Answer:
See explaination
Explanation:
See attachment for the detailed step by step solution of the given problem.
In this exercise we have to use the knowledge of force to calculate the required energies, so we have to:
a) 19094 N
b) 110.055 kPa
c) 1222 J
In the field of physics, force is a physical action that causes deformation or that changes the state of rest or movement of a given object.
a) Knowing that the force formula is defined by:
b) Knowing that the force exerted by an area is equal to the pressure in that area, we have:
c)So calculating the potential energy we have:
See more about force at brainly.com/question/26115859
Answer:
a) 19094 N
b) 110.055 kPa
c) 1222 J
Explanation:
The force on the gas is the weight plus the atmospheric pressure multiplied by the piston area
F = P + p * A
F = m * g + p * π/4 * d^2
F = 150 * 9.813 + 101570 * π/4 * 0.47^2 = 19094 N
The pressure is the force divided by the area of the piston
p1 = F / A
p1 = F / (π/4 * d^2)
p1 = 19094 / (π/4 * 0.47^2) = 110055 Pa = 110.055 kPa
variation of gravitational potential energy is defined as
ΔEp = m * g * Δh
ΔEp = 150 * 9.813 * 0.83 = 1222 J
193GPa, a = 17x10^-6/°C) carriage bolts of the clamp each have a diameter of 10mm, and they hold
the cylinder snug with a negligible force against the rigid jaws, determine the temperature at which
the average normal stress in either the magnesium or steel becomes 12 MPa.
Answer:
......................
Explanation:
Answer: the thermal conductivity of the second material is 125.9 W/m.k
Explanation:
Given that;
The two rods could be approximated as a fins of infinite length.
TA = 75°C, θA = (TA - T∞) = 75 - 25 = 50°C
TB = 70°C θB = (TB - T∞) = 70 - 25 = 45°C
Tb = 100°C θb = (Tb - T∞) = (100 - 25) = 75°C
T∞ = 25°C
KA = 200 W/m · K, KB = ?
Now
The temperature distribution for the infinite fins are given by
θ/θb = e^(-mx)
θA/θb= e^-√(hp/A.kA) x 1 --------------1
θB/θb = e^-√(hp/A.kB) x 1---------------2
next we take the natural logof both sides,
ln(θA/θb) = -√(hp/A.kA) x 1 ------------3
In(θB/θb) = -√(hp/A.kB) x 1 ------------4
now we divide 3 by 4
[ ln(θA/θb) /in(θB/θb)] = √(KB/KA)
we substitute
[ In(50/75) /In(45/75)] = √(KB/200)
In(0.6666) / In(0.6) = √KB / √200
-0.4055/-0.5108 = √KB / √200
0.7938 = √KB / 14.14
√KB = 11.22
KB = 125.9 W/m.k
So the thermal conductivity of the second material is 125.9 W/m.k
Explanation:
Step1
Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.
The expression for absolute pressure is given as follows:
Here, is absolute pressure,
is gauge pressure and
is atmospheric pressure.
Step2
Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.
Answer:
73.24 K byte
Explanation:
Assuming that
N = total number of samples
N = 10 * 5kHz
N = 50*10^3
Also, the total number of bits, T
T = 12 * N
T = 12 * 50*10^3
T = 600 * 10^3
And then, finally, the total number of byte,
B = 600*10^(3/8)
B = 75*10^3 byte
75*10^3 byte = 75*10^3/1024 kilo byte
And on converting to decimal, we will have
= 73.24 K byte
Therefore, the memory required = 73.24 K byte
Answer:
temperature of first extraction 330.8°C
temperature of second extraction 140.8°C
power output=3168Kw
Explanation:
Hello!
To solve this problem we must use the following steps.
1. We will call 1 the water vapor inlet, 2 the first extraction at 100kPa and 3 the second extraction at 200kPa
2. We use the continuity equation that states that the mass flow that enters must equal the two mass flows that leave
m1=m2+m3
As the problem says, 20% of the flow represents the first extraction for which 5 * 20% = 1kg / s
solving
5=1+m3
m3=4kg/s
3.
we find the enthalpies and temeperatures in each of the states, using thermodynamic tables
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties
4.we find the enthalpy and entropy of state 1 using pressure and temperature
h1=Enthalpy(Water;T=T1;P=P1)
h1=3457KJ/kg
s1=Entropy(Water;T=T1;P=P1)
s1=7.234KJ/kg
4.
remembering that it is a reversible process we find the enthalpy and the temperature in the first extraction with the pressure 1000 kPa and the entropy of state 1
h2=Enthalpy(Water;s=s1;P=P2)
h2=3116KJ/kg
T2=Temperature(Water;P=P2;s=s1)
T2=330.8°C
5.we find the enthalpy and the temperature in the second extraction with the pressure 200 kPav y the entropy of state 1
h3=Enthalpy(Water;s=s1;P=P3)
h3=2750KJ/kg
T3=Temperature(Water;P=P3;s=s1)
T3=140.8°C
6.
Finally, to find the power of the turbine, we must use the first law of thermodynamics that states that the energy that enters is the same that must come out.
For this case, the turbine uses a mass flow of 5kg / s until the first extraction, and then uses a mass flow of 4kg / s for the second extraction, taking into account the above we infer the following equation
W=m1(h1-h2)+m3(h2-h3)
W=5(3457-3116)+4(3116-2750)=3168Kw