Answer:
Explanation:
In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:
In this problem the flexural strength is defined with the following formula:
where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.
The force is then defined as:
Answer:
Explanation:
There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.
We are given the following information:
y₁ = 3 m
y₂ = 3 m − 2.4 m = 0.6 m
t₂ − t₁ = 0.4 s
a = -9.8 m/s²
t₀ = 0 s
v₀ = 0 m/s
We need to find y₀.
Use a constant acceleration equation:
y = y₀ + v₀ t + ½ at²
Evaluated at point 1:
3 = y₀ + (0) t₁ + ½ (-9.8) t₁²
3 = y₀ − 4.9 t₁²
Evaluated at point 2:
0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²
0.6 = y₀ − 4.9 t₂²
Solve for y₀ in the first equation and substitute into the second:
y₀ = 3 + 4.9 t₁²
0.6 = (3 + 4.9 t₁²) − 4.9 t₂²
0 = 2.4 + 4.9 (t₁² − t₂²)
We know t₂ = t₁ + 0.4:
0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)
0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))
0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)
0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)
0 = 2.4 − 3.92 t₁ − 0.784
0 = 1.616 − 3.92 t₁
t₁ = 0.412
Now we can plug this into the original equation and find y₀:
3 = y₀ − 4.9 t₁²
3 = y₀ − 4.9 (0.412)²
3 = y₀ − 0.83
y₀ = 3.83
Rounded to two significant figures, the height of the tree is 3.8 meters.
Answer:
Explanation:
Find attached the solution to the question
Answer:
Input Power = 6.341 KW
Explanation:
First, we need to calculate enthalpy of the water at inlet and exit state.
At inlet, water is at 20° C and 100 KPa. Under these conditions from saturated water table:
Since the water is in compresses liquid state and the data is not available in compressed liquid chart. Therefore, we use approximation:
h₁ = hf at 20° C = 83.915 KJ/kg
s₁ = sf at 20° C = 0.2965 KJ/kg.k
At the exit state,
P₂ = 5 M Pa
s₂ = s₁ = 0.2965 K J / kg.k (Isentropic Process)
Since Sg at 5 M Pa is greater than s₂. Therefore, water is in compresses liquid state. Therefore, from compressed liquid property table:
h₂ = 88.94 KJ/kg
Now, the total work done by the pump can be calculated as:
Pump Work = W = (Mass Flow Rate)(h₂ - h₁)
W = (53 kg/min)(1 min/60 sec)(88.94 KJ/kg - 83.915 KJ/kg)
W = 4.438 KW
The efficiency of pump is given as:
efficiency = η = Pump Work/Input Power
Input Power = W/η
Input Power = 4.438 KW/0.7
<u>Input Power = 6.341 KW</u>
