Answer:
<u>No</u>.
Explanation:
They are not all the same. Moreover, using a fluid that is not approved by the vehicle manufacturer will void the transmission warranty.
Answer:
(a) The stress on the steel wire is 19,000 Psi
(b) The strain on the steel wire is 0.00063
(c) The modulus of elasticity of the steel is 30,000,000 Psi
Explanation:
Given;
length of steel wire, L = 100 ft
cross-sectional area, A = 0.0144 in²
applied force, F = 270 lb
extension of the wire, e = 0.75 in
<u>Part (A)</u> The stress on the steel wire;
δ = F/A
= 270 / 0.0144
δ = 18750 lb/in² = 19,000 Psi
<u>Part (B)</u> The strain on the steel wire;
σ = e/ L
L = 100 ft = 1200 in
σ = 0.75 / 1200
σ = 0.00063
<u>Part (C)</u> The modulus of elasticity of the steel
E = δ/σ
= 19,000 / 0.00063
E = 30,000,000 Psi
Answer:
15.24°C
Explanation:
The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as
![COP=\frac{Q_{in}}{W}](https://tex.z-dn.net/?f=COP%3D%5Cfrac%7BQ_%7Bin%7D%7D%7BW%7D)
Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat
You can think of this quantity as similar to heat engine's efficiency
In our case, the COP of our heater is
![COP_{heater} = \frac{\frac{4500\ kJ}{3600\ s} *(T_{house}-T_{out})}{4\ kW}](https://tex.z-dn.net/?f=COP_%7Bheater%7D%20%3D%20%5Cfrac%7B%5Cfrac%7B4500%5C%20kJ%7D%7B3600%5C%20s%7D%20%2A%28T_%7Bhouse%7D-T_%7Bout%7D%29%7D%7B4%5C%20kW%7D)
Where T_{house} = 24°C and T_{out} is temperature outside
To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump
Which has COP of:
![COP_{carnot}=\frac{T_{house}}{T_{house}-T_{out}}](https://tex.z-dn.net/?f=COP_%7Bcarnot%7D%3D%5Cfrac%7BT_%7Bhouse%7D%7D%7BT_%7Bhouse%7D-T_%7Bout%7D%7D)
So we equate the COP of our heater with COP of Carnot heater
![\frac{1.25 *(T_{house}-T_{out})}{4}=\frac{T_{house}}{T_{house}-T_{out}}](https://tex.z-dn.net/?f=%5Cfrac%7B1.25%20%2A%28T_%7Bhouse%7D-T_%7Bout%7D%29%7D%7B4%7D%3D%5Cfrac%7BT_%7Bhouse%7D%7D%7BT_%7Bhouse%7D-T_%7Bout%7D%7D)
Rearrange the equation
![\frac{1.25}{4}(24-T_{out})^2-24=0](https://tex.z-dn.net/?f=%5Cfrac%7B1.25%7D%7B4%7D%2824-T_%7Bout%7D%29%5E2-24%3D0)
Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be
15.24°C
Answer:
d= 4.079m ≈ 4.1m
Explanation:
calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}
Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).
r = Radius of the shaft.
T = Twisting Moment or Torque.
J = Polar moment of inertia.
C = Modulus of rigidity for the shaft material.
l = Length of the shaft.
θ = Angle of twist in radians on a length.
Maximum Torque, ζ= τ × \frac{ π}{16} × d³
τ= 60 MPa
ζ= 800 N·m
800 = 60 × \frac{ π}{16} × d³
800= 11.78 × d³
d³= 800 ÷ 11.78
d³= 67.9
d= \sqrt[3]{} 67.9
d= 4.079m ≈ 4.1m
Answer:
the current consumed is 3.3 A
Explanation:
Given;
resistance, R = 30 ohms
inductance, L = 200 mH
Voltage supply, V = 230 V
frequency of the coil, f = 50 Hz
impedance, Z = 69.6 Ohms
The current consumed is calculated as;
![I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BZ%7D%20%5C%5C%5C%5CI%20%3D%20%5Cfrac%7B230%7D%7B69.6%7D%20%5C%5C%5C%5CI%20%3D%203.3%20%5C%20A)
Therefore, the current consumed is 3.3 A