Answer:
T=5.3° C
Explanation:
Given that
Power input to the pump = 2 KW
Building loses heat rate = 0.5 KW/C
So rate of heat transfer = 0.5(273+30-T)
rate of heat transfer = 0.5(303-T)
T=Ambient temperature
Building temperature = 30° C
We know that ,heat pump is used to heat the building.
COP of pump = 0.5 COP of Carnot heat pump
[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{273+30}{303-T}[/tex]
[tex]COP\ of\ Carnot\ heat\ pump=\dfrac{303}{303-T}[/tex]
[tex]COP\ of\ pump=\dfrac{303-T}{Power}[/tex]
[tex]COP\ of\ pump=0.5\times \dfrac{303-T}{2}[/tex] -----1
And also
[tex]COP\ of\ pump=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex] ----2
So from now equation 1 and 2
[tex]\dfrac{303-T}{4}=\dfrac{1}{2}\times \dfrac{303}{303-T}[/tex]
So T= 278.38 K=5.3° C
T=5.3° C
Ambient temperature =5.3° C.
Pressurized water ( 10 bar, 110°C) enters the bottom of an 10-m-long vertical tube of diameter 63 mm at a mass flow rate of 1.5 kg/s. The tube is located inside a combustion chamber, resulting in heat transfer to the tube. Superheated steam exits the top of the tube at 7 bar, 600°C. Determine the change in the rate at which the following quantities enter and exit the tube: (1) the combined thermal and flow work, (2) the mechanical energy, and (3) the total energy of the water. Also, (4) determine the heat transfer rate, . Hint: Relevant properties may be obtained from a thermodynamics text.
Answer:
(1) [tex]\Delta E = 4845.43 kW[/tex]
(2) [tex]\Delta E_{m} = 5.7319 kW[/tex]
(3) [tex]\Delta E_{t} = 4839.69 kW[/tex]
(4) q = 4839.69 kW[/tex]
Solution:
Using Saturated water-pressure table corresponding to pressure, P = 10 bar:
At saturated temperature, Specific enthalpy of water, [tex]h_{ws} = h_{f} = 762.5 kJ/kg[/tex]
At inlet:
Saturated temperature of water, [tex]T_{sw} = 179.88^{\circ}C[/tex]
Specific volume of water, [tex]V_{wi} = V_{f} = 0.00127 m^{3}/kg[/tex]
Using super heated water table corresponding to a temperature of [tex]600^{\circ}C[/tex] and at 7 bar:
At outlet:
Specific volume of water, [tex]V_{wso} = 0.5738 m^{3}/kg[/tex]
Specific enthalpy of water, [tex]h_{wo} = 3700.2 kJ/kg[/tex]
Now, at inlet, water's specific enthalpy is given by:
[tex]h_{i} = C_{p}(T - T_{sw}) + h_{ws}[/tex]
[tex]h_{i} = 4.187(110^{\circ} - 179.88^{\circ}) + 762.5[/tex]
[tex]h_{i} = -292.587 + 762.5= 469.912 kJ/kg[/tex]
(1) Now, the change in combined thermal energy and work flow is given by:
[tex]\Delta E = E_{o} - E_{i}[/tex]
[tex]\Delta E = m(h_{wo} - h_{i})[/tex]
[tex]\Delta E = 1.5(3700.2 - 469.912) = 4845.43 kW[/tex]
(2) The mechanical energy can be calculated as:
velocity at inlet, [tex]v_{i} = \rho A V_{wi}[/tex]
[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]
[tex]v_{i} = \frac{mV_{wi}}{frac{\pi d^{2}}{4}}[/tex]
[tex]v_{i} = \frac{1.5\times 0.00127}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]
[tex]v_{i} = 0.542 m/s[/tex]
Similarly,, the velocity at the outlet,
[tex]v_{o} = \frac{1.5\times 0.57378}{frac{\pi (63\times 10^{- 3})^{2}}{4}}[/tex]
[tex]v_{o} = 276.099 m/s[/tex]
Now, change in mechanical energy:
[tex]\Delta E_{m} = E_{mo} - E_{mi}[/tex]
[tex]\Delta E_{m} = m[(\frac{v_{o}^{2}}{2} + gz_{o}) - (\frac{v_{i}^{2}}{2} + gz_{i})][/tex]
[tex]\Delta E_{m} = 1.5[(\frac{276.099^{2}}{2} + 9.8(z_{o} - z_{i}) - (\frac{0.542^{2}}{2}][/tex]
[tex]\Delta E_{m} = 57319 J = 5.7319 kW[/tex]
(3) The total energy of water is given by:
[tex]\Delta E_{t} = E - E_{m} = 4845.43 - 5.7319 = 4839.69 kW[/tex]
(4) The rate of heat transfer:
q = [tex]\Delta E_{t} = 4839.69 kW[/tex]
Air enters an insulated turbine operating at steady state at 8 bar, 500K, and 150 m/s. At the exit the conditions are 1 bar, 320 K, and 10 m/s. There is no in elevation. Determine the work developed and the exergy destruction, each in kJ/kg of air flowing. Let To=300K and po=1bar significant change
Answer
given,
P₁ = 8 bar T₁ = 500 K V₁ = 150 m/s
P₂ = 1 bar T₂ = 320 K V₂ = 10 m/s
writing energy equation
h₁ + (KE)₁ + (PE)₁ + Q m = h₂ + (KE)₂ + (PE)₂ + W
[tex]W = (h_1 - h_2 ) + \dfrac{V_1^2-V_2^2}{2000}[/tex]
ideal gas property of air
T₁ = 500 K h₁ = 503.02 KJ/kg S₁ = 2.21952 kJ/kgK
T₂ = 320 K h₂ = 320.29 KJ/kg S₂ = 1.7679 kJ/kgK
[tex]W = (503.02-320.29) + \dfrac{150^2-10^2}{2000}[/tex]
W = 193.93 KJ/Kg
calculation of energy destruction
= [tex]T_0(S_2-S_1-Rln(\dfrac{P_2}{P_1}))[/tex]
= [tex]T_0(S_2-S_1+Rln(\dfrac{P_1}{P_2}))[/tex]
= [tex]300(1.7679-2.21952-\dfrac{8.314}{28.97}ln(\dfrac{8}{1}))[/tex]
=[tex]300 \times 0.145152[/tex]
=43.54 KJ/Kg
A rigid tank holds 22 kg of 127 °C water. If 9 kg of that is liquid water what is the pressure in the tank and volume of the tank?
Answer:
The pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.
Explanation:
Volume is constant as the tank is rigid. Take the saturation condition of water from the steam table for pressure at 127°C.
Given:
Total mass of water is 22 kg.
Mass of liquid water is 9 kg.
Temperature of water is 127°C.
From steam table at 127°C:
The pressure in the tank is 246.878 Kpa.
Specific volume of saturated water is 0.00106683 m³/kg.
Specific volume of saturated steam is 0.72721 m³/kg.
Calculation:
Step1
From steam table at 127°C:
The pressure in the tank is 246.878 Kpa.
Step2
Dryness fraction is calculated as follows:
[tex]x=\frac{m_{v}}{m_{t}}[/tex]
Here, dyness fraction is x, mass of vapor is [tex]m_{v}[/tex]and total mass is [tex]m_{t}[/tex].
Substitute the values in the above equation as follows:
[tex]x=\frac{m_{v}}{m_{t}}[/tex]
[tex]x=\frac{22-9}{22}[/tex]
x = 0.59
Step3
Specific volume of tank is calculated as follows:
[tex]v=v_{f}+x(v_{g}-v_{f})[/tex]
[tex]v=0.00106683+0.59(0.72721-0.00106683)[/tex]
[tex]v=0.00106683+0.42842447[/tex]
v=0.4295 m³/kg.
Step4
Volume is calculated as follows:
[tex]V=v\times m_{t}[/tex]
[tex]V=0.4295 \times22[/tex]
V=9.449 m³.
Thus, the pressure and volume of the tank are 246.878 Kpa and 9.449 m³ respectively.
Convert 250 lb·ft to N.m. Express your answer using three significant figures.
Answer:
It will be equivalent to 338.95 N-m
Explanation:
We have to convert 250 lb-ft to N-m
We know that 1 lb = 4.45 N
So foe converting from lb to N we have to multiply with 4.45
So 250 lb = 250×4.45 =125 N
And we know that 1 feet = 0.3048 meter
Now we have to convert 250 lb-ft to N-m
So [tex]250lb-ft=250\times 4.45N\times 0.348M=338.95N-m[/tex]
So 250 lb-ft = 338.95 N-m
The position of a particle along a straight-line path is defined by s = (t3 - 6t2 - 15t + 7) ft, where t is in seconds. When t = 8 s, determine the particle’s (a) instantaneous velocity and instantaneous acceleration, (b) average velocity and average speed
To determine the instantaneous velocity and acceleration of a particle described by the position function s(t), one needs to calculate the first and second derivatives of the position function and evaluate them at the given time, t = 8 s. The average velocity and speed require the change in position over a specific time interval, which is not provided in the question.
Explanation:The position of a particle along a straight-line path is given by the equation s = (t3 - 6t2 - 15t + 7) feet, where t is time in seconds. To find the particle's instantaneous velocity and instantaneous acceleration at t = 8 s, we need to take the first and second derivatives of the position function with respect to time, respectively.
The first derivative of s with respect to t gives us the velocity v(t), and the second derivative gives us the acceleration a(t). At t = 8 s, the velocities and accelerations can be calculated by plugging in the value of t into those derivatives.
To determine the average velocity and average speed, we take the change in position over the change in time interval for the specific time range provided.
Note: The question lacks sufficient specific information to calculate these values as the time interval for the average velocity and average speed is not provided. However, the general process of calculation has been explained.
Acceleration, instantaneous velocity, and particle position at different times are essential concepts in physics and particle motion analysis.
Acceleration is the rate of change of velocity with respect to time. In the given scenarios, acceleration is provided in terms of a function of time or as a constant value.
Instantaneous velocity is the velocity of the particle at a specific moment. It can be calculated by taking the derivative of the position function with respect to time.
Position of the particle at different times can be found by substituting the respective time values into the position function.
A particle moves with a constant speed of 6 m/s along a circular path of a radius of 4 m. What is the magnitude of its acceleration. Do not include units in your answer, assumed unit are m/s2.
Answer:
Acceleration in circular path will be 9
Explanation:
We have given speed of the particle in circular path = 6 m/sec
Radius of the circular path = 4 m
We have to find the centripetal acceleration [tex]a_c[/tex]
We know that centripetal acceleration is given by [tex]a_c=\frac{v^2}{r}=\frac{6^2}{4}=9[/tex]
As in question it is given that don't include the unit
So acceleration will be 9
Calculate the angle of banking on a bend of 100m radius so that vehicles can travel round the bend at 50km/hr without side thrust on the tyres.
Answer:
11.125°
Explanation:
Given:
Radius of bend, R = 100 m
Speed around the bend = 50 Km/hr = [tex]\frac{5}{18}\times50[/tex] = 13.89 m/s
Now,
We have the relation
[tex]\tan\theta=\frac{v^2}{gR}[/tex]
where,
θ = angle of banking
g is the acceleration due to gravity
on substituting the respective values, we get
[tex]\tan\theta=\frac{13.89^2}{9.81\times100}[/tex]
or
[tex]\tan\theta=0.1966[/tex]
or
θ = 11.125°
A square isothermal chip is of width w = 5 mm on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at T[infinity] = 15°C. From reliability considerations, the chip temperature must not exceed T = 85°C.f the coolant is air and the corresponding convection 200 W/m2 K, what is the maximum coefficient is h allowable chip power? If the coolant is a dielectric liquid for which h 3000 W/m2 K, what is the maxi- mum allowable power?
Answer:
Q(h=200)=0.35W
Q(h=3000)=5.25W
Explanation:
first part h=200W/Km^2
we must use the convection heat transfer equation for the chip
Q=hA(Ts-T∞)
h= convective coefficient=200W/m2 K
A=Base*Leght=5mmx5mm=25mm^2
Ts=temperature of the chip=85C
T∞=temperature of coolant=15C
Q=200x2.5x10^-5(85-15)=0.35W
Second part h=3000W/Km^2
Q=3000x2.5x10^-5(85-15)=5.25W
A 1/4th scale car is to be tested in a wind tunnel. If the full scale speed of the car is 30m/s, what should be the wind tunnel speed for Reynolds number similarity a) 30m/s b) 6m/s c) 7.5m/s d) 150m/s e) 120m/s
Answer:
e)v=120 m/s
Explanation:
Given that
Scale ratio = 1/4
Speed of car =30 m/s
lets wind tunnel speed is v
We know that Reynolds number given as
[tex]Re=\dfrac{\rho\ L\ V}{\mu }[/tex]
If all conditions taken as similar then
[tex](L\ V)_c=(L\ V)_w[/tex]
Given that
[tex]\dfrac{L_w}{L_c}=\dfrac{1}{4}[/tex]
So we can say that
4 x 30 = v x 1
v=120 m/s
A rectangular sheet of 120 mm x 160 mm can be used to develop the lateral surface of (a) A cylinder of radius 80/π (b) A square prism of side 40 mm (c) A hexagonal prism of side 20 mm (d) All of the above
Answer:
option D is correct
Explanation:
1) for a cylinder with radius is[tex]\frac{80}{\pi}[/tex]
lateral surface area of cylinder is [tex]2\pi rh[/tex]
[tex]= 2\pi \frac{80}{\pi}*120[/tex]
= 160* 120
2) fro square prism with side 40 mm
lateral surface area = 4ah
= 4*40 * 120
= 160*120
3)for hexagonal with side 20 mm
lateral surface area = 6ah
= 6*20*160
=120* 160
therefore option D is correcrt
Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The compressor pressure ratio is 7, and the turbine inlet temperature is 1200 K. For constant specific heats with k = 1.4 and Cp = 1.005 kJ/kg, calculate the percent thermal efficiency (enter a number only)
Answer:
The thermal efficiency of cycle is 42.6%.
Explanation:
Given that
[tex]T_1=300 K[/tex]
[tex]P_1=100KPa[/tex]
mass flow rate = 6 kg/s
Compression ratio = 7
Turbine inlet temperature = 1200 K
γ=1.4
We know that thermal efficiency of Brayton cycle given as
[tex]\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}[/tex]
Now by putting the values
[tex]\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}[/tex]
[tex]\eta=1-\dfrac{1}{7^{\frac{1.4-1}{1.4}}}[/tex]
η=0.426
So the thermal efficiency of cycle is 42.6%.
A substance temperature was 62 deg R. What is the temperature in deg C? A.) 50.7 B.) 45.54 C) 80.0 D) 94.4
Answer:
The temperature in degree Celsius will be -238.7055°C
Explanation:
We have given the substance temperature = 62°R
We have to convert degree Rankine to degree Celsius
For conversion from Rankine to Celsius we use formula
[tex]T_C=(T_R-491.67)\times\frac{5}{9}[/tex]
So [tex]T_C=(62-491.67)\times\frac{5}{9}[/tex]
[tex]T_C=-238.7055^{\circ}C[/tex]
So temperature in degree Celsius will be -238.7055°C
After calculation i got -238.7055°C but in option this is not given
The temperature of the substance will be "94.4°C". To understand the calculation, check below.
TemperatureAccording to the question,
Substance temperature, T°R = 62
or,
T°C = (T°R - 491.67) × [tex]\frac{5}{9}[/tex]
By substituting the values,
= -238.706
If we take the value,
T°C = (662 - 491.67) × [tex]\frac{5}{9}[/tex]
= 94.62°C or,
= 94.4°C
Thus the above response "Option D" is correct.
Find out more information about temperature here:
https://brainly.com/question/16559442
The shaft of a vacuum cleaner motor rotates with an angular acceleration of four times the shaft’s angular velocity raised to the ¾ power. The vacuum beater bar is attached to the motor shaft with pulley through a drive belt. The radii of the motor pulley and the beater bar are 0.25 in and 1.0 in respectively. Determine the angular velocity of the beater bar when t = 4 s, given that omega_0 is 1 rad/s when theta = 0.
Answer:
470 rad/s
Explanation:
The acceleration of the motor shaft is:
γ1 = 4*w1^(3/4)
When connected by a belt the pulleys have the same tangential speed
vt = w * r
vt1 = vt2
w1 * r1 = w2 * r2
w2 = w1 * r1/r2
Therefore:
γ2 = 4 * (w1 * r1/r2)^(3/4)
d(w1 * r1/r2)/dt = 4 * (w1 * r1/r2)^(3/4)
(r1/r2) * dw1/dt = 4 * (r1/r2)^(3/4) * (w1 * r1/r2)^(3/4)
dw1/dt = 4 * (r1/r2)^(-1/4) * (w1)^(3/4)
This is a differential equation.
Solving it through Wolfram Alpha:
w1(t) = (1 / 256) * (4 * (r1/r2)^(-1/4) * t - 4)^4
w1(4) = (1 / 256) * (4 * (0.25 / 1)^(-1/4) * 4 - 4)^4 = 470 rad/s
The angular velocity of the beater bar at t=4s is approximately 6.37 rad/s, based on the given angular acceleration equation and initial angular velocity.
Explanation:The angular velocity of the beater bar can be found using the relationship between angular acceleration and angular velocity. The given equation states that the angular acceleration is four times the angular velocity raised to the 3/4 power. Therefore, we can write:
α = 4 * ω^(3/4)
To find the angular velocity at t=4s, we can integrate the equation to get:
ω = 4/7 * t^7/4 + C
When t = 0, ω = ω_0 = 1 rad/s. Substituting these values, we can solve for C:
1 = 0 + C
Therefore, C = 1. Finally, we can substitute t = 4s into the equation to get the angular velocity:
ω = 4/7 * 4^7/4 + 1
Calculating this expression, we find that the angular velocity of the beater bar at t=4s is approximately 6.37 rad/s.
The Phoenix with a mass of 390 kg was a spacecraft used for exploration of Mars. Determine the weight of the Phoenix, in N, (a) on the surface of Mars where the acceleration of gravity is 3.73 m/s2 and (b) on Earth where the acceleration of gravity is 9.81 m/s2.
Answer:
a) on mars W=1454.7N
b)on earth W=3825.9N
Explanation:
The weight of any body with mass is given by the following equation
W=mg
where
m=mass
g=gravity
W=weight
Remember that the weight is expresed in Newton and the units are kgm/s^2
A)weight on the surface of mars
W=(390kg)(3.73m/s^2)=1454.7N
b) on earth
W=(390kg)(9.81m/s^2)=3825.9N
A train which is traveling at 70 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its decreased velocity is observed to be 52 mi/hr as it passes a point 1/2 mi beyond A. A car moving at 52 mi/hr passes point B at the same instant that the train reaches point A. In an unwise effort to beat the train to the crossing, the driver steps on the gas.
(a) Calculate the constant acceleration a that the car must have in order to beat the train to the crossing by 4.3 sec.
(b) find the velocity v of the car as it reaches the crossing.
Answer:
a) 0 mi/s^2
b) 52 mi/s
Explanation:
Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)
The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h
The equation for position under constant acceleration is:
X(t) = X0 + V0 * t + 1/2 * a * t^2
I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.
Also the equation for speed under constant acceleration is:
V(t) = V0 + a * t
Replacing
52 = 70 + a * t
Rearranging
a * t = 52 - 70
a = -18/t
I can then calculate the time it will take it to reach the crossing
1/2 * a * t^2 + V0 * t - X(t) = 0
Replacing
1/2 (-18/t) * t^ + 70 * t - 1/2 = 0
-9 * t + 70 * t = 1/2
61 * t = 1/2
t = (1/2)/61 = 0.0082 h = 29.5 s
And the acceleration is:
a = -18/0.0082 = -2195 mi/(h^2)
To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s
X(t) = X0 + V0 * t + 1/2 * a * t^2
52 mi/h = 0.0144 mi/s
1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2
1/2 = 0.363 + 317.5 * a
317.5 * a = 0.5 - 0.363
a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)
The car should remain at about constant speed.
It will be running at the same speed.
A 29-mm-diameter copper rod is 1.1 m long with a yield strength of 73 MPa. Determine the axial force necessary to cause the diameter of the rod to reduce by 0.01 percent, assuming elastic deformation. Check that the elastic deformation assumption is valid by comparing the axial stress to the yield strength. The axial force necessary to cause the diameter of the rod to reduce by 0.01 percent is ____kN.
Answer:
32.96 MPa
Explanation:
The Poisson ratio of copper is:
μ = 0.355
The Young's modulus of copper is:
E = 117 GPa
The equation for reduction of diameter of a rod is:
D = D0 * (1 - μ*σ/E)
Rearranging:
D = D0 - D0*μ*σ/E
D0*μ*σ/E = D0 - D
D0*μ*σ = E*(D0 - D)
σ = E*(D0 - D) / (D0*μ)
If the diameter is reduced by 0.01 percent
D = 0.9999*D0
σ = E*(D0 - 0.9999*D0) / (D0*μ)
σ = E*(0.0001*D0) / (D0*μ)
σ = 0.0001*E / μ
σ = 0.0001*117*10^9 / 0.355 = 32.96 MPa
This value is below the yield strength, therefore it is valid.
An alloy has a yield strength of 818 MPa and an elastic modulus of 104 GPa. Calculate the modulus of resilience for this alloy [in J/m3 (which is equivalent to Pa)] given that it exhibits linear elastic stress-strain behavior.
Answer:
Modulus of resilience will be [tex]3216942.308j/m^3[/tex]
Explanation:
We have given yield strength [tex]\sigma _y=818MPa[/tex]
Elastic modulus E = 104 GPa
We have to find the modulus
Modulus of resilience is given by
Modulus of resilience [tex]=\frac{\sigma _y^2}{2E}[/tex], here [tex]\sigma _y[/tex] is yield strength and E is elastic modulus
Modulus of resilience [tex]=\frac{(818\times 10^6)^2}{2\times 104\times 10^9}=3216942.308j/m^3[/tex]
Find the error in the following proof that 2 = 1. Consider the equation a = b. Multiply both sides by a to obtain a 2 = ab. Subtract b 2 from both sides to get a 2 − b 2 = ab − b 2 . Now factor each side, (a + b)(a − b) = b(a − b), and divide each side by (a − b) to get a + b = b. Finally, let a and b equal 1, which shows that 2 = 1.
Answer:
You can't divide by zero
Explanation:
The error appears when you divide each side by (a - b). If a = b, then (a - b) = 0 and you can't divide each side by 0. Moreover, the equation before division, that is, (a + b)(a − b) = b(a − b) is true after replacing (a - b) = 0 because it gives 0 = 0.
The error in the proof lies in the step where we divided both sides by (a - b) . Since [tex]\( a = b \), \( (a - b) \)[/tex] becomes 0, making the division by 0 undefined.
The error occurs when dividing both sides by (a - b) . Since ( a = b ), ( (a - b) becomes 0. Division by 0 is undefined, leading to the invalid conclusion.
To elaborate, dividing both sides by (a - b) in the equation (a + b)(a - b) = b(a - b) results in:
[tex]\[ \frac{(a + b)(a - b)}{(a - b)} = \frac{b(a - b)}{(a - b)} \]\[ \frac{(a + b)\cancel{(a - b)}}{\cancel{(a - b)}} = \frac{b\cancel{(a - b)}}{\cancel{(a - b)}} \]\[ a + b = b \][/tex]
This step assumes (a - b) is not equal to zero, leading to the false conclusion that ( a + b = b ), which is only true when a and ( b ) are equal.
Therefore, the proof incorrectly concludes that ( 2 = 1 ) due to the division by zero, highlighting the importance of avoiding such mathematical errors.
A reversible cyclic device does work while exchanging heat with three constant temperature reservoirs. The three reservoirs (1,2, and 3) are at temperatures of 1000K, 300K, and 500K. Four hundred kilojoules of heat are transferred from reservoir 1 to the device, and the total work done is 100 kJ. Find the magnitude and direction of the heat transfer from the other two reservoirs.
Answer:
Lets take [tex]Q_2[/tex] heat transfer take place from 500 K reservoir to device and [tex]Q_3[/tex] from device to 300 K reservoir.
From the energy conservation we can say that
[tex]400+Q_2=100+Q_3[/tex]
[tex]300=Q_3-Q_2[/tex] -----1
For reversible process
[tex]\dfrac{400}{1000}+\dfrac{Q_2}{500}-\dfrac{Q_3}{300}=0[/tex]
[tex]5Q_3-3Q_2=600[/tex] ----2
By solving above two equation
[tex]Q_3=-150 KJ,Q_2=-450KJ[/tex]
But here sign come negative it means that
[tex]Q_2[/tex] heat transfer take place from device to 500 K reservoir and [tex]Q_3[/tex] from 300 K reservoir to device.
In this exercise we have to use the knowledge of heat transfer to calculate the heat transferred to the other two reservoirs will be:
The magnitude of the Q2 is -450 in the out direction while the Q3 is -150 in the inward direction.
From the information given in the statement, we have that:
Temperatures of 1000K, 300K, and 500K. Four hundred kilojoules of heat are transferred Total work done is 100 kJ.knowing that from conservation we can say:
[tex]400+Q_2=100+Q_3\\300=Q_3-Q_2\\\frac{400}{1000}+\frac{Q_2}{500}-\frac{Q_3}{300}=0\\[/tex]
So solving we have:
[tex]Q_3=-150KJ\\Q_2=-450KJ[/tex]
See more about heat transfer at brainly.com/question/12107378
What colour is best for radiative heat transfer? a. Black b. Brown c. Blue d. White
Answer:
The correct answer is option 'a': Black
Explanation:
As we know that for an object which is black in color it absorbs all the electromagnetic radiation's that are incident on it. Thus if we need to transfer energy to an object by radiation the most suitable color for the process is black.
In contrast to black color white color is an excellent reflector, reflecting all the incident radiation that may be incident on it hence is the least suitable material for radiative heat transfer.
Find the Hooke's law of orthotropic and transverse isotropic material in matrix form
Answer:
Hooke's law is a principle of physics that states that the force F needed to extend or compress a spring by some distance X is proportional to that distance. That is: F = kX, where k is a constant factor characteristic of the spring: its stiffness, and X is small compared to the total possible deformation of the spring.
Explanation:
Explain the following boundary layer concepts (i) Boundary layer thickness (ii) Boundary layer transition
Answer with Explanation:
i) Boundary layer thickness: It is the thickness of the boundary layer formed around an object that is placed in the path of a flowing viscous fluid.The boundary layer thickness is the thickness up to which the effect of the object on the flow can be felt. When a viscous flowing fluid encounters an object in it's path of flow, the flowing fluid forms a thin layer of fluid over the object and this layer of fluid is known as boundary layer. This is a phenomenon only observed in the viscous fluids. As shown in the below figure a uniform flow of a viscous fluid encounters a plate, as we can see the thickness of the boundary layer goes on increasing as we move away from the leading edge of the plate the thickness of the boundary layer at any position is termed as boundary layer thickness.
ii) Boundary layer transition: It is the transition of the flow from a laminar nature to fully developed turbulent flow as it moves over an object. It occurs due to change in the Reynolds number of the flow as the effect of boundary layer increases as we move away from the leading edge of the object.
The absolute pressure in water at a depth of 9 m is read to be 185 kPa. Determine: a. The local atmospheric pressure b. The absolute pressure at a depth of 5 m in liquid whose specific gravity is 0.8 at the same location.
Answer:
a)Patm=135.95Kpa
b)Pabs=175.91Kpa
Explanation:
the absolute pressure is the sum of the water pressure plus the atmospheric pressure, which means that for point a we have the following equation
Pabs=Pw+Patm(1)
Where
Pabs=absolute pressure
Pw=Water pressure
Patm= atmospheric pressure
Water pressure is calculated with the following equation
Pw=γ.h(2)
where
γ=especific weight of water=9.81KN/M^3
H=depht
A)
Solving using ecuations 1 y 2
Patm=Pabs-Pw
Patm=185-9.81*5=135.95Kpa
B)
Solving using ecuations 1 y 2, and atmospheric pressure
Pabs=0.8x5x9.81+135.95=175.91Kpa
The A-36 steel pipe has a 6061-T6 aluminum core. It issubjected to a tensile force of 200 kN. Determine the averagenormal stress in the aluminum and the steel due to thisloading.The pipe has an outer diameter of 80 mm and aninner diameter of 70mm.
Answer:
In the steel: 815 kPa
In the aluminum: 270 kPa
Explanation:
The steel pipe will have a section of:
A1 = π/4 * (D^2 - d^2)
A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core:
A2 = π/4 * d^2
A2 = π/4 * 0.7^2 = 0.3848 m^2
The parts will have a certain stiffness:
k = E * A/l
We don't know their length, so we can consider this as stiffness per unit of length
k = E * A
For the steel pipe:
E = 210 GPa (for steel)
k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum:
E = 70 GPa
k2 = 70*10^9 * 0.3848 = 2.69*10^10 N
Hooke's law:
Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:
ε = f / k
When the force is distributed between both materials will stretch the same length:
f = f1 + f2
f1 / k1 = f2/ k2
Replacing:
f1 = f - f2
(f - f2) / k1 = f2 / k2
f/k1 - f2/k1 = f2/k2
f/k1 = f2 * (1/k2 + 1/k1)
f2 = (f/k1) / (1/k2 + 1/k1)
f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KN
f1 = 200 - 104 = 96 kN
Then we calculate the stresses:
σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPa
σ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
The answer is: within the steel: 815 kPa
When within the aluminum: 270 kPa
The aluminumWhen The steel pipe will have a piece of:Then A1 = π/4 * (D^2 - d^2)After that A1 = π/4 * (0.8^2 - 0.7^2) = 0.1178 m^2
The aluminum core is: Now A2 = π/4 * d^2Then A2 = π/4 * 0.7^2 = 0.3848 m^2
After that The parts will have a particular stiffness:k = E * A/l
We don't know their length, so we are able to consider this as stiffness per unit of length k = E * A
For the steel pipe: E = 210 GPa (for steel) k1 = 210*10^9 * 0.1178 = 2.47*10^10 N
For the aluminum: E = 70 GPak2 = 70*10^9 * 0.3848 = 2.69*10^10 NHooke's law:Δd = f / k
Since we are using stiffness per unit of length we use stretching per unit of length:ε = f / k
When the force is distributed between both materials will stretch the identical length:f = f1 + f2f1 / k1 = f2/ k2
Replacing: f1 = f - f2(f - f2) / k1 = f2 / k2f/k1 - f2/k1 = f2/k2f/k1 = f2 * (1/k2 + 1/k1)f2 = (f/k1) / (1/k2 + 1/k1)f2 = (200000/2.47*10^10) / (1/2.69*10^10 + 1/2.47*10^10) = 104000 N = 104 KNf1 = 200 - 104 = 96 kN
Then we calculate the stresses:σ1 = f1/A1 = 96000 / 0.1178 = 815000 Pa = 815 kPaσ2 = f2/A2 = 104000 / 0.3848 = 270000 Pa = 270 kPa
Find out more information about Aluminum here:
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The typical area of a commercial airplane's passenger window is 80.0 in^2 . At an altitude of 3.00 × 104 ft above the sea level, the atmospheric pressure is 0.350 atm. Determine the net force on the passenger window during flight at that altitude for both the English Engineering (EE) and SI unit systems. Use appropriate units and unit conversions in all steps of your calculations.
Answer:
The force over the plane windows are 764 lbf in the EE unit system and 3398 N in the international unit system.
Explanation:
The net force over the window is calculated by multiplying the difference in pressure by the area of the window:
F = Δp*A
The pressure inside the plane is around 1 atm, hence the difference in pressure is:
Δp = 1atm - 0.35 atm = 0.65 atm
Expressing in the EE unit system:
Δp = 0.65 atm * 14.69 lbf/in^2 = 9.55 lbf/in^2
Replacing in the force:
F = 9.55 lbf/in^2 * 80 in^2 = 764 lbf
For the international unit system, we re-calculate the window's area and the difference in pressure:
A = 80 in^2 * (0.0254 m/in)^2 = 0.0516 m^2
Δp = 0.65 atm * 101325 Pa = 65861 Pa = 65861 N/m^2
Replacing in the force:
F = 65861 N/m^2 *0.0516 m^2 = 3398 N
Engineering controls are the physical changes that employers make to the work environment or to equipment that make it safer to use.
A) True B) False
Answer:
True
Explanation:
Engineering controls are those techniques used to reduce or eliminate hazards of any condition, thereby protecting the workers.
These are mostly products that act as barriers between the worker and the hazard. This may include machinery or equipment. The common engineering controls used are glovebox, biosafety cabinet, fume hood, vented balance safety enclosure, HVAC system, lockout-tagout, sticky mat and rupture disc.
A 200L tank is evacuated and then filled through a valve connected to an air reservoir at 1 MPa and 20 °C. The valve is shut off when the pressure in the tank reaches 0.5 MPa. What is the mass of air in the tank?
Answer:
The air mass in the tank is 23.78 kg
Solution:
As per the question:
Volume of the tank, [tex]V_{t} = 200 l = 2\times 10^{- 3} m^{3}[/tex]
Pressure, P = 1 MPa = [tex]1\times 10^{6} Pa[/tex]
Temperature, T = [tex]20^{\circ}C[/tex] = 273 + 20 = 293 K
Pressure, P' = 0.5 MPa = [tex]0.5\times 10^{6} Pa[/tex]
Now,
To calculate the air mass, [tex]m_{a}[/tex] we use:
[tex]PV_{t} = m_{a}RT[/tex]
where
R = Rydberg constant = 0.287 J/kg.K
[tex]1\times 10^{6}\times 2\times 10^{- 3} = m_{a}0.287\times 293[/tex]
[tex]m_{a} = 23.78 kg[/tex]
Explain the difference between planning and shaping by the help of sketch
Explanation:
In shaping work piece will be stationary and tool will reciprocates,but on the other hand in planning work piece will reciprocates and tool will be stationary.Shaping is used for small work piece and planning is used for large work piece.Both shaping and planning are not continuous cutting process,cutting action take place in forward stroke and return stroke is idle stroke.The velocity of return stroke is much more than the forward stroke.
A worker's hammer is accidentally dropped from the 20th floor of a building under construction. With what velocity does it strike the pavement 304 ft below, and what time t is required?
Answer:
Final Velocity (Vf)= 139.864 ft/s
Time (t)= 4,34 s
Explanation:
This is a free fall problem, to solve it we will apply free fall concepts:
In a free fall the acceletarion is gravity (g) = 9,81 m/s2, if we convert it to ft/s^2 = g= 32.174 ft/s^2
Final velocity is Vf= Vo+ g*t[tex]Vf^{2} = Vo^{2} +2*g*hwhere h is height (304 ft in this case).
Vo =0 since the hammer wasn't moving when it stared to fall
Then Vf^2= 0 + 2* 32.174 ft/s^2 *304 ft
Vf^2= 19,561.8224 ft^2/s^2
Vf=[sqrt{19561.8224 ft^2/s^2}
Vf=139.864 ft/s
Time t= (Vf-Vo)/g => (139.864 ft/s-0)/32.174 ft/s^2 = 4.34 sec
Good luck!
A 0.5 m^3 container is filled with a mixture of 10% by volume ethanol and 90% by volume water at 25 °C. Find the weight of the liquid.
Answer:
total weight of liquid = 4788.25 N or 488.09 kg
Explanation:
given data
total volume = 0.5 m³
volume of ethanol = 10 % of volume = 0.10 × 0.5 = 0.05 m³
volume of water = 90 % at 25 °C of volume = 0.90 × 0.5 = 0.45 m³
to find out
weight of the liquid
solution
we know that density of water at 25 is 997 kg/m³
and density of ethanol is 789 kg/m³
so weight of water is = density × volume × g
put here value and we take g = 9.81
weight of water is = 997 × 0.45 × 9.81
weight of water = 4401.25 N ......................1
weight of ethanol is = density × volume × g
put here value and we take g = 9.81
weight of ethanol is = 789 × 0.05 × 9.81
weight of ethanol = 387.00 N ...............2
so total weight of liquid = sum of equation 1 add 2
total weight of liquid = 4401.25 + 387
total weight of liquid = 4788.25 N or 488.09 kg