at sea level. The gross weight of the vehicle is 160,000 N, wing area is 50 m^{2}, and $$C_{d} = 0.02 + 0.04C_{l}^{2}L$$ . Find (a) the thrust developed by the engine. (b) climb angle and rate of climb when flying at 75% maximum aerodynamic efficiency. (c) maximum rate of climb and velocity at which it occurs.

Adhil M.

Answer:

Given, $$ V_{max} = 790$$ kmph = $$218.44$$ m/s $$ W = 160,000$$ N $$ S = 50 $$ $$$$ $$ C_{d } = 0.02 + 0.04\times{C_{l}^{2}} $$ To be obtained (a) Thrust (T) (b) Climb angle and rate of climb (c) Maximum rate of climb and velocity Answers (a) $$ L = W$$ $$ C_{l} =\frac{L}{ 1/2 \times {\rho} \times {V^{2}} \times {S}}$$ $$ C_{l} =\frac{W}{ 1/2 \times {\rho} \times {V^{2}} \times {S}}$$ $$ C_{l} =\frac{160,000}{ 1/2 \times {1.225} \times {218.44^{2}} \times {50}}$$ $$ C_{l} = 0.10849$$ $$ C_{d} = 0.02 + 0.04 \times{0.10849^{2}}$$ $$ C_{d} = 0.0204708 $$ $$ T = D = 1/2 \times {\rho} \times {V^{2}} \times {S} \times {C_{d}}$$ $$ T = 1/2 \times {1.225} \times {218.44^{2}} \times {50} \times {0.0204708}$$ $$ T = 30189.81$$ N (b) $$E_{m} = \frac{1}{2 \times \sqrt{K \times C_{d_{0}}}}$$ $$ = \frac{1}{2 \times \sqrt{0.02 \times 0.04}}$$ $$ = 17.677 $$ $$ E = 0.75 \times E_{m} $$ $$ = 13.258 $$ $$ \gamma = \frac{T}{W} - \frac{1}{E} $$ $$ \gamma = \frac{30189.81}{160000} - \frac{1}{13.758} $$ $$ = 0.11325 $$ rad $$ = 6.484^{\circ}$$ $$ 13.258 = \frac{C_{l}}{0.02 + 0.04 \times C_{l}^{2}}$$ On solving this , $$ C_{l} = 1.566 $$ or $$0.319 $$ Taking minimum value of C_{l} $$ V = \sqrt {\frac{2 \times L}{\rho \times S \times C_{l}}}$$ $$ = \sqrt {\frac{2 \times 160000}{1.225 \times 50 \times 0.319}} $$ $$ = 127.576 $$ m/s Rate of climb , $$ \frac{dh}{dt} = V \times \gamma $$ $$ = 127.526 \times 0.1132 $$ $$ = 14.44$$ m/s (c) $$ V_{FC} = \sqrt{\frac{T/W \times W/S}{3 \times \rho \times C_{d_{0}}} \times ( 1 + \sqrt{1 + \frac{3}{E_{m}^{2} \times (T/W)^{2}}} ) } $$ $$ V_{FC} = 132.177 $$ m/s $$ \gamma_{FC} = T/W \times (1 - \frac{\gamma}{6}) - \frac{3}{2 \times E_{m} \times (T/W)^{2}} ) $$ $$ = 0.1098 $$ rad Rate of climb , $$ \frac{dh}{dt}_{FC} = V_{FC} \times \gamma_{FC} $$ $$ = 132.177 \times 0.1098 $$ $$ = 14.513 $$ m/s

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