Answer:
Work done = 125π J
Explanation:
Given:
P = P_i * ( 1 - (x/d)^2 / 25)
d = 5.0 cm
x = 5 * d cm = 25d
Pi = 12 bar
Work done = integral ( F . dx )
F (x) = P(x) * A
F (x) = (πd^2 / 4) * P_i * (1 - (x/d)^2 / 25)
Work done = integral ((πd^2 / 4) * P_i * (1 - (x/d)^2 / 25) ) . dx
For Limits 0 < x < 5d
Work done = (πd^2 / 4) * P_i integral ( (1 - (x/d)^2) / 25)) . dx
Integrate the function wrt x
Work done = (πd^2 / 4) * P_i * ( x - d*(x/d)^3 / 75 )
Evaluate Limits 0 < x < 5d :
Work done = (πd^2 / 4) * P_i * (5d - 5d / 3)
Work done = (πd^2 / 4) * P_i * (10*d / 3)
Work done = (5 π / 6)d^3 * P_i
Input the values:
Work done = (5 π / 6)(0.05)^3 * (1.2*10^6)
Work done = 125π J
Answer:
165 mm
Explanation:
The mass on the piston will apply a pressure on the oil. This is:
p = f / A
The force is the weight of the mass
f = m * a
Where a in the acceleration of gravity
A is the area of the piston
A = π/4 * D1^2
Then:
p = m * a / (π/4 * D1^2)
The height the oil will raise is the heignt of a colum that would create that same pressure at its base:
p = f / A
The weight of the column is:
f = m * a
The mass of the column is its volume multiplied by its specific gravity
m = V * S
The volume is the base are by the height
V = A * h
Then:
p = A * h * S * a / A
We cancel the areas:
p = h * S * a
Now we equate the pressures form the piston and the pil column:
m * a / (π/4 * D1^2) = h * S * a
We simplify the acceleration of gravity
m / (π/4 * D1^2) = h * S
Rearranging:
h = m / (π/4 * D1^2 * S)
Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is
h2 = h + h1
h2 = h1 + m / (π/4 * D1^2 * S)
h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm
Answer:
Rate of internal heat transfer = 23.2 Btu/Ibm
mass flow rate = 21.55 Ibm/s
Explanation:
using given data to obtain values from table F7.1
Enthalpy of water at temperature of 100 F = 68.04Btu/Ibm
Enthalpy of water at temperature of 50 F = 18.05 Btu/Ibm
from table F.3
specific constant of glycerin 
<u>The rate of internal heat transfer ( change in enthalpy ) </u>
h4 - h3 = Cp ( T4 - T3 ) --------------- ( 1 )
where ; T4 = 50 F
T3 = 10 F
Cp = 0.58 Btu/Ibm-R
substitute given values into equation 1
change in enthalpy ( h4 - h3 ) = 23.2 Btu/Ibm
<u>Determine mass flow rate of glycol</u>
attached below is the detailed solution
mass flow rate of glycol = 21.55 Ibm/s
Answer:
true
Explanation:
Equipment that are "locked-out/tagged-out" <em>prevent the electrician from being electrocuted</em> or attaining a serious injury in relation to it. Locking out an equipment prevents it from releasing its energy because such energy can be <em>hazardous</em> to the electrician. There are instances when the equipment accidentally starts up, thus, it is essential that the equipment's source of energy is<em> isolated.</em>
Answer:
my_mul:
.globl my_mul
my_mul:
//Multiply X0 and X1
// Does not handle negative X1!
// Note : This is an in efficient way to multipy!
SUB SP, SP, 16 //make room for X19 on the stack
STUR X19, [SP, 0] //push X19
ADD X19, X1, XZR //set X19 equal to X1
ADD X9 , XZR , XZR //set X9 to 0
mult_loop:
CBZ X19, mult_eol
ADD X9, X9, X0
SUB X19, X19, 1
B mult_loop
mult_eol:
LDUR X19, [SP, 0]
ADD X0, X9, XZR // Move X9 to X0 to return
ADD SP, SP, 16 // reset the stack
BR X30
Explanation:
