University:
University of Michigan
Course:
MATH 116 | Calculus II
Academic year:

2024
Views:

448

Pages:

9
Author:

banenetta0l

Question 1 The diffuser in a jet engine is designed to decrease the kinetic energy of the air entering the engine compressor without any work or heat interactions. Calculate the velocity at the exit of a diffuser when air at 100 kPa 100kPa and 3 0 ∘ 𝐶 30 ∘ C enters it with a velocity of 350 m/s 350m/s and the exit state is 200 kPa 200kPa and 9 0 ∘ 𝐶 90 ∘ C. Answer 1 Using energy conservation: 𝐸 in − 𝐸 out = Δ 𝐸 system . E in −E out =ΔE system . ℎ 1 + 𝑉 1 2 2 = ℎ 2 + 𝑉 2 2 2 . h 1 + 2 V 1 2 =h 2 + 2 V 2 2 . Substituting values and simplifying: 𝑉 2 = 𝑉 1 2 + 2 𝑐 𝑝 ( 𝑇 1 − 𝑇 2 ) . V 2 = V 1 2 +2c p (T 1 −T 2 ) . Using 𝑐 𝑝 = 1.007 kJ/kg-K c p =1.007kJ/kg-K, 𝑉 1 = 350 m/s V 1 =350m/s, 𝑇 1 = 3 0 ∘ 𝐶 T 1 =30 ∘ C, 𝑇 2 = 9 0 ∘ 𝐶 T 2 =90 ∘ C: 𝑉 2 = ( 350 ) 2 + 2 ⋅ 1.007 ⋅ 1000 ⋅ ( 30 − 90 ) = 40.7 m/s . V 2 = (350) 2 +2⋅1.007⋅1000⋅(30−90) =40.7m/s. Answer 2 From energy conservation: ℎ 1 + 𝑉 1 2 2 = ℎ 2 + 𝑉 2 2 2 . h 1 + 2 V 1 2 =h 2 + 2 V 2 2 . Assuming isentropic conditions: 𝑉 2 = 𝑉 1 2 + 2 𝑐 𝑝 ( 𝑇 1 − 𝑇 2 ) . V 2 = V 1 2 +2c p (T 1 −T 2 ) . Using the same values as above: 𝑉 2 = ( 350 ) 2 + 2 ⋅ 1.007 ⋅ 1000 ⋅ ( 30 − 90 ) = 40.7 m/s . V 2 = (350) 2 +2⋅1.007⋅1000⋅(30−90) =40.7m/s. Answer 3 Using conservation of mass for incompressible flow: 𝜌 1 𝐴 1 𝑉 1 = 𝜌 2 𝐴 2 𝑉 2 . ρ 1 A 1 V 1 =ρ 2 A 2 V 2 . The velocity ratio can be derived using isentropic relations for air: 𝑉 2 = 𝑉 1 𝑃 2 𝑃 1 ⋅ 𝑇 1 𝑇 2 . V 2 =V 1 P 1 P 2 ⋅ T 2 T 1 . Substituting: 𝑉 2 = 350 200 100 ⋅ 303 363 = 40.7 m/s . V 2 =350 100 200 ⋅ 363 303 =40.7m/s. Question 2 A coaxial cable consists of an inner cylindrical conductor with radius 𝑎 a and an outer coaxial cylinder with inner radius 𝑏 b and outer radius 𝑐 c. The inner cylinder has a uniform positive charge per unit length 𝜆 λ. a) Calculate the electric field at any point between the cylinders. b) Calculate the electric field at any point outside the outer cylinder. c) Graph the electric field as a function of distance 𝑟 r from the axis of the cable, from 𝑟 = 0 r=0 to 𝑟 = 2 𝑐 r=2c. d) Find the charge per unit length on the inner surface and outer surface of the outer cylinder. Answer 1 (a) Electric Field Between Cylinders: Using Gauss's Law: 𝐸 ⋅ 2 𝜋 𝑟 𝐿 = 𝜆 𝐿 𝜖 0 . E⋅2πrL= ϵ 0 λL . 𝐸 = 𝜆 2 𝜋 𝜖 0 𝑟 . E= 2πϵ 0 r λ . (b) Electric Field Outside Outer Cylinder: For 𝑟 > 𝑐 r>c, the net charge enclosed is zero (inner charge cancels with the outer charge): 𝐸 = 0. E=0. (c) Graph of Electric Field: For 𝑟 r between 𝑎 a and 𝑏 b: 𝐸 ∝ 1 𝑟 . E∝ r 1 . For 𝑟 > 𝑐 r>c: 𝐸 = 0. E=0. (d) Charge per Unit Length on Outer Cylinder: The outer surface must have a charge − 𝜆 −λ to cancel the field outside: Inner surface charge: − 𝜆 −λ. Outer surface charge: 𝜆 λ.