Consider the adiabatic compressor from a refrigerator that uses refrigerant R-134a as the working fluid, flowing at 0.05 kg/s. The refrigerant enters the compressor as a saturated vapor at a pressure of 0.14 MPa. The pressure of the refrigerant at the exit of the compressor is 0.8 MPa. Please interpolate for this problem where needed.
a. If the compressor is isentropic, what is the exit temperature?
b. What is the minimum amount of power (work) that this compressor might theoretically use?
c. If the isentropic efficiency of the actual, non-ideal, compressor is 0.85, what is the actual amount of power (work) that this compressor requires?
d. What is the entropy generated by this device? Assume that the compressor is adiabatic.
Answer:
Please see the attached file for the complete answer.
Explanation:
The spacing of rafters in a roof is 48-in o.c. Roof dead load = 5 psf. Snow load=30 psf. Roof sheathing is to be a sheathing grade of plywood, and panels are oriented in the strong direction. Deflection limits are L/240 for snow load and L/180 for total load. Find: The minimum span rating, thickness, and edge support requirements for the roof sheathing using ASD procedures.
Answer:
Explanation:
Step by step solution is found in the attachment.
Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.
(λf.λx.f(f x))(λy.y≠3) 2
Answer:
Decrease to typical from utilizing lambda-decrease:
The given lambda - math terms is, (λf.λx.f(f(fx)))(λy.y×3)2
The of taking the terms is significant in lambda - math,
For the term, (λy, y×3)2, we can substitute the incentive to the capacity.
Therefore apply beta-decrease on “(λy, y×3)2,“ will return 2 × 3 = 6
Presently the tem becomes, (λf λx f(f(fx)))6
The main term, (λf λx f(f(fx))) takes a capacity and a contention and substitute the contention in the capacity.
Here it is given that it is conceivable to substitute, the subsequent increase in the outcome.
In this way by applying next level beta - decrease, the term becomes f(f(f(6))), which is in ordinary structure.
A silicon pn junction diode at t has a cross sectional area of cm the length of the p region is and the length of the n region is the doping concentrations are determine approximately the series resistance of the diode and the current through the diode that will produce a drop across this series resistance.
Answer:
Explanation:
r=72.3 is my thought
I hope it is helpful
Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow rate Q and tube diameter D and (b) mass flow rate mp and tube diameter. The Reynolds number is 1800 in a section where the tube diameter is 6 mm. (c) Find the Reynolds number for the same flow rate in a section where the tube diameter is 6 mm.
Answer:
a) [tex]Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}[/tex], b) [tex]Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}[/tex], c) 1600
Explanation:
a) The Reynolds Number is modelled after the following formula:
[tex]Re = \frac{\rho \cdot v \cdot D}{\mu}[/tex]
Where:
[tex]\rho[/tex] - Fluid density.
[tex]\mu[/tex] - Dynamics viscosity.
[tex]D[/tex] - Diameter of the tube.
[tex]v[/tex] - Fluid speed.
The formula can be expanded as follows:
[tex]Re = \frac{\rho \cdot \frac{4Q}{\pi\cdot D^{2}}\cdot D }{\mu}[/tex]
[tex]Re = \frac{4\cdot \rho \cdot Q}{\pi\cdot \mu\cdot D}[/tex]
b) The Reynolds Number has this alternative form:
[tex]Re = \frac{4\cdot \dot m}{\pi\cdot \mu\cdot D}[/tex]
c) Since the diameter is the same than original tube, the Reynolds number is 1600.
A 35 ft simply supported beam is loaded with concentrated loads 15 ft in from each support. On one end, the dead load is 8.0 kips and the live load is 18.0 kips. At the other end, the dead load is 4.0 kips and the live load is 9.0 kips. Include the self-weight of the beam in the design. Lateral supports are provided at the supports and the load points. Determine the least-weight W-shape to carry the load. Use A992 steel and Cb
Answer:
From the load equation
F=stress*Area
Given stresses are 8 kips and 9 kips.
Hence the minimum weight supported=6.695 lbs.
Answer:
ASD = 306 kips-feet
LRSD = 1387.5 kips-feet
Explanation:
a step by step process to solving this problem.
ΣM at A = 0
where;
RB * 35 - (8+18)15 - (4+9)20 = 0
RB = 18.57k
also E y = 0;
RA + RB = 18 + 8 + 9 +4 = 20.43 k
taking the maximum moment at mid point;
Mc = RA * 35/2 - (8 +18) (35/2 -15)
Mc = 292.525
therefore, MD = RA * 15 = 20.43 * 15 = 306.45 kips-feet
MD = 306.45 kip-feet
ME = 279 kip-feet .IE 18.57 * 15
considering the unsupported length; 35 - (15*2 = 5ft )
now we have that;
L b = L p = 5ft
where L p = 1.76 r y(√e/f y)
L p = 1.76 r y √29000/50
r y = 1.4 inch
so we have that M r = M p for L b = L p where
M p = 2 F y ≤ 1.5 s x F y
Recall from the expression,
RA + RB = (8+4) * 1.2 + (18+9) * 1.6 = 57.6
RA * 35 = 4 * 1.2 * 15 + 9 *1.6 * 15 + 8 * 1.2 * 20 + 18 * 1.6 * 20
RA = 30.17 k
the maximum moment at D = 30.17 * 15 = 452.55 kips-feet
Z required = MD / F y = 452.55 * 12 / 50 = 108.61 inch³
so we have S x = 452.55 * 12 / 1.5 * 50 = 72.4 inch³
also r = 1.41 in
Taking LRFD solution:
where the design strength ∅ M n = 0.9 * Z x * F y
given r = 2.97
Z x = 370 and S x = 81.5, we have
∅ M n = 0.9 * 370 * 50 = 16650 k-inch = 1387.5 kips-feet
this tells us it is safe.
ASD solution:
for L b = L p, and where M n = M p = F c r S x
we already have value for S x as 81.5 so
F c r = Z x times F y divide by S x
F c r = 370 * 50 / 81.5 = 227 kips per sq.in
considering the strength;
Strength = M n / Ωb = (0.6 * 81.5 * 50) * (1.5) / 12 = 306 kips-feet
This justifies that it is safe because is less than 306
Water is flowing in a pipe. Which is the correct statement about the effect of an increase in the Reynolds number of the flow:
a.if the flow is laminar it cannot become turbulent if the wall is smooth.
b.if the flow is turbulent it could become laminar.
c.if the flow is laminar it could become turbulent.
d.if the flow is laminar it cannot become turbulent unless the wall is rough.
Answer:
c.if the flow is laminar it could become turbulent.
Explanation:
The Reynolds number (Re) is a dimensionless quantity used to help predict flow patterns in different fluid flow situations. At low Reynolds numbers of below 2300 flows tend to be dominated by laminar, while at high Reynolds numbers above 4000, turbulence results from differences in the fluid's speed and direction. In between these values is the transition region of flow.
In practice, fluid flow is generally chaotic, and very small changes to shape and surface roughness of bounding surfaces can result in very different flows.
For a steel alloy it has been determined that a carburizing heat treatment of 7 hour duration will raise the carbon concentration to 0.38 wt% at a point 3.8 mm from the surface. Estimate the time (in hours) necessary to achieve the same concentration at a 6.2 mm position for an identical steel and at the same carburizing temperature.
Answer:
18.6h
Explanation:
To solve this Duck's second law in form of Diffusion will be used.
Also note that since the temperature is constant D (change) will also be constant.
Please go through the attached files for further explanation and how the answer Is gotten.
A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron casting and oil can be modeled as incompressible with specific heats 0.10 Btu/lbm o R, and 0.45 Btu/lbm o R, respectively. For the iron casting and oil as the system, determine: a) The final equilibrium temperature (o F) b) The total entropy change for this process (Btu/ o R) (Hint: Total entropy change is the sum of entropy change of iron casting and oil.)
Answer:
a) The final equilibrium temperature is 83.23°F
b) The entropy production within the system is 1.9 Btu/°R
Explanation:
See attached workings
a) Equilibrium temp. ≈ 77.01°F.
b) Total entropy change ≈ 104.58 Btu/°R.
To solve this problem, we can apply the principle of energy conservation and the definition of entropy change.
a) The final equilibrium temperature can be found using the principle of energy conservation, which states that the heat lost by the hot object (iron casting) equals the heat gained by the cold object (oil) during the process.
The equation for energy conservation is:
[tex]\[ m_{\text{iron}} \times C_{\text{iron}} \times (T_{\text{final}} - T_{\text{initial, iron}}) = m_{\text{oil}} \times C_{\text{oil}} \times (T_{\text{final}} - T_{\text{initial, oil}}) \][/tex]
Where:
- [tex]\( m_{\text{iron}} \)[/tex] = mass of iron casting = 50 lbm
- [tex]\( C_{\text{iron}} \)[/tex] = specific heat of iron casting = 0.10 Btu/lbm °R
- [tex]\( T_{\text{initial, iron}} \)[/tex] = initial temperature of iron casting = 700 °F
- [tex]\( m_{\text{oil}} \)[/tex] = mass of oil = 2121 lbm
- [tex]\( C_{\text{oil}} \)[/tex] = specific heat of oil = 0.45 Btu/lbm °R
- [tex]\( T_{\text{initial, oil}} \)[/tex] = initial temperature of oil = 80 °F
- [tex]\( T_{\text{final}} \)[/tex] = final equilibrium temperature (unknown)
Now, let's solve for [tex]\( T_{\text{final}} \)[/tex]:
[tex]\[ 50 \times 0.10 \times (T_{\text{final}} - 700) = 2121 \times 0.45 \times (T_{\text{final}} - 80) \][/tex]
[tex]\[ 5(T_{\text{final}} - 700) = 954.45(T_{\text{final}} - 80) \][/tex]
[tex]\[ 5T_{\text{final}} - 3500 = 954.45T_{\text{final}} - 76356 \][/tex]
[tex]\[ 0 = 949.45T_{\text{final}} - 72856 \][/tex]
[tex]\[ T_{\text{final}} = \frac{72856}{949.45} \][/tex]
[tex]\[ T_{\text{final}} \approx 77.01 \, ^\circ F \][/tex]
So, the final equilibrium temperature is approximately [tex]\( 77.01 \, ^\circ F \).[/tex]
b) The total entropy change for the process can be calculated using the formula:
[tex]\[ \Delta S = \Delta S_{\text{iron}} + \Delta S_{\text{oil}} \][/tex]
Where:
- [tex]\( \Delta S_{\text{iron}} = \frac{Q_{\text{iron}}}{T_{\text{initial, iron}}} \)[/tex]
- [tex]\( \Delta S_{\text{oil}} = \frac{Q_{\text{oil}}}{T_{\text{initial, oil}}} \)[/tex]
- [tex]\( Q_{\text{iron}} \) = heat lost by the iron casting[/tex]
- [tex]\( Q_{\text{oil}} \) = heat gained by the oil[/tex]
Let's calculate:
[tex]\[ Q_{\text{iron}} = m_{\text{iron}} \times C_{\text{iron}} \times (T_{\text{final}} - T_{\text{initial, iron}}) \][/tex]
[tex]\[ Q_{\text{iron}} = 50 \times 0.10 \times (77.01 - 700) \][/tex]
[tex]\[ Q_{\text{iron}} \approx -3175.495 \, \text{Btu} \][/tex]
[tex]\[ Q_{\text{oil}} = m_{\text{oil}} \times C_{\text{oil}} \times (T_{\text{final}} - T_{\text{initial, oil}}) \][/tex]
[tex]\[ Q_{\text{oil}} = 2121 \times 0.45 \times (77.01 - 80) \][/tex]
[tex]\[ Q_{\text{oil}} \approx 8729.535 \, \text{Btu} \][/tex]
Now, calculate entropy changes:
[tex]\[ \Delta S_{\text{iron}} = \frac{-3175.495}{700} \][/tex]
[tex]\[ \Delta S_{\text{iron}} \approx -4.5364 \, \text{Btu/°R} \][/tex]
[tex]\[ \Delta S_{\text{oil}} = \frac{8729.535}{80} \][/tex]
[tex]\[ \Delta S_{\text{oil}} \approx 109.118 \, \text{Btu/°R} \][/tex]
[tex]\[ \Delta S = -4.5364 + 109.118 \][/tex]
[tex]\[ \Delta S \approx 104.5816 \, \text{Btu/°R} \][/tex]
So, the total entropy change for this process is approximately [tex]\( 104.5816 \, \text{Btu/°R} \).[/tex]
Amplifiers are extensively used in the baseband portion of a radio receiver system to condition the baseband signal to produce an output signal ready for digital sampling and storage. Some of the key design features of baseband amplifiers include
i. DC gain,
ii. output swing,
iii. power consumption, and
iv. bandwidth.
Answer:
Please see the attached file for the complete answer.
Explanation:
Compute the acceleration of gravity for a given distance from the earth's center, distCenter, assigning the result to accelGravity. The expression for the acceleration of gravity is: (G * M) / (d2), where G is the gravitational constant 6.673 x 10-11, M is the mass of the earth 5.98 x 1024 (in kg) and d is the distance in meters from the earth's center (stored in variable distCenter).Sample program:#include int main(void) { const double G = 6.673e-11; const double M = 5.98e24; double accelGravity = 0.0; double distCenter = 0.0; distCenter = 6.38e6; printf("accelGravity: %lf\n", accelGravity); return 0;
Answer:
See Explaination
Explanation:
#include <stdio.h>
int main(void)
{
const double G = 6.673e-11;
const double M = 5.98e24;
double accelGravity = 0.0;
double distCenter = 0.0;
distCenter = 6.38e6;
//<StudentCode>
accelGravity = (G * M) / (distCenter * distCenter);
printf("accelGravity: %lf\n", accelGravity);
return 0;
}
Consider a C.T. system in s-plane below. Draw DF1 (Direct Form 1) realization. (by hand) Perform system realization using MATLAB Simulink. (Use same parameters as [Q 11] for "Step Function" and "Delay Block".) Use the "step function", "delay", and "summer" to build the input, X(s). Use a "scope" as the output, Y(s). Include screen shot of the SIMULINK schematic model page, input scope trace, output scope trace in report.Discuss the results in your own words. s +5 H(S) = 52 + 11s + 10
Answer:
See the attached file for the answer.
Explanation:
See the attached file for the explanation
A circular specimen of MgO is loaded in three-point bending. Calculate the minimum possible radius of the specimen without fracture, given that: the applied load is 5560 N the flexural strength is 105 MPa the separation between the supports is 45 mm Input your answer as X.XX mm, but without the unit of mm.
Answer:
radius = 9.1 × [tex]10^{-3}[/tex] m
Explanation:
given data
applied load = 5560 N
flexural strength = 105 MPa
separation between the support = 45 mm
solution
we apply here minimum radius formula that is
radius = [tex]\sqrt[3]{\frac{FL}{\sigma \pi}}[/tex] .................1
here F is applied load and is length
put here value and we get
radius = [tex]\sqrt[3]{\frac{5560\times 45\times 10^{-3}}{105 \times 10^6 \pi}}[/tex]
solve it we get
radius = 9.1 × [tex]10^{-3}[/tex] m
1. A four-lane freeway (two lanes in each direction) is located on rolling terrain and has 12-ft lanes, no lateral obstructions within 6 ft of the pavement edges, and there are two ramps within three miles upstream of the segment midpoint and three ramps within three miles downstream of the segment midpoint. A weekday directional peak-hour volume of 1800 vehicles (familiar users) is observed, with 700 arriving in the most congested 15-min period. If a LOS no worse than C is desired, determine the maximum number of heavy vehicles that can be present in the peak-hour traffic stream.
Answer:
Maximum number of vehicle = 308
Explanation:
See the attached file for the calculation.
A prototype of a part is to be fabricated using stereolithography. The part is shaped like a right triangle whose base = 36 mm, height 48mm, and thickness = 30 mm. In the stereolithography process, the layer thickness = 0.15 mm. Diameter of the laser beam spot = 0.40 mm, and the beam is moved across the surface of the photopolymer at a velocity of 2200 mm/s. Compute the minimum possible time (use units of hours, 3 significant figs) required to build the part, if 25 s are lost each layer to lower the height of the platform that holds the part. Neglect the time for setup and post-processing.
Answer:
1.443hrs
Explanation:
Please kindly check attachment for the detailed and step by step solution to the problem.
The reel has a mass of 30 kg and a radius of gyration about A of kA = 120 mm. The suspended cylinder has a mass of 40 kg. Starting from rest, the motor M exerts a constant force of P = 300 N on the cable. Determine the mass moment of inertia of the reel. Determine the velocity of the cylinder after it has traveled upward 2 m. Determine the time that was taken to travel the 2 m distance.
Answer:
Explanation:
Mass of the reel is given as,
M = 30kg
Radius of gyration about A is given as,
Ka = 120mm = 0.12m
Mass of the cylinder suspended
Mc = 40kg
Then, weight of the cylinder is
W = mg = Mc × g
W = 40 × 9.81 = 392.4 N
The exert force
P = 300N
Radius of the reel
R = 150mm = 0.15m
r = 75mm = 0.075m
Check attached for diagram of this problem
A. Moment of inertial of the reel?
We will assume the reel is a thin hoop shape, so we will use the thin hoop shape formula
I = M•Ka²
I = 30 × 0.12²
I = 0.432 kgm²
Therefore, Moment of inertia of reel is 0.432 kg.m²
B. Velocity if the cylinder after it moves 2m?
Applying torque equation
Στ = Iα
Where
α is angular acceleration
I is moment of inertia
τ is the torque..
Where τ = F × r
The two force acting on the reel is the weight of the cylinder and it is acting downward and the force applied
Then, taking torque about point A
Στ = Iα.
P × R — W × r = Iα
300 × 0.15 - 392.4 × 0.075 = 0.432 × α
45 — 29.43= 0.432α
15.57 = 0.432α
α = 15.57 / 0.432
α = 36.042 rad/s²
Now, The angle turned by the reel when the cylinder moves 2m above can be determined using
S= rθ
Then, θ = S/r
θ = 2 / 0.075
θ = 26.667 rad
Initially, the reel is at rest, then, the initial angular velocity is 0
ωo = 0 rad/s
Now, apply the kinematic equation and calculate the final angular velocity of the reel,
ω² = ωo² + 2αθ
ω² = 0² + 2 × 36.042 × 26.667
ω² = 0 + 1922.24
ω = √1922.24
ω = 43.84 rad/s
Now, using the relationship between linear velocity and angular velocity
Then, the velocity of cylinder
V = rω
V = 0.075 × 43.84
V = 3.2883 m/s
Therefore, Velocity of Cylinder is 3.288 m/s.
C. Time taken to travel 2m
From equation of circular motion
ω = ωo + αt
43.84 = 0 + 36.042t
43.84 = 36.042t
t = 43.84 / 36.042
t = 1.216 seconds
The time taken to travel 2m is 1.216 seconds
Construction of a reservoir behind a dam Group of answer choices
a. stabilizes the sides of reservoir walls; the water buoys them up.
b. destabilizes the slopes because of the weight of the dam.
c. can destablize the slopes by increasing pore fluid pressure.
d. has no effect on slope stability.
Answer:
c. can destablize the slopes by increasing pore fluid pressure.
Explanation: A reservoir is a term used in Agriculture or geography to describe a location where water is either collected artificially or naturally for later use in farming or to act as dam for the supply of water in large communities. Most countries of the world make use of reservoirs to store water for future use.
Problem 1: Energy from Flow. Before the advent of the steam-powered engine, most mechanical processes were driven by extracting power from a nearby river using a water wheel. A miller wants to construct a water wheel to grind grain into an hour. He needs a total power output of 0.5 kW to meet the demand of his mill. If a nearby waterfall will flow at a rate of 400 liters/minute onto the top of the wheel, what is the required diameter of the water wheel to achieve the desired power output?
Answer:
Diameter will be 27394.76 m
Explanation:
Power P = 0.5 kW = 500 W
Time t required for grinding = 1 hr = 3600 sec
Energy required E = P x t
E = 500 x 3600 = 1800000 J
Flow rate of water Q = 400 ltr/min
We convert to m3/sec
400 ltr/min = 400/(1000 x 60) m3/ses
Q = 0.0067 m3/sec
Energy provided by flow will be
E = pgQd
Where p = density of water = 1000 kg/m3
g = acceleration due to gravity 9.81 m/s2
d = diameter of wheel.
Equating both energy, we have,
1800000 = 1000 x 9.81 x 0.0067 x d
1800000 = 65.73d
d = 1800000/65.73
d = 27394.76 m
An air-conditioner with refrigerant-134a as the working fluid is used to keep a room at 23°C by rejecting the waste heat to the outdoor air at 37°C. The room gains heat through the walls and the windows at a rate of 250 kJ/min while the heat generated by the computer, TV, and lights amounts to 900 W. The refrigerant enters the compressor at 400 kPa as a saturated vapor at a rate of 100 L/min and leaves at 1200 kPa and 70°C. Determine (a) the actual COP, (b) the maximum COP, and (c) the minimum volume flow rate of the refrigerant at the compressor inlet for the same compressor inlet and exit conditions.
Answer:
(a) 3.455
(b) 21.143
(c) 16.36L/min
Explanation:
In this question, we’d be providing solution to the working process of a refrigerator given the data in the question.
Please check attachment for complete solution and step by step explanation
a steam coil is immersed in a stirred heating tank. Saturated steam at 7.50 bar condenses within the coil , and the condensate emerges at at its saturation temperature. A solvent with a heat capacity of 2.30 kJ is fed to the tank at a steady rate of 12.0 kg/min and a temperature of 25°C, and the heated solvent is discharged at the same flow rate. The tank is initially filled with 760 kg of solvent at 25°C, at which point the flows of both steam and solvent are commenced. The rate at which heat is transferred from the steam coil to the solvent is given by the expression where UA (the product of a heat transfer coefficient and the area through which the heat is transferred) equals 11.5 kJ/min·°C. The tank is well stirred, so that the temperature of the contents is spatially uniform and equals the outlet temperature.
Write a differential energy balance on the tank contents.
Answer:
d/dt[mCp(Ts-Ti)] = FCp(Ts-Ti) - FoCp(Ts-Ti) + uA(Ts-Ti)
Explanation:
Differential balance equation on the tank is given as;
Accumulation = energy of inlet steam - energy of outlet steam+ heat transfer from the steamwhere;Accumulation = d/dt[mcp(Ts-Ti)]
Energy of inlet steam = FCp(Ts-Ti)
Energy of outlet steam = FoCp(Ts-Ti)
Heat transfer from the steam = uA(Ts-Ti)
Substituting into the formula, we have;
Accumulation = energy of inlet steam - energy of outlet steam+ heat transfer from the steamd/dt[mCp(Ts-Ti)] = FCp(Ts-Ti) - FoCp(Ts-Ti) + uA(Ts-Ti)The differential energy balance is [tex]\(\frac{dT}{dt} = \frac{11.5 T_s + 690 - 39.1 T}{1748}\).[/tex]
To write a differential energy balance on the tank contents, we need to consider the energy entering and leaving the system. The system in question is the well-stirred heating tank. Here are the steps to formulate the energy balance:
1. Define the system and parameters:
- The solvent enters the tank at a flow rate of 12.0 kg/min and at a temperature of 25°C.
- The solvent has a heat capacity [tex]\( C_p = 2.30 \text{ kJ/kg°C} \).[/tex]
- The tank is initially filled with 760 kg of solvent at 25°C.
- Heat transfer rate from the steam coil to the solvent is given by
[tex]\( UA(T_s - T) \)[/tex] where [tex]\( UA = 11.5 \text{ kJ/min·°C} \), \( T_s \)[/tex] is the steam temperature, and ( T ) is the solvent temperature.
- The tank is well-stirred, ensuring uniform temperature throughout, and the outlet temperature equals the tank temperature.
2. Energy balance:
The energy balance for a well-stirred tank in differential form is given by:
[tex]\[ \frac{d(U)}{dt} = \dot{Q}_{\text{in}} + \dot{m} C_p T_{\text{in}} - \dot{m} C_p T_{\text{out}} \][/tex]
Where:
- ( U ) is the internal energy of the tank contents.
[tex]- \( \dot{Q}_{\text{in}} \)[/tex] is the heat transfer rate from the steam coil.
[tex]- \( \dot{m} \)[/tex] is the mass flow rate of the solvent.
[tex]- \( T_{\text{in}} \)[/tex] is the inlet temperature of the solvent.
[tex]- \( T_{\text{out}} \)[/tex] of the solvent (equal to tank temperature ( T ).
Since [tex]\( \dot{m}_{\text{in}} = \dot{m}_{\text{out}} = \dot{m} \) and \( T_{\text{out}} = T \):[/tex]
[tex]\[ \frac{d(U)}{dt} = UA(T_s - T) + \dot{m} C_p (T_{\text{in}} - T) \][/tex]
3. Internal energy change:
The internal energy change of the tank contents can be expressed as:
[tex]\[ \frac{d(U)}{dt} = m C_p \frac{dT}{dt} \][/tex]
Where ( m ) is the mass of the solvent in the tank (760 kg) and ( T ) is the solvent temperature in the tank.
4. Combine the equations:
Substituting [tex]\( \frac{d(U)}{dt} = m C_p \frac{dT}{dt} \)[/tex] into the energy balance:
[tex]\[ m C_p \frac{dT}{dt} = UA(T_s - T) + \dot{m} C_p (T_{\text{in}} - T) \][/tex]
5. Simplify the equation:
Substitute the given values:
[tex]- \( m = 760 \text{ kg} \)\\ - \( C_p = 2.30 \text{ kJ/kg°C} \)\\ - \( \dot{m} = 12.0 \text{ kg/min} \) - \( UA = 11.5 \text{ kJ/min·°C} \)\\ - \( T_{\text{in}} = 25 \text{ °C} \)[/tex]
[tex]\[ 760 \cdot 2.30 \frac{dT}{dt} = 11.5 (T_s - T) + 12.0 \cdot 2.30 (25 - T) \][/tex]
Simplify to:
[tex]\frac{dT}{dt} = 11.5 (T_s - T) + 27.6 (25 - T) \]\[ 1748[/tex]
Further simplify:
[tex]\[ 1748 \frac{dT}{dt} = 11.5 T_s - 11.5 T + 690 - 27.6 T \][/tex]
Combine like terms:
[tex]\[ 1748 \frac{dT}{dt} = 11.5 T_s + 690 - 39.1 T \][/tex]
Finally:
[tex]\[ \frac{dT}{dt} = \frac{11.5 T_s + 690 - 39.1 T}{1748} \][/tex]
This is the differential energy balance equation for the tank contents.
Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg, and the air experiences an entropy decrease of 0.40 kJ/kg·K. Using constant specific heats, determine
(a) the exit temperature of the air,
(b) the work input to the compressor, and
(c) the entropy generation during this process.
Answer:
a) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air = [tex] c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}[/tex]
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
[tex] -0.40= 1.005(ln T_2 - 5.68697)- 0.5968[/tex]
Solving for T2 we have:
[tex] T_2 = 5.8828[/tex]
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
[tex] w_in = c_p(T_2 - T_1)+q_out[/tex]
[tex] w_in = 1.005(358.8 - 295)+120[/tex]
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es = [tex] \frac{q_out}{T_1}[/tex]
[tex] \frac{120kJ/kg.k}{295K}[/tex]
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k
Consider the expression for the change in entropy of the air.
[tex]\to \Delta S_{air}=c_p \ \In \frac{T_2}{T_1} - R \In \frac{P_2}{P_1} \\\\[/tex]
Here, change in entropy of air in a compressor is[tex]\Delta S_{air}[/tex], specific heat at constant pressure is [tex]c_p[/tex], the inlet temperature is [tex]T_1[/tex], outlet temperature is [tex]T_2[/tex], the gas constant is R, inlet pressure is [tex]P_1[/tex], and outlet pressure is[tex]P_2[/tex].
From the ideal gas specific heats of various common gases table, select the specific heat at constant pressure[tex]c_p[/tex] and gas constant (R) at air and temperature:
[tex]\to 22\ C \ \ or \ \ 295\ K\ \ as 1.005 \ \frac{kJ}{kg \cdot K} \ \ and \ \ 0.287\ \frac{kJ}{kg \cdot K}[/tex]
Substituting
Take exponential on both sides of the equation.
[tex]\to T_2 = exp(5.8828) = 358.8\ K \\\\[/tex]
Hence, the exit temperature of the air is [tex]358.8\ K[/tex]. Apply the energy balance to calculate the work input.
[tex]\to W_{in}=c_p(T_2-T_1 )+q_{out}[/tex]
Here, work input is [tex]w_{in}[/tex] initial temperature is [tex]T_1[/tex], exit temperature is [tex]T_2[/tex], and heat transfer outlet is [tex]q_{out}[/tex]
Substituting
Hence, the work input to the compressor is
Express the entropy generated in the process.
[tex]\to S_{gen} =\Delta S_{air}+\Delta S_{swr}[/tex]
Here, entropy generated is [tex]S_{gen}[/tex], change in entropy of air in a compressor is [tex]\Delta S_{air}[/tex], and change in entropy in the surrounding is [tex]\Delta S_{swr}[/tex].
Finding the changes into the surrounded entropy.
[tex]\to \Delta S_{swr} = \frac{q_{out}}{T_{swr}}[/tex]
Here, the heat transfer outlet is [tex]q_{out}[/tex] and the surrounded temperature is [tex]T_{swr}[/tex].
Substituting
[tex]120\ \frac{kJ}kg } \ for\ q_{out} \ and\ 22\ C \ for\ T_{swr}[/tex]
[tex]\Delta S_{swr} = \frac{ 120\frac{kJ}{kg}}{(22+273)\ K} =0.4068 \frac{kJ}{kg\cdot K}[/tex]
Finding the entropy generated process.
[tex]S_{gen} = \Delta S_{air} +\Delta S_{swr}\\[/tex]
Substituting
[tex]-0.40 \frac{kJ}kg\cdot K} \ for \ \Delta S_{air},\ and\ 0.4068 \frac{kJ}{kg\cdot K}\ for\ \Delta S_{swr}\\\\[/tex]
[tex]S_{gen}=-0.40 \frac{kJ}{kg\cdot K} +0.4068 \frac{kJ}{kg\cdot K} = 0.0068 \frac{kJ}{kg\cdot K}[/tex]
Therefore, the entropy generated in the process is [tex]0.0068\ \frac{kJ}{kgK}[/tex].
Learn more:
brainly.com/question/16392696
Given an unsorted array of distinct positive integers A[1..n] in the range between 1 and 10000 and an integer i in the same range. Here n can be arbitrary large. You want to find out whether there are 2 elements of the array that add up to i. Give an algorithm that runs in time O(n).
Answer:
Explanation:
Arbitrary means That no restrictions where placed on the number rather still each number is finite and has finite length. For the answer to the question--
Find(A,n,i)
for j =0 to 10000 do
frequency[j]=0
for j=1 to n do
frequency[A[j]]= frequency[A[j]]+1
for j =1 to n do
if i>=A[j] then
if (i-A[j])!=A[j] and frequency[i-A[j]]>0 then
return true
else if (i-A[j])==A[j] and frequency[j-A[j]]>1 then
return true
else
if (A[j]-i)!=A[j] and frequency[A[j]-i]>0 then
return true
else if (A[j]-i)==A[j] and frequency[A[j]-i]>1 then
return true
return false
simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The beam has a length L 5 4 m and rectangular cross section with a width of 200 mm and height of 300 mm. Determine the maximum permissible value for the maximum inten- sity, 0 q , if the allowable normal stresses in tension and compression are 120 MPa.
Answer:
q₀ = 350,740.2885 N/m
Explanation:
Given
[tex]q(x)=\frac{x}{L} q_{0}[/tex]
σ = 120 MPa = 120*10⁶ Pa
[tex]L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\[/tex]
We can see the pic shown in order to understand the question.
We apply
∑MB = 0 (Counterclockwise is the positive rotation direction)
⇒ - Av*L + (q₀*L/2)*(L/3) = 0
⇒ Av = q₀*L/6 (↑)
Then, we apply
[tex]v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x[/tex]
Then, we can get the maximum bending moment as follows
[tex]M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m[/tex]
then we get
[tex]M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}[/tex]
We get the inertia as follows
[tex]I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}[/tex]
We use the formula
σ = M*y/I
⇒ M = σ*I/y
where
[tex]y=\frac{h}{2} =\frac{0.3m}{2}=0.15m[/tex]
If M = Mmax, we have
[tex](\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}[/tex]
From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displayed by (b) dispersion strengthening. (1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles. (2) The hardening/strengthening effect is not retained at elevated temperatures for this process (3) The hardening/strengthening effect is retained at elevated temperatures for this process (4) The strength is developed by a heat treatment (5) The strength is developed without a heat treatment
Answer:
(a) Precipitation hardening
(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.
(2) The hardening/strengthening effect is not retained at elevated temperatures for this process.
(4) The strength is developed by a heat treatment.
(b) Dispersion strengthening
(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.
(3) The hardening/strengthening effect is retained at elevated temperatures for this process.
(5) The strength is developed without a heat treatment.
Water in a household plumbing system originates at the neighborhood water main where the pressure is 480 kPa, the velocity is 5 m/s, and the elevation is 2.44 m. A 19-mm (3/4-in) copper service line supplies water to a two-story residence where the faucet in the master bedroom is 40 m (of pipe) away from the main and at an elevation of 7.62 m. If the sum of the minor-loss coefficients is 3.5, estimate the maximum (open faucet) flow. How would this flow be affected by the operation of other faucets in the house?
Answer:
1. Maximum flow = 0.7768 L/s
2. The flow would reduced if other faucets were open. This is due to increase pipe flow and frictional resistance between the water main and the faucets.
Explanation:
See the attached file for the calculation.
A voltage regulator is to provide a constant DC voltage Vl=10V to a load Rl from a nominal Vcc=15V supply voltage. The load can vary from 20Ω to 1KΩ. The supply voltage Vcc can vary from 13V to 16V. The op-amp can provide a maximum output current of 20mA. a)Find the βnecessary for the transistor to provide the needed current. b)Find the maximum power the transistor must dissipate.
Answer:
Beta values can be from the equation=change in Vcc/nominal Vcc
Beta=16-3/15=3/15=1/5=0.20
Maximum power=I^2*R=40 W
Consider a venture with a small hole drilled in the side of the throat. This hole is connected via a tube to a closed reservoir. The purpose of the venture is to create vacuum in the reservoir when the venture is placed in an airstream. (The vacuum is defined as the pressure difference below the outside ambient pressure.) The venture has a throat to inlet area ratio of 0.85. Calculate the maximum vacuum obtainable in the reservoir when the venturi is placed in an airstream of 90 m/s at standard sea level conditions.
Answer:
1913meter per second square.
Explanation:
From the Context the vacuum can be said be the presence of 5e difference below the outside ambient temperature.
For the Venturi.
Please go through the attached file for the rest of the solutions and the answer.
a hollow shaft is required to transmit 600kw at 110 rpm, the maximum torque being 20% greater than the mean. the shear stress is not to exceed 63 mpa and twist in a length of 3 meters not t exeed 1.4 degrees Calculate the minimum external diameter satisfying these conditions.
Answer:
175.5 mm
Explanation:
a hollow shaft of diameter ratio 3/8 is required to transmit 600kw at 110 rpm, the maximum torque being 20% greater than the mean. the shear stress is not to exceed 63 mpa and twist in a length of 3 meters not t exeed 1.4 degrees Calculate the minimum external diameter satisfying these conditions. G = 80 GPa
Let
D = external diameter of shaft
Given that:
d = internal diameter of the shaft = 3/8 × D = 0.375D,
Power (P) = 600 Kw, Speed (N) = 110 rpm, Shear stress (τ) = 63 MPa = 63 × 10⁶ Pa, Angle of twist (θ) = 1.4⁰, length (l) = 3 m, G = 80 GPa = 80 × 10⁹ Pa
The torque (T) is given by the equation:
[tex]T=\frac{60 *P}{2\pi N}\\ Substituting:\\T=\frac{60*600*10^3}{2\pi*110} =52087Nm[/tex]
The maximum torque ([tex]T_{max[/tex]) = 1.2T = 1.2 × 52087 =62504 Nm
Using Torsion equation:
[tex]\frac{T}{J} =\frac{\tau}{R}\\ J=\frac{T.R}{\tau} \\\frac{\pi}{32}[D^4-(0.375D)^4]=\frac{62504*D}{2(63*10^6)} \\D^3(0.9473)=0.00505\\D=0.1727m=172.7mm[/tex]
[tex]\theta=1.4^0=\frac{1.4*\pi}{180}rad[/tex]
From the torsion equation:
[tex]\frac{T}{J} =\frac{G\theta }{l}\\ J=\frac{T.l}{G\theta} \\\frac{\pi}{32}[D^4-(0.375D)^4]=\frac{62504*3}{84*10^9*\frac{1.4*\pi}{180} } \\D=0.1755m=175.5mm[/tex]
The conditions would be satisfied if the external diameter is 175.5 mm
Alloy parts were cast with a sand mold that took 160 secs for a cube-shaped casting to solidify. The cube was 50 mm on a side. (a) Determine the value of the Chvorinov's mold constant. (b) If the same alloy and mold type were used, find the total solidification time for a cylindrical casting in which the diameter = 25 mm and length = 50 mm.
Answer:
a) k = 6.4 s/cm²
b) Te = 160 s
Explanation:
a) Given
Solidification time in s: Te = 160 s
L = 50 mm = 5 cm
We use the formula
Te = k*(V/A)ⁿ
k = Te*(A/V)ⁿ
where
k is the Chvorinov's mold constant
V is casting volume in cm³ = V = L³ = (5 cm)³ = 125 cm³
A is the surface area of the casting in cm² = A = L² = (5 cm)² = 25 cm²
n = 2 (assumed)
⇒ k = 160 s*(25 cm²/125 cm³)²
⇒ k = 6.4 s/cm²
b) Given
D = 25 mm = 2.5 cm ⇒ R = D/2 = 2.5 cm/2 = 1.25 cm
h = 50 mm = 5 cm
k = 6.4 s/cm²
n = 2
We find the volume as follows
V = π*R²*h ⇒ V = π*(1.25 cm)²*(5 cm) = 24.5436 cm³
and the surface area
A = π*R² = π*(1.25 cm)² = 4.9087 cm²
We apply the equation
Te = k*(V/A)ⁿ
⇒ Te = (6.4 s/cm²)*(24.5436 cm³/4.9087 cm²)²
⇒ Te = 160 s
consider a household that uses 23.8 kw-hour of electricity per day on average. (kw-hours is a measure of energy that will be discussed in detail in a later chapter. at this point we want to establish estimations.) most of that electricity is supplied by fossil fuels. to reduce their carbon footprint, the household wants to install solar panels, which receive on average 336 w/m2 from the sun each day. if the solar panels are 19.0% efficient (fraction of solar energy converted into useable electrical energy), what area of solar panels is needed to power the household
Answer:
15.53 m2
Explanation:
Energy needed = 23.8 kW-hr
Solar intensity of required panel = 336 W/m2
Efficiency of panels = 19%
Power needed = 23.8kW-hr = 23800 W-h,
In one day there are 24 hr, therefore power required = 23800/24 = 991.67 W
991.67 = 19% of incident power
991.67 = 0.19x
x = incident power = 991.67/0.19 = 5219.32 W
Surface area required = 5219.32/336
= 15.53 m2