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Suemon Kwok
New Zealand
เข้าร่วมเมื่อ 2 ต.ค. 2011
Engineering tutorials and guides. Vlogging about engineering, engineering journey and life in general.
Thermodynamics 6-35C A refrigerator has a COP of 1.5. That is, the refrigerator removes 1.5 kWh of
Thermodynamics 6-35C
A refrigerator has a COP of 1.5. That is, the refrigerator removes 1.5 kWh of energy from the refrigerated space for each 1 kWh of electricity it consumes. Is this a violation of the first law of thermodynamics? Explain.
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#Suemon #Thermodynamics #ThermodynamicsTutorial
A refrigerator has a COP of 1.5. That is, the refrigerator removes 1.5 kWh of energy from the refrigerated space for each 1 kWh of electricity it consumes. Is this a violation of the first law of thermodynamics? Explain.
Subscribe to my channel
th-cam.com/users/SuemonKwok
#Suemon #Thermodynamics #ThermodynamicsTutorial
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Thermodynamics 6-34C A heat pump that is used to heat a house has a COP of 2.5. That is, the heat
3 ชั่วโมงที่ผ่านมา
Thermodynamics 6-34C A heat pump that is used to heat a house has a COP of 2.5. That is, the heat pump delivers 2.5 kWh of energy to the house for each 1 kWh of electricity it consumes. Is this a violation of the first law of thermodynamics? Explain. Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-33C Define the coefficient of performance of a heat pump in words. Can it be
3 ชั่วโมงที่ผ่านมา
Thermodynamics 6-33C Define the coefficient of performance of a heat pump in words. Can it be greater than unity? Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-32C Define the coefficient of performance of a refrigerator in words. Can it be
4 ชั่วโมงที่ผ่านมา
Thermodynamics 6-32C Define the coefficient of performance of a refrigerator in words. Can it be greater than unity? Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-31C What is the difference between a refrigerator and an air conditioner?
5 ชั่วโมงที่ผ่านมา
Thermodynamics 6-31C What is the difference between a refrigerator and an air conditioner? Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-30C What is the difference between a refrigerator and a heat pump?
5 ชั่วโมงที่ผ่านมา
Thermodynamics 6-30C What is the difference between a refrigerator and a heat pump? Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-29 Repeat Prob. 6-27 for a simple payback period of three years instead of five
มุมมอง 16 ชั่วโมงที่ผ่านมา
Thermodynamics 6-29 Repeat Prob. 6-27 for a simple payback period of three years instead of five years. Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-28 Reconsider Prob. 6-27. Using appropriate software, investigate the price of
มุมมอง 16 ชั่วโมงที่ผ่านมา
Thermodynamics 6-28 Reconsider Prob. 6-27. Using appropriate software, investigate the price of coal for varying simple pay-back periods, plant construction costs, and operating efficiency. Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-27 A country needs to build new power plants to meet the increasing demand for
มุมมอง 17 ชั่วโมงที่ผ่านมา
Thermodynamics 6-27 A country needs to build new power plants to meet the increasing demand for electric power. One possibility is to build coal-fired power plants, which cost $1300 per kW to construct and have an efficiency of 40 percent. Another possibility is to build clean-burning Integrated Gasification Combined Cycle (IGCC) plants where the coal is subjected to heat and pressure to gasify...
Thermodynamics 6-26 An Ocean Thermal Energy Conversion (OTEC) power plant built in Hawaii in
มุมมอง 124 ชั่วโมงที่ผ่านมา
Thermodynamics 6-26E An Ocean Thermal Energy Conversion (OTEC) power plant built in Hawaii in 1987 was designed to operate between the temperature limits of 86°F at the ocean surface and 41°F at a depth of 2100 ft. About 13,300 gpm of cold seawater was to be pumped from deep ocean through a 40-in diameter pipe to serve as the cooling medium or heat sink. If the cooling water experiences a tempe...
Robotics 2.25 Show that the eigenvalues of a rotation matrix are 1, e^αi, and e^(-αi), where
มุมมอง 54 ชั่วโมงที่ผ่านมา
Robotics 2.25 Show that the eigenvalues of a rotation matrix are 1, e^αi, and e^(-αi), where i = √(-1). Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Robotics #RoboticsTutorial
Thermodynamics 6-25 A coal-burning steam power plant produces a net power of 300 MW with an
มุมมอง 247 ชั่วโมงที่ผ่านมา
Thermodynamics 6-25 A coal-burning steam power plant produces a net power of 300 MW with an overall thermal efficiency of 32 percent. The actual gravimetric air-fuel ratio in the furnace is calculated to be 12 kg air/kg fuel. The heating value of the coal is 28,000 kJ/kg. Determine (a) the amount of coal consumed during a 24-hour period and (b) the rate of air flowing through the furnace. Subsc...
Thermodynamics 6-24 Solar energy stored in large bodies of water, called solar ponds, is being used
มุมมอง 167 ชั่วโมงที่ผ่านมา
Thermodynamics 6-24 Solar energy stored in large bodies of water, called solar ponds, is being used to generate electricity. If such a solar power plant has an efficiency of 3 percent and a net power output of 150 kW, determine the average value of the required solar energy collection rate, in Btu/h. Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTuto...
Thermodynamics 6-23 An automobile engine consumes fuel at a rate of 22 L/h and delivers 55 kW
มุมมอง 309 ชั่วโมงที่ผ่านมา
Thermodynamics 6-23 An automobile engine consumes fuel at a rate of 22 L/h and delivers 55 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm3, determine the efficiency of this engine. Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-22 A steam power plant with a power output of 150 MW consumes coal at a rate of
มุมมอง 5512 ชั่วโมงที่ผ่านมา
Thermodynamics 6-22 A steam power plant with a power output of 150 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,000 kJ/kg, determine the overall efficiency of this plant. Subscribe to my channel th-cam.com/users/SuemonKwok #Suemon #Thermodynamics #ThermodynamicsTutorial
Thermodynamics 6-21 A heat engine that propels a ship produces 500 Btu/lbm of work while rejecting
มุมมอง 1912 ชั่วโมงที่ผ่านมา
Thermodynamics 6-21 A heat engine that propels a ship produces 500 Btu/lbm of work while rejecting
Thermodynamics 6-20 A heat engine with a thermal efficiency of 45 percent rejects 500 kJ/kg of heat.
มุมมอง 2412 ชั่วโมงที่ผ่านมา
Thermodynamics 6-20 A heat engine with a thermal efficiency of 45 percent rejects 500 kJ/kg of heat.
Robotics 2.24 Prove Cayley's formula for proper orthonormal matrices.
มุมมอง 512 ชั่วโมงที่ผ่านมา
Robotics 2.24 Prove Cayley's formula for proper orthonormal matrices.
Thermodynamics 6-19 A 600-MW steam power plant, which is cooled by a nearby river, has a thermal
มุมมอง 4914 ชั่วโมงที่ผ่านมา
Thermodynamics 6-19 A 600-MW steam power plant, which is cooled by a nearby river, has a thermal
What targets would nuclear strikes hit in the U.S.
มุมมอง 2214 ชั่วโมงที่ผ่านมา
What targets would nuclear strikes hit in the U.S.
Thermodynamics 6-18 A heat engine has a heat input of 3 × 10^4 Btu/h and a thermal efficiency of
มุมมอง 2516 ชั่วโมงที่ผ่านมา
Thermodynamics 6-18 A heat engine has a heat input of 3 × 10^4 Btu/h and a thermal efficiency of
Thermodynamics 6-17 A steam power plant receives heat from a furnace at a rate of 280 GJ/h. Heat
มุมมอง 3616 ชั่วโมงที่ผ่านมา
Thermodynamics 6-17 A steam power plant receives heat from a furnace at a rate of 280 GJ/h. Heat
Robotics 2.23 Give an algorithm to construct the definition of a frame (_A^U)T from three points
มุมมอง 419 ชั่วโมงที่ผ่านมา
Robotics 2.23 Give an algorithm to construct the definition of a frame (_A^U)T from three points
Thermodynamics 6-16 A heat engine has a total heat input of 1.3 kJ and a thermal efficiency of 35
มุมมอง 3921 ชั่วโมงที่ผ่านมา
Thermodynamics 6-16 A heat engine has a total heat input of 1.3 kJ and a thermal efficiency of 35
Thermodynamics 6-15C Consider a pan of water being heated (a) by placing it on an electric range
มุมมอง 1121 ชั่วโมงที่ผ่านมา
Thermodynamics 6-15C Consider a pan of water being heated (a) by placing it on an electric range
Thermodynamics 6-14C Baseboard heaters are basically electric resistance heaters and are frequently
มุมมอง 821 ชั่วโมงที่ผ่านมา
Thermodynamics 6-14C Baseboard heaters are basically electric resistance heaters and are frequently
Thermodynamics 6-13C Are the efficiencies of all the work-producing devices, including the
มุมมอง 9วันที่ผ่านมา
Thermodynamics 6-13C Are the efficiencies of all the work-producing devices, including the
Thermodynamics 6-12C In the absence of any friction and other irreversibilities, can a heat engine
มุมมอง 30วันที่ผ่านมา
Thermodynamics 6-12C In the absence of any friction and other irreversibilities, can a heat engine
Thermodynamics 6-11C Does a heat engine that has a thermal efficiency of 100 percent necessarily
มุมมอง 31วันที่ผ่านมา
Thermodynamics 6-11C Does a heat engine that has a thermal efficiency of 100 percent necessarily
Thermodynamics 6-10C s it possible for a heat engine to operate without rejecting any waste heat to
มุมมอง 11วันที่ผ่านมา
Thermodynamics 6-10C s it possible for a heat engine to operate without rejecting any waste heat to
Disclaimers code has been uploaded to anti cheat software such as Turnitin. Change it up if you are planning on using it. "Given" W_dot=15E7 [kW] Cost_coal=1300 [$/kW] eta_coal=0.40 Cost_IGCC=1500 [$/kW] eta_IGCC=0.48 HV_coal=28000 [kJ/kg] PaybackYears=5 [yr] "Analysis" time=PaybackYears*Convert(yr, h) ConstCost_coal=W_dot*Cost_coal ConstCost_IGCC=W_dot*Cost_IGCC ConstCostDif=ConstCost_IGCC-ConstCost_coal W_e=W_dot*time m_coal_coal=W_e/(eta_coal*HV_coal)*Convert(kWh, kJ) m_coal_IGCC=W_e/(eta_IGCC*HV_coal)*Convert(kWh, kJ) DELTAm_coal=m_coal_coal-m_coal_IGCC UnitCost_coal=ConstCostDif/DELTAm_coal*1000
Thank you! This is right
Thanks for watching and checking. The first 500 videos are designed for 2nd and 3rd year university students with the expectation of having mastered foundations and attempted the question. It's designed as a review material and check if the process is correct. From 5-31 onward the teaching style shifts towards high school and 1st year university students. If you have any questions regarding this or any other tutorial feel free to post in the comments. Best of luck in your studies
How different is the electrical code / system from Canada?
In terms of voltages and frequencies Canada In Canada, the standard voltage is 120/240 volts with a frequency of 60 Hz. This means that typical household outlets supply 120 volts, and high-power applications (like ovens and dryers) use 240 volts. New Zealand In New Zealand, the standard voltage is 230/400 volts with a frequency of 50 Hz. Most household appliances, electronics, and lighting are designed to operate at 230 volts, similar to many other countries in Europe and Asia. ______________________________________________________________________ In terms of code and legislation Canada In Canada, the electrical code is known as the Canadian Electrical Code (CEC), or CSA C22.1. Published by the Canadian Standards Association, the Code outlines standards for the installation and maintenance of electrical equipment. The CEC is periodically updated, with changes reviewed on a three-year cycle. This code applies across various regions, but local authorities often adopt and sometimes amend it to fit local conditions. New Zealand In New Zealand, the equivalent regulations are encapsulated in the New Zealand Electrical Code of Practice (ECP). Issued by WorkSafe New Zealand under the Electricity Act 1992, these codes include detailed guidelines for safe electrical distances, power systems earthing, and working standards for electricians. The framework is designed to cover everything from homeowner electrical work in domestic installations to high voltage live line work. Comparison While both systems aim to ensure electrical safety and standardization, they are developed independently and adapted to the respective regional requirements: CEC (Canada): Emphasizes a prescriptive model but recognizes alternative methods for safe installations. It covers general and specific installation rules across various industries. ECP (New Zealand): Focuses on detailed practical standards for safety distances, harmonic levels, and live line work. It is tailored to the unique electrical layout and requirements within New Zealand. Sources: 1580WKS-17-energy-safety-COP-for-domestic-wiring-work.pdf sac-ace.ca/resources/electric-sign-codes-and-standards/canadian-electrical-code/ www.worksafe.govt.nz/laws-and-regulations/electrical-and-gas-codes-of-practice/electricity-codes-of-practice/
plss upload other chapter 6 examples we have final for thermodynmicsss <3 <3 <3 <3
Thanks for watching, more to come later today it takes time to solve and edit. Good luck in your studies
Wp we need more questions ty ❤
Thanks for watching, more to come later today it takes time to solve and edit
when will the new videos come. i have a final in 3 days ):
Thanks for watching, more to come later today it takes time to solve and edit. Good luck in your studies
Thank you, sir, for all the videos you make. I hope my remedial test of thermodynamics tomorrow will go well. 😔🙏🏼
Thanks for wawtching, good luck in your studies
Any requirement of 2year diploma optometrist Experience 5 year
See website for requirements if you're have overseas qualifications and converting to NZ optometrist odob.health.nz/document/6783/Registration%20Pathways%20(December%202022).pdf Alternatively search on google Registration Pathways from accredited prescribed qualifications for Optometrists in NZ
Hi!, Would you be willing to try a thermodynamics question?
Hi if you post it here in the comments I can have a look, When it comes to thermodynamics I'm novice myself while have mechanical engineering degree I don't specialize thermodynamics. I mostly design small component and factory assembly and maintenance. I can't accept requests for videos my focus at the moment is working through the textbook to help struggling students get the basic foundations in thermodynamics. I have a back log of 30 years worth of questions from thermodynamics, heat transfer, solid mechanics, system dynamics, mechatronics, mechanism and dynamics, Mechanical design, fluid mechanics and robotics
Thank you very much for the expalination
Thanks for watching
I want this book
Yunus and Cengal 9th, 10th and 11th.
I didn't know how to get it @@SuemonKwok
@ I just googled it and it's 1st search result. iunajaf.edu.iq/Gradual/Publicationoflectures/uploadsPdf/pdfcoffee.com_engineering-thermodynamics-by-cengel-boles-and-kanoglu-9th-edition-pdf-free.pdf%20-%202023.01.13%20-%2006.32.12pm.pdf
Hi, why do we calculate change in temperature of the room through enthalpy? It is a rigid room, so no work is possible and we could as well use internal energy?
Both U and H approaches are valid. Here is the explanation below 1. Using Internal Energy (ΔU): - For an ideal gas at constant volume: ΔU = mCvΔT - The work term from the fan would add directly to ΔU - So: Qin + Wfan = mCvΔT 2. Using Enthalpy (ΔH): - For an ideal gas: ΔH = mCpΔT - The work term from the fan would add to ΔH - So: Qin + Wfan = mCpΔT The key point is that for an ideal gas: - Cp - Cv = R - At constant pressure (which we have in this case), the PV work is zero - The fan work goes entirely into increasing the internal energy of the gas Both approaches will give us the same temperature change because: - The room is at constant pressure (specified in problem) - The PV work is zero (rigid room) - All the fan work goes into increasing the internal energy of the air In fact, if we were to solve this using Cv instead of Cp: Cv = Cp - R = 1.005 - 0.287 = 0.718 kJ/(kg·K) The equation would become: Qin + Wfan = mCvΔT + mRΔT = m(Cv + R)ΔT = mCpΔT Which gives us exactly the same result. So while you're correct that we could use internal energy, using enthalpy is equally valid and perhaps simpler since we're given Cp directly. The constant pressure condition ensures both approaches are equivalent.
@ thank you so much
For the density we can do this way Total mass / the Tank Volume [1.8]
Yes it's valid ρ = m_total/V_total =518.1 kg/1.8 m^3 = 287.85 kg/m^3 ρ = 1/v = 1/0.003474 = 287.83 kg/m^3 small difference (0.02 kg/m³) is likely due to rounding in the calculations
Code for visualization import numpy as np import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D def construct_frame(P1, P2, P3): # Define vectors X = P2 - P1 XY = P3 - P1 # Normalize X to get unit vector X̂ X_hat = X / np.linalg.norm(X) # Project XY onto the plane orthogonal to X̂ Y = XY - np.dot(XY, X_hat) * X_hat # Normalize Y to get unit vector Ŷ Y_hat = Y / np.linalg.norm(Y) # Compute Ẑ as the cross product of X̂ and Ŷ Z_hat = np.cross(X_hat, Y_hat) return X_hat, Y_hat, Z_hat def plot_frame(P1, X_hat, Y_hat, Z_hat, P2, P3): fig = plt.figure() ax = fig.add_subplot(111, projection='3d') # Plot original points ax.scatter(*P1, color='r', label='P1 (origin)') ax.scatter(*P2, color='g', label='P2 (on X axis)') ax.scatter(*P3, color='b', label='P3 (near Y axis)') # Plot vector from P1 to P3 as a solid line with reduced thickness ax.plot([P1[0], P3[0]], [P1[1], P3[1]], [P1[2], P3[2]], color='b', linestyle='-', linewidth=1, label='Vector P1 to P3') # Plot frame axes with reduced arrow size arrow_length = 0.5 # Adjust this value to control arrow size ax.quiver(*P1, *X_hat, color='r', length=arrow_length, normalize=True, linestyle='--', label='Unit Vector X̂ (dashed)') ax.quiver(*P1, *Y_hat, color='g', length=arrow_length, normalize=True, linestyle='--', label='Unit Vector Ŷ (dashed)') ax.quiver(*P1, *Z_hat, color='b', length=0.2, normalize=True, linestyle='--', label='Unit Vector Ẑ (dashed)') # Annotate points ax.text(*P1, 'P1', color='red') ax.text(*P2, 'P2', color='green') ax.text(*P3, 'P3', color='blue') # Label axes ax.set_xlabel('X') ax.set_ylabel('Y') ax.set_zlabel('Z') plt.legend() plt.show() # Define points P1, P2, and P3 P1 = np.array([0, 0, 0]) P2 = np.array([1, 0, 0]) P3 = np.array([0.5, 1, 0]) # Construct frame X_hat, Y_hat, Z_hat = construct_frame(P1, P2, P3) # Plot frame plot_frame(P1, X_hat, Y_hat, Z_hat, P2, P3)
👌👌
Thanks for watching
love this man
Thanks for watching. The first 500 videos are designed for 2nd and 3rd year university students with the expectation of having mastered foundations and attempted the question. It's designed as a review material and check if the process is correct. From 5-31 onward the teaching style shifts towards high school and 1st year university students. If you have any questions regarding this tutorial feel free to post in the comments. Best of luck in your studies
thank you, it helped me a lot.
Thanks for watching. The first 500 videos are designed for 2nd and 3rd year university students with the expectation of having mastered foundations and attempted the question. It's designed as a review material and check if the process is correct. From 5-31 onward the teaching style shifts towards high school and 1st year university students. If you have a question regarding this tutorial feel free to post in the comments. Best of luck in your studies
thank youuu but why is the quality be 0.5? no need to find it first?
From the question " An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. If one-half of entire tank. the liquid is evaporated during " We know it initially contains saturated liquid and we know the half the tank is evaporating so that means half is liquid and half is vapor
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. T1=40 [C] m_dot=0.20 [kg/s] h_f=ENTHALPY(Steam_IAPWS,T=T1,x=0) h_g=ENTHALPY(Steam_IAPWS,T=T1,x=1) h_fg=h_g-h_f Q_dot=m_dot*h_fg "Some Wrong Solutions with Common Mistakes:" W1_Q=m_dot*h_f "Using hf" W2_Q=m_dot*h_g "Using hg" W3_Q=h_fg "not using mass flow rate" W4_Q=m_dot*(h_f+h_g) "Adding hf and hg" This is the step by step. 1) First, let's understand what we're looking for: - We need to find the rate of heat transfer from the tube - The water enters as saturated vapor at 40°C - It leaves as saturated liquid at 40°C - Mass flow rate is 0.20 kg/s 2) The heat transfer will be equal to the latent heat released during condensation: Q = ṁ × hfg where: - Q = rate of heat transfer - ṁ = mass flow rate = 0.20 kg/s - hfg = enthalpy of vaporization at 40°C 3) Looking up the enthalpy of vaporization (hfg) for water at 40°C: hfg at 40°C = 2406.7 kJ/kg 4) Now we can calculate: Q = 0.20 kg/s × 2406.7 kJ/kg Q = 481.34 kW Therefore, the rate of heat transfer from the tube is 481.34 kW (or 481,340 W). Note: The negative sign is often used to indicate heat leaving the system, so you could also express this as -481.34 kW to indicate heat being removed from the vapor.
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. T1=50 [C] m_dot=2 [kg/s] W_dot_e=8 [kJ/s] cp=1.005 [kJ/kg-K] W_dot_e=m_dot*cp*(T2-T1) "Some Wrong Solutions with Common Mistakes:" cv=0.718 [kJ/kg-K] W_dot_e=cp*(W1_T2-T1) "Not using mass flow rate" W_dot_e=m_dot*cv*(W2_T2-T1) "Using cv" W_dot_e=m_dot*cp*W3_T2 "Ignoring T1" This is the step by step 1) Let's identify what we know: * Power input (Q̇) = 8 kW = 8000 W * Mass flow rate (ṁ) = 2 kg/s * Inlet temperature (T₁) = 50°C * Need to find exit temperature (T₂) 2) For this problem, we can use the steady-flow energy balance equation: * Q̇ = ṁ × cp × (T₂ - T₁) * where cp is the specific heat of air at constant pressure * cp for air ≈ 1.005 kJ/kg·K 3) Rearranging the equation to solve for T₂: * T₂ = T₁ + Q̇/(ṁ × cp) 4) Let's substitute the values: * T₂ = 50°C + 8000 W/(2 kg/s × 1005 J/kg·K) * T₂ = 50°C + 8000/(2 × 1005) * T₂ = 50°C + 3.98°C * T₂ = 53.98°C Therefore, the exit temperature of the air is approximately 54°C.
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. P1=1000 [kPa] T1=300 [C] P2=400 [kPa] h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=h1 v2=VOLUME(Steam_IAPWS,h=h2,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_v2=VOLUME(Steam_IAPWS,T=T1,P=P2) "Assuming the volume to remain constant" u1=INTENERGY(Steam,T=T1,P=P1) W2_v2=VOLUME(Steam_IAPWS,u=u1,P=P2) "Assuming u=constant" W3_v2=VOLUME(Steam_IAPWS,T=T1,P=P2) "Assuming T=constant" This is the step by step. 1) First, let's recall key principles for throttling process: - It's adiabatic (no heat transfer) - Work done is zero - Change in kinetic energy is negligible (given) - Therefore, according to First Law of Thermodynamics, enthalpy remains constant (h₁ = h₂) 2) Given conditions: - Initial state (1): P₁ = 1 MPa, T₁ = 300°C - Final state (2): P₂ = 0.4 MPa - h₁ = h₂ (constant enthalpy process) 3) Let's find initial enthalpy (h₁): - At P₁ = 1 MPa, T₁ = 300°C - Using steam tables for superheated steam - h₁ ≈ 3052.1 kJ/kg 4) Since h₁ = h₂: - h₂ = 3052.1 kJ/kg at P₂ = 0.4 MPa 5) Now, at P₂ = 0.4 MPa and h₂ = 3052.1 kJ/kg: - Using steam tables or interpolation - The specific volume v₂ ≈ 0.635 m³/kg Therefore, the specific volume of steam after throttling is 0.635 m³/kg. Note: The exact value might vary slightly depending on the steam tables used, but this should be very close to the correct answer.
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. T1=27 [C] P1=5 [atm] P2=1 [atm] T2=T1 "The temperature of an ideal gas remains constant during throttling, and thus T2=T1" "Some Wrong Solutions with Common Mistakes:" W1_T2=T1*P1/P2 "Assuming v=constant and using C" W2_T2=(T1+273)*P1/P2-273 "Assuming v=constant and using K" W3_T2=T1*P2/P1 "Assuming v=constant and pressures backwards and using C" W4_T2=(T1+273)*P2/P1 "Assuming v=constant and pressures backwards and using K" T2=T1 The temperature of an ideal gas remains constant during throttling, and thus T2=T1
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. Vel_1=0 [m/s] Vel_2=280 [m/s] m=2.5 [kg/s] T2=400 [C] P2=2000 [kPa] "The rate form of energy balance is E_dot_in - E_dot_out = DELTAE_dot_cv" v2=VOLUME(Steam_IAPWS,T=T2,P=P2) m=(1/v2)*A2*Vel_2 "A2 in m^2" "Some Wrong Solutions with Common Mistakes:" R=0.4615 [kJ/kg-K] P2*v2ideal=R*(T2+273) m=(1/v2ideal)*W1_A2*Vel_2 "assuming ideal gas" P1*v2ideal=R*T2 m=(1/v2ideal)*W2_A2*Vel_2 "assuming ideal gas and using C" m=W3_A2*Vel_2 "not using specific volume" 1. First, let's identify the given information: o Exit velocity (V₂) = 280 m/s o Mass flow rate (ṁ) = 2.5 kg/s o Exit temperature (T₂) = 400°C = 673.15 K o Exit pressure (P₂) = 2 MPa = 2 × 10⁶ Pa 2. To find the exit area, we can use the continuity equation: o ṁ = ρ₂A₂V₂ o where A₂ is the exit area and ρ₂ is the exit density 3. We need to find ρ₂ using the ideal gas equation for steam: o P = ρRT o where R for steam = 461.5 J/kg·K 4. Rearranging the ideal gas equation to find density: o ρ₂ = P₂/(RT₂) o ρ₂ = (2 × 10⁶)/(461.5 × 673.15) o ρ₂ = 6.43 kg/m³ 5. Now we can find the area using the continuity equation: o A₂ = ṁ/(ρ₂V₂) o A₂ = 2.5/(6.43 × 280) o A₂ = 0.00139 m² o A₂ = 13.9 cm² Therefore, the exit area of the nozzle is 13.9 cm².
If u write kelvin temperature and add 273 in 2.4.so our answer totally change.. It will be 590.62 m/s.........
Greetings I believe I referring to C_p unit symbol being K and swapping it with degrees Celsius. Not referring 2.4 degrees
@@SuemonKwok ok thanks
love the videos helping me out so much
Thanks for watching. These are my older videos, my teaching style has improved in newer videos. Videos from 5-31 onward I solve as if I was solving it for the first time, shifts towards high school and 1st year university students. The first 500 videos were designed for 2nd and 3rd year university students with the expectation of having mastered foundations and attempted the question. It's designed as a review material and check if the process is correct. This is supplementary material to student's learning not a direct copy and paste although it could used as such, the grade will be lower.
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. Vel_1=90 [m/s] m=3.5 [kg/s] T1=300 [C] P1=500 [kPa] "The rate form of energy balance is E_dot_in - E_dot_out = DELTAE_dot_cv" v1=VOLUME(Steam_IAPWS,T=T1,P=P1) m=(1/v1)*A*Vel_1 "Some Wrong Solutions with Common Mistakes:" R=0.4615 [kJ/kg-K] P1*v1ideal=R*(T1+273) m=(1/v1ideal)*W1_A*Vel_1 "assuming ideal gas" P1*v2ideal=R*T1 m=(1/v2ideal)*W2_A*Vel_1 "assuming ideal gas and using C" m=W3_A*Vel_1 "not using specific volume" we will use the continuity equation and the ideal gas law to find the inlet area of the diffuser. Given: • Mass flow rate (ẍ{m}) = 3.5 kg/s • Inlet pressure ({P}) = 0.5 MPa = 500,000 Pa • Inlet temperature ({T}) = 300°C = 573.15 K (since {T} in Kelvin = {T} in °C + 273.15) • Inlet velocity ({V}) = 90 m/s We need to find the inlet area ({A}). Step 1: Use the Continuity Equation The continuity equation for steady flow is given by: ẍ{m} = ρ {A} {V} where: • ẍ{m} is the mass flow rate • ρ is the density of the steam • {A} is the inlet area • {V} is the inlet velocity Step 2: Determine the Density of the Steam To find the density (ρ), we can use the ideal gas law: ρ = {P} / ({R} {T}) where: • {P} is the pressure • {R} is the specific gas constant for steam • {T} is the temperature For steam, the specific gas constant ({R}) is approximately 0.4615 kJ/(kg⋅K). Step 3: Calculate the Density ρ = {P} / ({R} {T}) = 500,000 Pa / (0.4615 kJ/(kg⋅K) ⋅ 573.15 K) First, convert the specific gas constant to consistent units: {R} = 0.4615 kJ/(kg⋅K) = 461.5 J/(kg⋅K) Now, calculate the density: ρ = 500,000 Pa / (461.5 J/(kg⋅K) ⋅ 573.15 K) = 500,000 / (461.5 ⋅ 573.15) = 500,000 / 264,677.225 ρ ≈ 1.89 kg/m³ Step 4: Solve for the Inlet Area Using the continuity equation: ẍ{m} = ρ {A} {V} 3.5 kg/s = 1.89 kg/m³ ⋅ {A} ⋅ 90 m/s {A} = 3.5 / (1.89 ⋅ 90) {A} = 3.5 / 170.1 {A} ≈ 0.02057 m² Conclusion The inlet area of the diffuser is approximately 0.02057 m².
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. P1=1400 [kPa] T1=70 [C] P2=600 [kPa] h1=ENTHALPY(R134a,T=T1,P=P1) T2=TEMPERATURE(R134a,h=h1,P=P2) "Some Wrong Solutions with Common Mistakes:" W1_T2=T1 "Assuming the temperature to remain constant" W2_T2=TEMPERATURE(R134a,x=0,P=P2) "Taking the temperature to be the saturation temperature at P2" u1=INTENERGY(R134a,T=T1,P=P1) W3_T2=TEMPERATURE(R134a,u=u1,P=P2) "Assuming u=constant" v1=VOLUME(R134a,T=T1,P=P1) W4_T2=TEMPERATURE(R134a,v=v1,P=P2) "Assuming v=constant" • Enthalpy (h1) at 1400 kPa and 70°C ≈ 435 kJ/kg • Temperature (T2) at 600 kPa and 435 kJ/kg ≈ 57°C
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. P1=140 [kPa] x1=1 P2=900 [kPa] T2=60 [C] m_dot=0.108 [kg/s] Q_dot_loss=1.10 [kJ/s] h1=ENTHALPY(R134a,x=x1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" W_dot_in-Q_dot_loss=m_dot*(h2-h1) "Some Wrong Solutions with Common Mistakes:" W1_Win+Q_dot_loss=m_dot*(h2-h1) "Wrong direction for heat transfer" W2_Win =m_dot*(h2-h1) "Not considering heat loss" u1=INTENERGY(R134a,x=x1,P=P1) u2=INTENERGY(R134a,T=T2,P=P2) W3_Win-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy" W4_Win+Q_dot_loss=u2-u1 "Using internal energy and wrong direction for heat transfer" 6.04 = 0.108*(h2-h1) = 55.93 kJ/kg Therefore: W_dot_in = 0.108 kg/s × 55.93 kJ/kg + 1.10 kJ/s W_dot_in = 6.04 + 1.10 W_dot_in = 7.14 kW
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. P1=200 [kPa] T1=150 [C] P2=800 [kPa] T2=350 [C] m_dot=1.30 [kg/s] Q_dot_loss=0 [kJ/s] h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" W_dot_in-Q_dot_loss=m_dot*(h2-h1) "Some Wrong Solutions with Common Mistakes:" W1_Win-Q_dot_loss=(h2-h1)/m_dot "Dividing by mass flow rate instead of multiplying" W2_Win-Q_dot_loss=h2-h1 "Not considering mass flow rate" u1=INTENERGY(Steam_IAPWS,T=T1,P=P1) u2=INTENERGY(Steam_IAPWS,T=T2,P=P2) W3_Win-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy" W4_Win-Q_dot_loss=u2-u1 "Using internal energy and ignoring mass flow rate" This is the step by step. 1) For an adiabatic compressor, we can use the energy balance equation: Ẇ = ṁ(h₂ - h₁) where: - Ẇ is power input - ṁ is mass flow rate = 1.30 kg/s - h₂ is outlet enthalpy - h₁ is inlet enthalpy 2) We need to find h₁ and h₂ using steam tables for: State 1: P₁ = 0.2 MPa, T₁ = 150°C State 2: P₂ = 0.8 MPa, T₂ = 350°C 3) From steam tables: At State 1 (0.2 MPa, 150°C): h₁ = 2771.1 kJ/kg At State 2 (0.8 MPa, 350°C): h₂ = 3164.6 kJ/kg 4) Now we can calculate the power input: Ẇ = ṁ(h₂ - h₁) Ẇ = 1.30 kg/s × (3164.6 - 2771.1) kJ/kg Ẇ = 1.30 × 393.5 kJ/s Ẇ = 511.55 kW Therefore, the power input to the compressor is 511.55 kW.
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. T1=500 [C] P1=4000 [kPa] T2=250 [C] P2=500 [kPa] m_dot=(1350/3600) [kg/s] Q_dot_loss=25 [kJ/s] h1=ENTHALPY(Steam_IAPWS,T=T1,P=P1) h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2) "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" W_dot_out+Q_dot_loss=m_dot*(h1-h2) "Some Wrong Solutions with Common Mistakes:" W1_Wout=m_dot*(h1-h2) "Disregarding heat loss" W2_Wout-Q_dot_loss=m_dot*(h1-h2) "Assuming heat gain instead of loss" u1=INTENERGY(Steam_IAPWS,T=T1,P=P1) u2=INTENERGY(Steam_IAPWS,T=T2,P=P2) W3_Wout+Q_dot_loss=m_dot*(u1-u2) "Using internal energy instead of enthalpy" W4_Wout-Q_dot_loss=m_dot*(u1-u2) "Using internal energy and wrong direction for heat" Problem Solution: Turbine Power Output To solve this problem, we will use the principles of thermodynamics and the steady-flow energy equation for a control volume, such as a turbine. Step 1: Convert the Mass Flow Rate to kg/s The mass flow rate is given as 1350 kg/h. Converting it to kg/s: ṁ = 1350 kg/h / 3600 s/h = 0.375 kg/s Step 2: Apply the Steady-Flow Energy Equation The steady-flow energy equation is: Q̇ - Ẇ = ṁ (h2 - h1) Where: • Q̇ is the heat transfer rate (-25 kJ/s, since heat is lost). • Ẇ is the work done by the turbine (the power output we need to find). • ṁ is the mass flow rate. • h1 and h2 are the specific enthalpies at the inlet and outlet, respectively. Step 3: Determine the Specific Enthalpies Using steam tables or thermodynamic property software, we find: • For steam at 4 MPa and 500°C: h1 ≈ 3408.1 kJ/kg • For steam at 0.5 MPa and 250°C: h2 ≈ 2924.7 kJ/kg Step 4: Solve for Ẇ Substitute the known values into the energy equation: -25 kJ/s - Ẇ = 0.375 kg/s × (2924.7 kJ/kg - 3408.1 kJ/kg) Simplify: -25 kJ/s - Ẇ = 0.375 kg/s × (-483.4 kJ/kg) -25 kJ/s - Ẇ = -181.275 kJ/s Ẇ = -181.275 kJ/s + 25 kJ/s Ẇ = -156.275 kJ/s Final Answer The power output of the turbine is: Ẇ ≈ -156.3 kW This indicates that the turbine produces approximately 156.3 kW of power.
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. P1=1000 [kPa] T1=1500 [K] P2=200 [kPa] T2=900 [K] m_dot=0.1 [kg/s] Q_dot_loss=15 [kJ/s] cp_air=1.005 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" W_dot_out+Q_dot_loss=m_dot*cp_air*(T1-T2) "Alternative: Variable specific heats - using EES data" W_dot_out_variable+Q_dot_loss=m_dot*(ENTHALPY(Air,T=T1)-ENTHALPY(Air,T=T2)) "Some Wrong Solutions with Common Mistakes:" W1_Wout=m_dot*cp_air*(T1-T2) "Disregarding heat loss" 1. We'll use the constant specific heat equation since we're given cp_air: W_dot_out + Q_dot_loss = m_dot × cp_air × (T1 - T2) 2. Plugging in the values: W_dot_out + 15 = 0.1 × 1.005 × (1500 - 900) 3. Simplify the right side: • Temperature difference = 1500 - 900 = 600 K • 0.1 × 1.005 × 600 = 60.3 kJ/s 4. Therefore: W_dot_out + 15 = 60.3 W_dot_out = 60.3 - 15 W_dot_out = 45.3 kW ≈ 45 kW
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. P1=1200 [kPa] T1=100 [C] P2=180 [kPa] T2=50 [C] m_dot=1.25 [kg/s] Q_dot_loss=0 [kJ/s] h1=ENTHALPY(R134a,T=T1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" -W_dot_out-Q_dot_loss=m_dot*(h2-h1) "Some Wrong Solutions with Common Mistakes:" -W1_Wout-Q_dot_loss=(h2-h1)/m_dot "Dividing by mass flow rate instead of multiplying" -W2_Wout-Q_dot_loss=h2-h1 "Not considering mass flow rate" u1=INTENERGY(R134a,T=T1,P=P1) u2=INTENERGY(R134a,T=T2,P=P2) -W3_Wout-Q_dot_loss=m_dot*(u2-u1) "Using internal energy instead of enthalpy" -W4_Wout-Q_dot_loss=u2-u1 "Using internal energy and ignoring mass flow rate" {\dot{W}}_{out}:\ -{\dot{W}}_{out}\ -\ 0\ =\ 1.25\ kg/s\ \ast\ (411.2\ -\ 447.2)\ kJ/kg\ -W_dot_out\ =\ 1.25\ \ast\ (-36)\ kJ/s\ W_dot_out\ =\ 45\ kW
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. Tcold_1=7 [C] m_dot_cold=4 [kg/min] Thot_1=70 [C] m_dot_hot=5 [kg/min] c_air=1.005 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_hot*c_air*Thot_1+m_dot_cold*c_air*Tcold_1=(m_dot_hot+m_dot_cold)*c_air*Tmix "Some Wrong Solutions with Common Mistakes:" W1_Tmix=(Tcold_1+Thot_1)/2 "Taking the average temperature of inlet fluids" This is step by step using conservation of mass and energy for adiabatic mixing. 1) For an adiabatic mixing process, the energy entering equals the energy leaving • Heat transfer = 0 (adiabatic) • No work is done • Only mass flows and their enthalpies matter 2) Let's use the conservation of energy equation: • m₁h₁ + m₂h₂ = m₃h₃ • Where h = cp × T (for constant specific heat) • For air, we can assume constant cp 3) Conservation of mass: • m₁ + m₂ = m₃ • 4 kg/min + 5 kg/min = 9 kg/min 4) Using conservation of energy with cp being constant: • m₁cp T₁ + m₂cp T₂ = m₃cp T₃ • cp can be cancelled out as it's the same for all streams 5) Therefore: • 4(7) + 5(70) = 9T₃ • 28 + 350 = 9T₃ • 378 = 9T₃ 6) Solving for T₃: • T₃ = 378/9 = 42°C The exit temperature of the mixture is 42°C. This makes sense because: - The result is between the two input temperatures (7°C and 70°C) - It's closer to 70°C because the hot stream has a higher mass flow rate
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. Tcold_1=10 [C] m_dot_cold=5 [kg/min] Thot_1=60 [C] m_dot_hot=2 [kg/min] c_w=4.18 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_hot*c_w*Thot_1+m_dot_cold*c_w*Tcold_1=(m_dot_hot+m_dot_cold)*c_w*Tmix "Some Wrong Solutions with Common Mistakes:" W1_Tmix=(Tcold_1+Thot_1)/2 "Taking the average temperature of inlet fluids" This is the step by step using conservation of mass and energy. 1) First, let's identify what we know: • Cold water: T₁ = 10°C, ṁ₁ = 5 kg/min • Hot water: T₂ = 60°C, ṁ₂ = 2 kg/min • We need to find: T_exit (mixture temperature) 2) For steady-state flow mixing: • Conservation of mass: ṁ₁ + ṁ₂ = ṁ_exit • Conservation of energy: ṁ₁T₁ + ṁ₂T₂ = (ṁ₁ + ṁ₂)T_exit 3) Substituting the values into the energy equation: • (5 kg/min × 10°C) + (2 kg/min × 60°C) = (5 + 2)kg/min × T_exit • 50 + 120 = 7T_exit • 170 = 7T_exit 4) Solving for T_exit: • T_exit = 170 ÷ 7 • T_exit = 24.29°C Therefore, the exit temperature of the mixture is approximately 24.3°C. This makes sense because: - The final temperature is between the two initial temperatures (10°C and 60°C) - It's closer to the cold water temperature because there's more cold water flow (5 kg/min vs 2 kg/min)
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. Tcold_1=15 [C] m_dot_cold=5 [kg/s] Thot_1=90 [C] Thot_2=50 [C] m_dot_hot=4 [kg/s] c_w=4.18 [kJ/kg-C] Q_loss=0 [kJ/s] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_hot*c_w*(Thot_1-Thot_2)=m_dot_cold*c_w*(Tcold_2-Tcold_1)+Q_loss "Some Wrong Solutions with Common Mistakes:" Thot_1-Thot_2=W1_Tcold_2-Tcold_1 "Equating temperature changes of fluids" W2_Tcold_2=90 "Taking exit temp of cold fluid=inlet temp of hot fluid" This is the step by step using energy balance principles. 1) For an adiabatic heat exchanger (no heat loss to surroundings): • Heat lost by hot water = Heat gained by cold water 2) Using the equation: Q = mċ × cp × ΔT • Where mċ = mass flow rate • cp = specific heat capacity of water (4.186 kJ/kg·°C) • ΔT = temperature change 3) For hot water (h): • mċh = 4 kg/s • Th,in = 90°C • Th,out = 50°C • Qh = 4 × 4.186 × (50 - 90) • Qh = -669.76 kW (negative because heat is lost) 4) For cold water (c): • mċc = 5 kg/s • Tc,in = 15°C • Tc,out = unknown • Qc = 5 × 4.186 × (Tc,out - 15) • Qc = 20.93(Tc,out - 15) kW 5) Since Qh = -Qc in an adiabatic system: • -669.76 = 20.93(Tc,out - 15) • -669.76 = 20.93Tc,out - 313.95 • -355.81 = 20.93Tc,out • Tc,out = 47°C Therefore, the exit temperature of cold water is 47°C. To verify this makes sense: - The cold water temperature increased (15°C → 47°C) - The hot water temperature decreased (90°C → 50°C) - The final temperature of cold water (47°C) is less than the final temperature of hot water (50°C), which is physically reasonable - Energy is conserved in the system
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. Tw1=15 [C] m_dot_w=2 [kg/s] Tair1=85 [C] Tair2=20 [C] m_dot_air=3 [kg/s] Q_loss=25 [kJ/s] c_w=4.18 [kJ/kg-C] cp_air=1.005 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_air*cp_air*(Tair1-Tair2)=m_dot_w*c_w*(Tw2-Tw1)+Q_loss "Some Wrong Solutions with Common Mistakes:" m_dot_air*cp_air*(Tair1-Tair2)=m_dot_w*c_w*(W1_Tw2-Tw1) "Not considering Q_loss" m_dot_air*cp_air*(Tair1-Tair2)=m_dot_w*c_w*(W2_Tw2-Tw1)-Q_loss "Taking heat loss as heat gain" (Tair1-Tair2)=(W3_Tw2-Tw1) "Equating temperature changes of fluids" cv_air=0.718 [kJ/kg-K] m_dot_air*cv_air*(Tair1-Tair2)=m_dot_w*c_w*(W4_Tw2-Tw1)+Q_loss "Using cv for air" This is step by step using conservation of energy and the given information. 1) First, let's identify the known variables: * Cold water inlet: ṁw = 2 kg/s, Tw,in = 15°C * Hot air inlet: ṁa = 3 kg/s, Ta,in = 85°C * Heat loss: Q̇loss = 25 kJ/s = 25,000 W * Hot air outlet: Ta,out = 20°C * Need to find: Tw,out 2) We'll need the specific heat capacity of: * Water (cp,w) = 4.186 kJ/kg·°C * Air (cp,a) = 1.005 kJ/kg·°C 3) Using conservation of energy: Energy lost by hot air = Energy gained by cold water + Heat loss to surroundings ṁa × cp,a × (Ta,in - Ta,out) = ṁw × cp,w × (Tw,out - Tw,in) + Q̇loss 4) Let's substitute the known values: 3 kg/s × 1.005 kJ/kg·°C × (85°C - 20°C) = 2 kg/s × 4.186 kJ/kg·°C × (Tw,out - 15°C) + 25 kJ/s 5) Solve the left side: 3 × 1.005 × 65 = 2 × 4.186 × (Tw,out - 15) + 25 195.975 = 8.372(Tw,out - 15) + 25 6) Solve for Tw,out: 195.975 - 25 = 8.372(Tw,out - 15) 170.975 = 8.372(Tw,out - 15) 170.975 = 8.372Tw,out - 125.58 296.555 = 8.372Tw,out Tw,out = 35.42°C Therefore, the exit temperature of cold water is 35.42°C. To verify this makes sense: - The cold water temperature increased (15°C → 35.42°C) - The hot air temperature decreased (85°C → 20°C) - The final temperature of cold water (35.42°C) is between the inlet temperatures - The energy balance is maintained considering the heat loss
Disclaimers code has been uploaded to anti cheat software such as Turnitin. If you plan on using it change it up. Tw1=15 [C] m_dot_w=5 [kg/s] Tair1=90 [C] Tair2=20 [C] m_dot_air=5 [kg/s] c_w=4.18 [kJ/kg-C] cp_air=1.005 [kJ/kg-C] "The rate form of energy balance for a steady-flow system is E_dot_in = E_dot_out" m_dot_air*cp_air*(Tair1-Tair2)=m_dot_w*c_w*(Tw2-Tw1) "Some Wrong Solutions with Common Mistakes:" (Tair1-Tair2)=(W1_Tw2-Tw1) "Equating temperature changes of fluids" cv_air=0.718 [kJ/kg-K] m_dot_air*cv_air*(Tair1-Tair2)=m_dot_w*c_w*(W2_Tw2-Tw1) "Using cv for air" W3_Tw2=Tair1 "Setting inlet temperature of hot fluid = exit temperature of cold fluid" W4_Tw2=Tair2 "Setting exit temperature of hot fluid = exit temperature of cold fluid" This is problem step by step using the principles of heat exchange and energy conservation. 1) For an adiabatic heat exchanger: - Heat lost by hot fluid = Heat gained by cold fluid - No heat is lost to the surroundings 2) Let's use the equation: ṁₕcₚₕ(Tₕᵢ - Tₕₒ) = ṁₖcₚₖ(Tₖₒ - Tₖᵢ) Where: - ṁₕ = mass flow rate of hot air = 5 kg/s - ṁₖ = mass flow rate of cold water = 5 kg/s - cₚₕ = specific heat capacity of air ≈ 1.005 kJ/kg·K - cₚₖ = specific heat capacity of water = 4.186 kJ/kg·K - Tₕᵢ = inlet temperature of hot air = 90°C - Tₕₒ = outlet temperature of hot air = 20°C - Tₖᵢ = inlet temperature of cold water = 15°C - Tₖₒ = outlet temperature of cold water (unknown) 3) Substituting the values: 5 × 1.005(90 - 20) = 5 × 4.186(Tₖₒ - 15) 4) Left side: 5 × 1.005 × 70 = 351.75 kJ/s 5) Right side: 5 × 4.186(Tₖₒ - 15) = 351.75 20.93(Tₖₒ - 15) = 351.75 Tₖₒ - 15 = 16.81 Tₖₒ = 31.81°C Therefore, the exit temperature of the cold water is 31.81°C. This makes sense because: - The water temperature increased but remained below the initial hot air temperature - The total energy is conserved - The temperature difference between the fluids decreased as expected in a heat exchanger
hi! thank you so much for this video, I'm curious between electrical, civil, mechanical, and industrial engineering, which one is better for color blind people, and will we able to get employed?
So studying any engineering is no problem. So the follow is the least difficulty to really difficult. However there are solutions to overcome the challenge. As for the job market I'm notsure it depends where you live. Mechanical Engineering Least impacted by colorblindness Most work involves physical components, CAD drawings, and mechanical systems Color coding is rarely critical for core job functions Civil Engineering Generally manageable with colorblindness Might face some challenges with: Reading colored maps and plans Identifying soil types Traffic signal design Most of these challenges can be overcome with tools and accommodations Industrial Engineering Generally manageable May face minor challenges with: Quality control tasks involving color inspection Color-coded process flows Safety signage design Many alternative methods exist for these tasks Electrical Engineering Most challenging for colorblind individuals Potential difficulties with: Wire color coding (critical for safety) Reading resistor color bands Circuit board inspection LED indicators However, there are workarounds like: Digital multimeters Color identification apps Assistance from colleagues Label systems
@@SuemonKwok oh my god this really clears out all my concerns. as a student who loves physics and math, knowing that youre colourblind is heartbreaking, i thought i was done with my dreams. i guess i should go with mechanical next year, i hope everything works out, huge thank you for all the explanation😮💨. i might have to watch your tutorials as i study later😃
@@farrelaps Good luck in your studies. I'm covering thermodynamics at moments note that the first 500 videos are designed for 2nd and 3rd university students so it may seem bad for 1st year students. From 5-31 onwards I adapted my teaching style for 1st year university students and high school students. When you get to the videos feel free to ask questions especially chapter 1 through 5. Up to 5-30. 5-31 should start getting better.
Correction the symbol should be Q not Q_dot this was copy and paste error
Thank you so much i don't usually comment in videos but I should show a bit of appreciation for the someone who thaught me ❤
Thanks for watching
Thank you, but dot that is top of the Q is wrong. The result is not in kj/s.
You are correct, should be Q. Probably just copy and paste error from previous questions. I'll add it to the pinned comment.
Sir videos of thermo problems have not good quality but you are great.
Thanks for watching. The first 500 videos are designed for 2nd and 3rd year university students with the expectation of having mastered foundations and attempted the question. It's designed as a review material and check if the process is correct. This is supplementary material to student's learning not a direct copy and paste although it could used as such, the grade will be lower. From 5-31 onward the teaching style shifts towards high school and 1st year university students. If you have a question regarding this tutorial feel free to post in the comments. Best of luck in your studies
Assalamualaikum sir
And how do we know that its constant volume process?
" A 10-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure inside rises to 20 psia. " initially at 14.7 psia until the pressure inside rises to 20 psia. Since the pressure is rising you can rule out constant pressure which (C_p) specific heat capacity with respect to constant pressure. Also there is no flow that crosses the boundary This is a rigid tank and so the is volume remains constant so that is why we use (C_v) specific heat capacity with respect to constant volume. It's something you just assume, you don't see tanks changing volume unless there was a piston or mechanism to reduce the volume
Why are we not using cp(enthalpy) here as there is no boundary work involved here?
This is a closed system since no mass enters or leaves we use CV because the tank is rigid and there is nothing flowing out. While the turbine stirs the oxygen it doesn't cross the boundary so it's internal energy not enthalpy because it's not flowing out.
Why cant we use enthalpy here instead of internal energy Like Qin-Wout=m(h2-h1) How do we know that the work done is at the boundary
Enthalpy is flowing energy, internal energy is when energy is static (resting) Here is the longer explanation The relationship between enthalpy and internal energy is: h = u + Pv The enthalpy equation (Q = m(h₂-h₁)) is specifically used for: Flow processes (like in pipes, turbines, nozzles) Constant pressure processes When flow work (Pv) is involved In this piston-cylinder problem: This is a closed system (no mass flow in/out) The pressure isn't constant (it changes from 250 kPa to 300 kPa) We're dealing with expansion work, not flow work Regarding boundary work: In this problem, the work is indeed boundary work because: It's performed by the system pushing against the piston (a moving boundary) It's a result of volume change against a pressure The force is acting at the system boundary (the piston surface) The work equation used (W = P∆V) is valid here because: The process from position 2 to 3 occurs at constant pressure (300 kPa) For constant pressure processes, boundary work simplifies to P∆V In general, for thermodynamic processes: Use internal energy (u) for closed systems where mass doesn't cross boundaries Use enthalpy (h) for open systems where mass flows across system boundaries
Alright tysm❤
Correction: Step 9 there is a typo T_in is 20 degrees not 160
very helpful
Thanks for watching
In our tables we do not have a pressure of 180 kPa, and if I wanted to make an approximation I don't know which temperature to choose.
You need to linear interpolate the values closest to 180 kPa. Above and below. You're finding specific volume in order to find mass flow rate then final enthalpy then temperature General for ratio method of linear interpolation (Max_A-Min_B)/(Max_A-X_(given_z ) )=(Max_(A_2 ) -Min_(B_2 ))/(Y_(unknown_w )-Min_(B_2 ) ) Example (200 kPa -160 kPa)/(200 kPa - 180 kPa)=(0.099951 m^3/kg - 0.12355 m^3/kg)/(v_(g@180 kPa) - 0.12355 m^3/kg) v_(g@180 kPa)=0.1117505 m^3/kg It's close enough to the table but not exact
at the last part of the question, why did we assume Tin=160 instead of Tin=20? air enters at 20 degree celcius , isn t it the air equation?
You are correct. Probably typo or copy and paste error. I'll put in the pinned comment.