Explanation:
The obtained data from water properties tables are:
Point 1 (condenser exit) @ 8 KPa, saturated fluid
[tex]h_{f} = 173.358 \\h_{fg} = 2402.522[/tex]
Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid
[tex]h_{2a} = 489.752\\h_{2b} = 313.2[/tex]
Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam
[tex]h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876[/tex]
Point 4 (Turbine exit) @ 8 KPa, mixed fluid
[tex]x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119[/tex]
Calculate mass flow rates
Part a) @ 18 MPa
mass flow
[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}[/tex]
Heat transfer rate through boiler
[tex]Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W[/tex]
Heat transfer rate through condenser
[tex]Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W[/tex]
Thermal Efficiency
[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485[/tex]
Part b) @ 4 MPa
mass flow
[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}[/tex]
Heat transfer rate through boiler
[tex]Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W[/tex]
Heat transfer rate through condenser
[tex]Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W[/tex]
Thermal Efficiency
[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275[/tex]
For (a) 18 MPa:
- The mass flow rate of steam is approximately 238,050 kg/h.
- The heat transfer rates are approximately 674,970 kW (boiler) and 481,360 kW (condenser).
- The thermal efficiency is approximately 14.81%.
For (b) 4 MPa:
- The mass flow rate of steam is approximately 187,320 kg/h.
- The heat transfer rates are approximately 480,040 kW (boiler) and 380,450 kW (condenser).
- The thermal efficiency is approximately 20.83%.
Explanation and Calculation
To analyze the ideal Rankine cycle, we need to follow these steps for each case:
Case (a): 18 MPa
1. Determine Specific Enthalpies
- Condenser exit: Saturated liquid at 8 kPa.
From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]
-Pump exit: Compressed liquid at 18 MPa.
Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]
[tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]
[tex]\[ h_2 \approx 191.81 + 0.001 \times (18000 - 8) \approx 209.81 \, \text{kJ/kg} \][/tex]
- Boiler exit: Saturated vapor at 18 MPa.
From steam tables: [tex]\( h_3 \approx 2821.6 \, \text{kJ/kg} \)[/tex]
-Turbine exit: Expanding to 8 kPa.
From steam tables: [tex]\( h_4 \approx 2201.4 \, \text{kJ/kg} \)[/tex]
2. Mass Flow Rate of Steam
Using the power output:
[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} (2821.6 - 2201.4 - (209.81 - 191.81)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} \times 420.2 \][/tex]
[tex]\[ \dot{m} \approx 238050 \, \text{kg/h} \][/tex]
3. Heat Transfer Rates
- Boiler:
[tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]
[tex]\[ \dot{Q}_{\text{in}} = 238050 \times (2821.6 - 209.81) \approx 674.97 \times 10^3 \, \text{kW} \][/tex]
- Condenser:
[tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]
[tex]\[ \dot{Q}_{\text{out}} = 238050 \times (2201.4 - 191.81) \approx 481.36 \times 10^3 \, \text{kW} \][/tex]
4. Thermal Efficiency
[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]
[tex]\[ \eta = \frac{100 \times 10^3}{674.97 \times 10^3} \approx 14.81\% \][/tex]
Case (b): 4 MPa
1. Determine Specific Enthalpies
- Condenser exit: Saturated liquid at 8 kPa.
From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]
- Pump exit: Compressed liquid at 4 MPa.
Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]
[tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]
[tex]\[ h_2 \approx 191.81 + 0.001 \times (4000 - 8) \approx 195.81 \, \text{kJ/kg} \][/tex]
- Boiler exit: Saturated vapor at 4 MPa.
From steam tables: [tex]\( h_3 \approx 2749.7 \, \text{kJ/kg} \)[/tex]
- Turbine exit: Expanding to 8 kPa.
From steam tables: [tex]\( h_4 \approx 2222.0 \, \text{kJ/kg} \)[/tex]
2. Mass Flow Rate of Steam
Using the power output:
[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} (2749.7 - 2222.0 - (195.81 - 191.81)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} \times 531.9 \][/tex]
[tex]\[ \dot{m} \approx 187320 \, \text{kg/h} \][/tex]
3. Heat Transfer Rates
- Boiler:
[tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]
[tex]\[ \dot{Q}_{\text{in}} = 187320 \times (2749.7 - 195.81) \approx 480.04 \times 10^3 \, \text{kW} \][/tex]
- Condenser:
[tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]
[tex]\[ \dot{Q}_{\text{out}} = 187320 \times (2222.0 - 191.81) \approx 380.45 \times 10^3 \, \text{kW} \][/tex]
4. Thermal Efficiency
[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]
[tex]\[ \eta = \frac{100 \times 10^3}{480.04 \times 10^3} \approx 20.83\% \][/tex]
At Westtown High School, the mean score on the French final examination was with a standard deviation of , while the mean score on the Spanish final examination was with a standard deviation of . To earn a language honor at graduation, students must score in the th percentile on all their language final exams. Brynne scored on both the French exam and the Spanish exam. Is Brynne qualified for honors?
Answer:
The score for both exams 88 is above the 90 percentile, so then Brynne qualified for honors. See the explanation below.
Explanation:
Assuming the following question:"At Westtown High School, the mean score on the French final examination was 81 with a standard deviation of 5, while the mean score on the Spanish final examination was 72 with a standard deviation of 12.
To earn a language honor at graduation, students must score in the 90th percentile on all their language final exams. Brynne scored 88 on both the French exam and the Spanish exam. Is Brynne qualified for honors?"
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
French case
Let X the random variable that represent the scores for the French exam of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(81,5)[/tex]
Where [tex]\mu=81[/tex] and [tex]\sigma=5[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.90[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-81}{5}[/tex]
And if we solve for a we got
[tex]a=81 +1.28*5=87.4[/tex]
So the value for the scores that separates the bottom 90% of data from the top 10% is 87.4 (90th percentile).
And since the score of Brynne is 88 is above the 90 percentile
Spanish case
Let X the random variable that represent the scores for the Spanish exam of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(72,12)[/tex]
Where [tex]\mu=72[/tex] and [tex]\sigma=12[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.90[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-72}{12}[/tex]
And if we solve for a we got
[tex]a=72 +1.28*12=87.36[/tex]
So the value for the scores that separates the bottom 90% of data from the top 10% is 87.36 (90th percentile).
And since the score of Brynne is 88 is above the 90 percentile
My Notes How many grams of perchloric acid, HClO4, are contained in 39.1 g of 74.9 wt% aqueous perchloric acid?
How many grams of water are in the same solution?
Answer:
a)29.9 b) 9.81
Explanation:
Wt% = mass of solute / mass of solvent × 100
0.749 = mass of solute / mass of solvent
a) Mass of perchloric acid = 0.749 × 39.1 = 29.29
b) Mass of water = 39.1 - 29.29 = 9.81
Determine the percent increase in the nominal moment capacity of the section in Problem 2 when including compression steel at top equal to 0.5 the area of the tension steel at the bottom.
Explanation:
Please kindly share your problem two with us as to know the actual problem we are dealing with, the question looks incomplete
The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.
a. What is the forward thrust of this engine?
b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?
c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?
Answer:
T = 5416.67 N
T = -2083.5 N
T = 0
Explanation:
Forward thrust has positive values and reverse thrust has negative values.
part a
Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s
The thrust force represents the horizontal or x-component of momentum equation:
[tex]T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N[/tex]
Answer: The thrust force T = 5416.67 N
part b
Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:
[tex]T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N[/tex]
Answer: The thrust force T = -2083.5 N reverse direction
part c
Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.
Answer: T = 0 N
Assume a strain gage is bonded to the cylinder wall surface in the direction of the hoop strain. The strain gage has nominal resistance R0 and a Gage Factor GF. It is connected in a Wheatstone bridge configuration where all resistors have the same nominal resistance; the bridge has an input voltage, Vin. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.)Calculate the voltage change ∆V across the Wheatstone bridge when the cylinder is pressurized to ∆P = 2.5 atm. Assume the vessel is made of 3004 aluminum with height h = 21 cm, diameter d = 9 cm, and thickness t = 65 µm. The Gage Factor is GF = 2 and the Wheatstone bridge has Vin = 6 V. The strain gage has nominal resistance R0 = 120 Ω.
Answer:
5.994 V
Explanation:
The pressure as a function of hoop strain is given:
[tex]P = \frac{4*E*t}{D}*\frac{e_{h} }{2-v}[/tex]
[tex]e_{h} = \frac{D*P*(2-v)}{4*E*t} .... Eq1[/tex]
For wheat-stone bridge with equal nominal resistance of resistors:
[tex]V_{out} = \frac{GF*e*V_{in} }{4} .... Eq2[/tex]
Hence, input Eq1 into Eq2
[tex]V_{out} = \frac{GF*e*V_{in}*D*P*(2-v) }{16*E*t} .....Eq3\\[/tex]
Given data:
P = 253313 Pa
D = d + 2t = 0.09013 m
t = 65 um
GF = 2
E = 75 GPa
v = 0.33
Use the data above and compute Vout using Eq3
[tex]V_{out} = \frac{2*6*0.09013*253313*(2-0.33) }{16*75*10^9*65*10^-6} \\\\V_{out} = 0.006285 V\\\\change in V = 6 - 0.006285 = 5.994 V[/tex]
(1 point) Consider the initial value problem 2????y′=4y, y(−1)=−2. Find the value of the constant ???? and the exponent ???? so that y=???????????? is the solution of this initial value problem. y= help (formulas) Determine the largest interval of the form ????<????<???? on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution. help (inequalities) What is the actual interval of existence for the solution (from part a)? help (inequalities)
Answer:
Here is the missing part of the question ;
find the value of the constant C and the exponent r so that y = Ctr is the solution of this initial value problem. Determine the largest interval of the form a<t<b
Explanation:
The step by step explanation is as given in the attachment. You will notice I used β as the exponential in place of r.A coal-fired power plant equipped with a SO2 scrubber is required to achieve an overall SO2 removal efficiency of 85%. The existing scrubber is 95% efficient. Rather than treating the entire gas stream to 95% removal, the plant proposes to treat part of the flue gas to 95% removal, and to bypass the remainder around the scrubber. Calculate the fraction of the flue gas stream that can be bypassed around the scrubber (i.e., Qbypass/Q) and still satisfy the regulatory requirement.
Answer:
bypassed fraction B will be B= 0.105 (10.5%)
Explanation:
doing a mass balance of SO₂ at the exit
total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂
F*(1-er) = Fs*(1-es) + Fb
where
er= required efficiency
es= scrubber efficiency
Fs and Fb = total mass inflow of SO₂ to the scrubber and to the bypass respectively
F= total mass inflow of SO₂
and from a mass balance at the inlet
F= Fs+ Fb
therefore the bypassed fraction B=Fb/F is
F*(1-er) = Fs*(1-es) + Fb
1-er= (1-B)*(1-es) +B
1-er = 1-es - (1-es)*B + B
(es-er) = es*B
B= (es-er)/es = 1- er/es
replacing values
B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)
A room is cooled by circulating chilled water through a heat exchanger located in the room. The air is circulated through the heat exchanger by a 0.25-hp (shaft output) fan. Typical efficiency of small electric motors driving 0.25-hp equipment is 60 percent. Determine the rate of heat supply by the fan–motor assembly to the room.
To determine the rate of heat supply by the fan-motor assembly, the electrical input power is calculated based on the 0.25-hp shaft output and 60% efficiency of the motor. The resulting heat supply to the room is the same as the electrical power input, which is 310.83 watts.
Explanation:The question asks to determine the rate of heat supply by a fan-motor assembly used to circulate chilled water through a heat exchanger for cooling a room. Given that the fan has a shaft output of 0.25 horsepower (hp) and that small electric motors driving such equipment typically have an efficiency of 60 percent, we can calculate the electrical power input needed to run the fan.
The electrical power input (Pinput) can be calculated as:
Pinput = Poutput / Efficiency
Where Poutput is the shaft output power (0.25 hp) and 'Efficiency' is the efficiency of the electric motor (60%, or 0.60 in decimal form).
Pinput = (0.25 hp) / 0.60
To convert horsepower to watts, we use the conversion factor 1 hp = 746 watts.
Pinput = (0.25 hp × 746 watts/hp) / 0.60
Pinput = 310.83 watts (rounded to two decimal places)
The rate of heat supply to the room will be equal to the electrical power input to the fan, which is 310.83 watts. This accounts for both the useful work done and all inefficiencies in the system that convert electrical energy into heat.
A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.
a. What is the energyon in terms of k adm
b. If the oscillator is a bonded atom with k = 15 N/m and m = 4 × 10-26 kg, what is the frequency (Hz) of the emitted photon? (Note: the energy of a photon is Ephoton= hf)
c. In which region of the electromagnetic spectrum (x-ray, visible, microwave, etc.) does this photon belong?
The answer explains the energy of a quantum harmonic oscillator, calculates the frequency of an emitted photon, and identifies the region of the electromagnetic spectrum it belongs to.
Explanation:a. Energy: The energy of a quantum harmonic oscillator can be represented as En = (n+1/2)h(sqrt(k/M)), where n = 0,1,2... and h represents Planck's constant.
b. Frequency Calculation: Using the given values of k = 15 N/m and m = 4 x 10^-26 kg, you can calculate the frequency of the emitted photon using the formula w = sqrt(k/M)/(2pi).
c. Electromagnetic Spectrum: To determine the region of the electromagnetic spectrum the photon belongs to, compare the frequency calculated to the known ranges of various regions like x-ray, visible, and microwave.
Write IEEE floating point representation of the following decimal number. Show your work.
1.25
Answer:
00111111101000000000000000000000
Explanation:
View Image
0 01111111 01000000000000000000000
The first bit is the sign bit. It's 0 for positive numbers and 1 for negative numbers.
The next 8-bits are for the exponents.
The first 0-126₁₀ (0-2⁷⁻¹) are for the negative exponent 2⁻¹-2⁻¹²⁶.
And the last 127-256₁₀ (2⁷-2⁸) are for the positive exponents 2⁰-2¹²⁶.
You have 1.25₁₀ which is 1.010₂ in binary. But IEEE wants it in scientific notation form. So its actually 1.010₂*2⁰
The exponent bit value is 127+0=127 which is 01111111 in binary.
The last 23-bits are for the mantissa, which is the fraction part of your number. 0.25₁₀ in binary is 010₂... so your mantissa will be:
010...00000000000000000000
Recycled materials content is environmental information that is typically: a. self-declared by the manufacturer. b. independently certified by third-party entities. c. listed in the building code. d. part of an environmental label. e. None of the above
Answer:
a. Self declared by the manufacturer
Explanation:
Recycled content refers to the portion of materials used in a product that have been diverted from the solid waste stream. If those materials are diverted during the manufacturing process, they are be referred to as pre-consumer recycled content (sometimes referred to as post-industrial). If they are diverted after consumer use, they are
post-consumer .
Check my work Check My Work button is now disabledItem 16Item 16 3 points As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.07 N/m.
Answer:
W = 1.7593 * 10 ^ (-7) KJ
Explanation:
The work done by the bubble is given:
[tex]W = sigma*\int\limits^2_1 {} \, dA \\\\W = sigma*( {A_{2} - A_{1} } ) \\\\A = pi*D^2\\\\W = sigma*pi*(D^2_{2} - D^2_{1})\\\\W = 0.07 * pi * (0.03^2 - 0.01^2)*10^(-3)\\\\W = 1.7593 *10^(-7) KJ[/tex]
Answer: W = 1.7593 * 10 ^ (-7) KJ
When generating a hierarchical cluster, a tree structure called a _______.
Answer:
Dendrogram
Explanation:
Dendrogram is referred to as the tree structure that represents the hierarchy between the object in a cluster. it is also referred to as an output that is drawn from clustering.
The dendrogram is interpreted by observing the object situated at the higher side in a scatter plot. By joining the object at the same level it represents the order of cluster. another purpose of modifying form of Dendrogram is to help in calculating the distance between the objects in clusters.
A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20 kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump.
Answer:
a) 69,630KW
b) 203 KW
Explanation:
The data obtained from Tables A-4, A-5 and A-6 is as follows:
[tex]h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2} = 255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\[/tex]
[tex]x_{4} = \frac{s_{4} - s_{f} }{s_{fg} } \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\[/tex]
The power produced and consumed by turbine and pump respectively are:
[tex]W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW[/tex]
A glass window pane, 1 m wide, 1.5 m high, and 5 mm thick, has a thermal conductivity of kg = 1.4 W/(m∙K). On a cold winter day, the indoor and outdoor temperatures are 15 °C and −25 °C respectively. (a) For a single-pane window at steady state, what is the rate of heat transfer through the glass? (10 pts) (b) To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by a dead-air space. The thermal conductivity of air is ka = 0.024 W/(m∙K). If the spacing between the two glasses is 10 mm. Calculate the temperatures of the glass surfaces in contact with the dead-air at the steady state. (10 pts) (c) Calculate the heat loss through the double-pane window in (b). (5 pts)
Answer:
a) Rate of heat transfer = 16.8KW
b) Temperature of glass surface = 15degree Celsius
c) Heat loss through frame = 141.34W
Explanation:
The concept used to approach this question is the Fourier's law of head conduction postulated by Joseph Fourier. it states that the rate of heat flow through a single homogeneous solid is directly proportional to the area and to the direction o heat flow and to the change in temperature with respect to the path length. Mathematically,
Q = -KA dt/dx
The detailed and step by step calculation is attached below
You are given a semiconductor resistor made from silicon with an impurity concentration of resistivity 1.00×10−3Ωm1.00×10−3Ωm. The resistor has a height of HH =0.5 mmmm, a length of LL = 2 mmmm, and a width of WW = 1.25 mmmm. The resistor can absorb (dissipate) up to PP = 7.81WW. What is the resistance of the resistor (RR), the maximum voltage (VV), and the maximum current (II)?
Answer:
The resistance (R) of the resistor is 2.4 ohm
The maximum voltage (V) is 4.33V
The maximum current (I) is 1.80A
Explanation:
Resistance (R) = resistivity×length/area
Resistivity = 0.003 ohm meter, length = 2mm = 2/1000 = 0.002m, width = 1.25mm = 1.25/1000 = 0.00125m, height = 0.5mm = 0.0005m, area = width × height = 0.00125m × 0.0005m = 6.25×10^-7m^2
R = 0.003×0.002/6.25×10^-7 = 3.2 ohm
Power (P) = V^2/R
V^2 = P × R = 7.81 × 3.2= 24.992
V = √24.992 = 4.99V
P = IV
I = P/V = 7.81/4.99 = 1.57A
assume a strain gage is bonded to the cylinder wall surface in the direction of the axial strain. The strain gage has nominal resistance R0 and a Gage Factor GF . It is connected in a Wheatstone bridge configuration where all resistors have the same nominal resistance; the bridge has an input voltage Vin. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.) Develop an expression for the voltage change ?V across the bridge if the cylinder pressure changes by ?P.
Explanation:
Note: For equations refer the attached document!
The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:
Eq 1
Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:
Eq 2
For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:
Eq 3
Finally, solving for unknown pressure as a function of hoop strain:
Eq 4
Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is
Eq 5
Consequently, a small differential change in ΔR/R can be expressed as
Eq 6
Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves
Eq 7
Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF
Eq 8
For Wheat stone bridge:
Eq 9
Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:
Eq9
Substituting e with respective stress-strain relation
Eq 10
Water flowing through both a small pipe and a large pipe can fill a water tank in 4 hours. Water flowing through the small pipe alone can fill the tank in 15 more hours. How many hours would it take to fill the tank using only the small pipe?
Answer with explanation:
As is the question the answer would be 19 hours, and the key to solving it is in the phrase in 15 more hours, basically what they are saying is that the small pipe takes 15 hours more than both the big and the small to fill the tank. Since both pipes working together can fill the tank in 4 hours we need to add 4 and 15 to solve the problem.
If the question is how many hours would it take to fill the tank using only the big pipe? Then we could solve t for the following equation:
[tex]\frac{1}{4+15} + \frac{1}{t} = \frac{1}{4}[/tex]
Getting as a result: 5.06
Note that the equation is the result of taking the rate of the small pipe (what we solved before), plus the unknown rate of the big one equals the rate of both.
What is the standard half-cell potential for the oxidation of methane under acidic conditions? The reaction for methane is as follows:
CH4(g) + 2H20(l) → CO2 +8H⁺ +8e⁻
What element is oxidized and how does its oxidation state change?
Answer:
The element that is oxidized is carbon.
Its oxidation state increased. It increased from -4 to +4
Explanation:
Oxidation is a process that involves increase in oxidation number.
The oxidation number of carbon in CH4 is -4
C + (1×4) = 0
C + 4 = 0
C = 0 - 4 = -4
The oxidation number of carbon in CO2 is +4
C + (2×-2) = 0
C - 4 = 0
C = 0+4 = 4
Increase in the oxidation number of carbon from -4 to +4 means carbon is oxidized
Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 300°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 3560 kJ and kinetic and potential energy effects are negligible.
Determine the final volume, in m3, and the work for the process, in kJ.
To determine the final volume and work done during the heating of water in a piston-cylinder assembly at constant pressure, tabulated data like steam tables are required since water is not an ideal gas. The work done is calculated using the formula W = PΔV.
Explanation:The student has been asked to find the final volume and the work done during a constant pressure process in which 5 kg of water is heated in a piston-cylinder assembly from an initial state of 5 bar and 300°C. To solve for the final volume and work done, one would typically use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat input minus the work output. However, since water at this state is not an ideal gas, tabulated data from steam tables or software would be used to determine the specific volume at the final state and then multiplied by the mass to find the total volume. The work done in a constant pressure process is equal to the pressure times the change in volume (W = PΔV). Without the final specific volume from the tables, we cannot compute the final volume or work directly.
Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) Cylindrical wall (b) Plane shell (c) Spherical shell (d) All of them (e) None of them
Answer:
Plane shell which is option b
Explanation:
The temperature in the case of a steady one-dimensional heat conduction through a plane wall is always a linear function of the thickness. for steady state dT/dt = 0
in such case, the temperature gradient dT/dx, the thermal conductivity are all linear function of x.
For a plane wall, the inner temperature is always less than the outside temperature.
why is the thermal conductivity of super insolation order of magnitude lower than the thermal conductivity of ordinary insulation?
Answer:
Super insulation are obtained by using layers of highly reflective sheets separated by glass fibers in an vacuumed space. Radiation heat transfer between any of the surfaces is inversely proportional to the number of sheets used and thus heat lost by radiation will be very low by using these highly reflective sheets which will an effective way of heat transfer.
Explanation:
In each case indicate whether the quantity in question increased, decreased or stayed the same when the string length is increased. Assume that the tension is unchanged. The function generator is kept at the same frequency, and the string is in resonance in all cases. Part A Number of antinodes ___ Number of antinodes ___ increased. decreased. stayed the same. Request Answer Part B Wavelength ___ Wavelength ___ increased. decreased. stayed the same. Request Answer Part C Fundamental frequency ___ Fundamental frequency ___ increased. decreased. stayed the same.
Answer:
Answer: No of anti-nodes increases
Answer: wavelength remains same.
Answer: fundamental frequency decreases
Explanation:
a)
The number of nodes (n) would have (n-1) anti-nodes.
The relation of Length of string with n is given below:
[tex]L = \frac{n*lambda}{2}[/tex]
Hence, n and L are directly proportional so as string length increases number of nodes and anti-nodes also increases.
Answer: No of anti-nodes increases
b)
Wavelength is dependent on the frequency:
[tex]lambda = \frac{v}{f}[/tex]
The speed v of the string remains same through-out and frequency of generator is unchanged!
Hence according to above relationship lambda is unchanged.
Answer: wavelength remains same.
C
Fundamental frequency equates to 1st harmonic that 2 nodes and 1 anti-node. Wavelength is = 2*L
Hence, if L increases wavelength increases
Using relation in part b
As wavelength increases fundamental frequency decreases
Answer: fundamental frequency decreases
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100), ( b ) (110), and ( c ) (111) planes.
Answer:
a)5.28 Å , b)3.73 Å , c)3.048 Å
Explanation:
the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.
Therefore, a particular unit cell consist only 1/8th part of an atom.
The lattice constant of a simple cubic primitive cell is 5.28 Å
We know formula of distance,
d = [tex]\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}[/tex]
a)(100)
a=5.28 Å
Distance = [tex]\frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}[/tex]=5.28 Å
b)(110)
Distance = [tex]\frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}}[/tex] = 3.73 Å
c)(111)
Distance= [tex]\frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}[/tex]= 3.048 Å
The gage pressure in a liquid at a depth of 3 m is read to be 48 kPa. Determine the gage pressure in the same liquid at a depth of 9 m.
The gage pressure in the same liquid at a depth of 9 m is__________ kPa.
Answer: 144kpa
Explanation:
pressure density =p
The Gage pressure (P1)=48kpa
The height (h1)=3m
The Gage pressure (P2)=?
The height (h2)=9m
P=p * g * h
P1/P2=p * g * h1/p * g * h2
Cancelling out similar terms:
Therefore, P1/P2=h1/h2
P2=P1*h2/h1
Hence, P2=48*9/3=144kpa.
Consider a point in a structural member that is subjected to plane stress. Normal and shear stress magnitudes acting on horizontal and vertical planes at the point are Sx = 175 MPa, Sy = 90 MPa, and Sxy = 75 MPa.
Answer:
C = 132.5 MPa
R = 86.20 MPa
Explanation:
Given
σx = 175 MPa
σy = 90 MPa
τxy = 75 MPa
For the given state of stress at a point in a structural member, determine the center C and the radius R of Mohr’s circle.
We apply the following equation for the center C
C = (σx + σy) / 2
C = (175 MPa + 90 MPa) / 2
C = 132.5 MPa
The Radius can be obtained as follows
R = √(((σx - σy) / 2)² + (τxy)²)
R = √(((175 MPa - 90 MPa) / 2)² + (75 MPa)²)
R = 86.20 MPa
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb.
To calculate the theoretical density of Niobium with a BCC structure, we use its atomic radius and atomic weight, converting these into the cube's edge length using the BCC relation, then apply the density formula.
Explanation:To calculate the theoretical density of Niobium (Nb), which has a body-centered cubic (BCC) crystal structure, we first use the known values: atomic radius = 0.143 nm (or 0.143 × 10-9 m) and atomic weight = 92.91 g/mol. The formula for the density (ρ) in a BCC structure is ρ = (2 × M) / (a3 × NA), where M is the atomic mass, a is the edge length of the cube, and NA is Avogadro's number (6.022 × 1023 atoms/mol).
Since it's a BCC structure, the atomic radius relates to the cube's edge length (a) as a = 4r / √3. Substituting the given atomic radius, we find a = 4 * 0.143 × 10-9 m / √3. Then, to find the density, we substitute M (92.91 g/mol), a, and NA into the density formula. This calculation will give us the theoretical density of Niobium in g/cm3.
If a barrel of oil weighs 1.5 kN, calculate the specific weight, density, and specific gravity of the oil. The barrel weighs 110 N
Answer
given,
oil barrel weight = 1.5 k N = 1500 N
weight of the barrel = 110 N
Assuming volume of barrel = 0.159 m³
weight of oil = 1500-110
= 1390 N
[tex]specific\ weight = \dfrac{weight}{volume}[/tex]
[tex]specific\ weight = \dfrac{1390}{0.159}[/tex]
= 8742.14 N/m³
[tex]mass = \dfrac{weight}{g}[/tex]
[tex]mass = \dfrac{1390}{9.8}[/tex]
= 141.84 kg
[tex]density = \dfrac{mass}{volume}[/tex]
[tex]density = \dfrac{141.84}{0.159}[/tex]
= 892.05 kg/m³
[tex]Specific\ gravity = \dfrac{density\ of\ oil}{density\ of\ water}[/tex]
[tex]Specific\ gravity = \dfrac{892.05}{1000}[/tex]
= 0.892
A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m^3, undergoes a constant-pressure expansion at 2 bar to a final volume of 0.12 m^3, while being slowly heated through the base. The change in internal energy of the gas is 0.25 kJ. the piston and cylinder walls are fabricated from heat-resistant material and the piston moves smoothly in the cylinder. The local atmosphere pressure is 1 bar.
a.) For the gas as the system, evaluate the work and the heat transfer, each in
b.) For the piston as a syatem, evautate the work and change in potential energy, in kJ.
Answer: (a). W = 4KJ and Q = 4.25KJ
(b). W = -2KJ and ΔPE = 2KJ
Explanation:
(a).
i. We are asked to calculate the work done during the expansion process considering gas as system.
from W = [tex]\int\limits^a_b {p} \, dV[/tex] where a = V₂ and b = V₁
so W = P(V₂-V₁)
W = (2 × 10²) (0.12 - 0.10)
W = 4 KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
since motion of the system is constrained, there is no change in both the potential and kinetic energy i.e. ΔPE = ΔKE = 0
∴ Q - W = ΔU
Q = ΔU + W
Q = 0.25 + 4
Q = 4.25 KJ
(b).
i. to calculate the work done during the expansion process considering piston as system;
W = [tex]\int\limits^a_b {(Patm - Pgas)} \, dV[/tex]where a and b represent V₂ and V₁ respectively.
W = (Patm - Pgas)(V₂ - V₁)
W = (1-2) ×10² × (0.12-0.1)
W = -2KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
Q = 0 since the piston and cylinder walls are perfectly insulated.
for piston, we neglect the change in internal energy and kinetic energy
ΔU = ΔKE = 0
from Q - W = ΔU + ΔKE + ΔPE
0 - (-2) = 0 + 0 + ΔPE
∴ ΔPE = 2KJ
If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the conductor.
Answer:
The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²
Explanation:
The electric field intensity due to a thin conducting sheet is given by the following formula:
Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)
From this formula:
Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)
We have the following data:
Electric Field Intensity = 1.5 N/C
Permittivity of free space = 8.85 x 10^-12 C²/N.m²
Therefore,
Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)
Surface Charge Density = 26.55 x 10^-12 C/m²
Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².
The surface charge density on the conductor is; σ = 13.275 × 10⁻¹² C/m²
What is the surface charge density?The formula for surface charge density on a conductor in an electric field just outside the surface of conductor is;
σ = E * ϵ₀
where;
E is electric field = 1.5 N/C
ϵ₀ is permittivity of space = 8.85 × 10⁻¹² C²/N.m²
Thus;
σ = 1.5 * 8.85 × 10⁻¹²
σ = 13.275 × 10⁻¹² C/m²
Read more about Surface Charge Density at; https://brainly.com/question/14306160