Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i
1 answer:
To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as
Where
T = Static temperature
V = Velocity of Fluid
Specific Heat
Re-arrange to find the static temperature we have that
Now the pressure of helium by using the Adiabatic pressure temperature is
Where,
= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that
Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>
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Answer:
N_c = 3.03 N
F = 1.81 N
Explanation:
Given:
- The attachment missing from the question is given:
- The given expressions for the radial and θ direction of motion:
r = 0.5*θ
θ = 0.5*t^2 ...... (correction for the question)
- Mass of peg m = 0.5 kg
Find:
a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.
b) Determine the magnitude of the normal force of the slot on the peg.
Solution:
- Determine the expressions for radial kinematics:
dr/dt = 0.5*dθ/dt
d^2r/dt^2 = 0.5*d^2θ/dt^2
- Similarly the expressions for θ direction kinematics:
dθ/dt = t
d^2θ/dt^2 = 1
- Evaluate each at time t = 2 s.
θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°
r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2
- Evaluate the angle ψ between radial and horizontal direction:
tan Ψ = r / (dr/dθ) = 1 / 0.5
Ψ = 63.43°
- Develop a free body diagram (attached) and the compute the radial and θ acceleration:
a_r = d^2r / dt^2 - r * dθ/dt
a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2
a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)
a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2
- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:
Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r
θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ
Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:
N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5
N_c = 3.03 N
F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5
F = 1.81 N
Answer: check answers in pictures (2 p)
Answer:
Both the technicians are correct.
Explanation:
As the pressure increases the temperature will also increase in accordance with the Boyle's law hence greater amount of heat is formed.
When the temperature increases the intermolecular spaces increase as the molecules of the fuel gain energy and their kinetic energy increases. This is limited to temperatures below dissociation/combustion temperature of diesel .