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Discuss the Principle of Arc Generation in Welding.
Principle of Arc Generation To generate the arc, at first a contact is made between the electrode and the workprece Due to which partial melting and evaporation occurs at the interface and the two ends are severaly heated due to which the electrons and ions at both the ends become electrically unstaRead more
Principle of Arc Generation
→When the electrode is in contact with the workpiece due to short circuit, electric arc will begenerated.
To continue the arc further some gap is maintained between the electrode and the workpiece, known as the Arc Length
Due to movement of electrons from Negative to (+ve) 2/3 rd of heat will be generated on Anode (+ve side), (Since the Kinetic energy of electron is more than that of ions.)
→ and 1/3rd of Heat will be generated on the cathode
→ Due to continuously changing the Polarity, uniform Heat will be generated on cathode and anode.
Note: To concentrate more heat on the electrode and workpiece, DC welding can be used.
See lessDefine Welding. Differentiate different types of welding.
Welding is a fabrication process which is used to produce the permanent joints between two similar or dissimilar materials with or without the application of heat, with or without application of pressure or with or with application of filler material for joining the materials. - For example - Arc WeRead more
– For example – Arc Welding & Gas Welding: Only Heat
– Friction & Resistance Welling: Both Heat & Pressure
– Explosion welding: Only Pressure
Classification of Welding
See lessA car of mass 1100 kg drives with a velocity such that it has a kinetic energy of 400 kJ (see Fig. 3.6). Find the velocity. If the car is raised with a crane, how high should it be lifted in the standard gravitational field to have a potential energy that equals the kinetic energy? g
See lessIn a single acting, two stage reciprocating air compressor 4.5kg of air per minute are compressed from 1.013 bar and 15 deg C through a pressure ratio of 9 to 1. Both stages have the same pressure ratio and the law of compression and expansion in both stages is following P1.3=constant. An intercooler is fitted in between stages and the cooling process is complete. The clearance volume of both stages is 5% of their respective swept volumes and the compressor speed is at 300rpm. a)Sketch the schematic diagram of the compressor and label clearly the TDC, BDC, Vs1, Vs2, Vc, d and L. b) Sketch the P-V diagram and label clearly the TDC, BDC, Vs1, Vs2, Vc, and Vd. c) Calculate the indicated power (kW) d) Determine the swept volume for each stage required (liter) e) Calculate the volumetric efficiency for the 1st stage (%) f) Determine the isothermal power (kW) g) Calculate the isothermal efficiency (%) *R = 0.287 kJ/kg.K
Given data, Air mass flow rate(m)= 4.5kg/min= 0.015kg/cycle No. of stages(n)=2 Compressor speed, N= 300RPM Initial condition: P1= 1.013bar, T1= 15°C= 288K Final pressure(P2) = 9P1 Clearance volume(Vc)= 5% of Vs, k= 1.3 R = 0.287 KJ/kgK Compression ratio of each stage for perfect intercooling (r) = 3Read more
Given data,
Air mass flow rate(m)= 4.5kg/min= 0.015kg/cycle
No. of stages(n)=2
Compressor speed, N= 300RPM
Initial condition: P1= 1.013bar, T1= 15°C= 288K
Final pressure(P2) = 9P1
Clearance volume(Vc)= 5% of Vs,
k= 1.3
R = 0.287 KJ/kgK
Compression ratio of each stage for perfect intercooling (r) = 3
P1V1 = mRT1
101m3 × Vs1 = 0.015 ×0.297 × 288
Vs1 = 0.01224 m^3
Vs1 = 12.24 L
Now, temperature of air after compression,
T2 = Ta × r^(k-1/k)
T2 = 288 × 3^(1.3-1/1.3)
T2 = 371.10 K
Vs2 = mR×(T2/P2)
Vs2 = 0.015×0.287×(371.01/911.7)
Vs2 = 1.752 L
Indiacted Power,
IP = (300×2×1.3/0.3×60) × (0.015×0.287×288) × {9^(0.3/2.6) -1}
IP = 15.50kW
Volumentric Efficiency For Stage 1,
ηV1 = (1 + C – C (r)^1/k)
ηV1 = (1 + 0.05 – 0.05 ×(3)^1/1.3) × (100)
ηV1 = 0.9336 × 100
ηV1 = 93.36 %
Now, if compression is isothermal then work done for each stage,
Wiso = mRT1 × ln (r)
Wiso = (4.5/60) × 0.287 × 288 × ln (3)
Wiso = 6.81 kN
Hence, Isothermal Power Consumption for both stage,
Piso = 2× Wiso
Piso = 2 × 6.81
Piso = 13.62 kW
Isothermal Efficiency,
ηiso = (Piso/Pactual) × 100
ηiso = (13.62/15.50) × 100
ηiso = 87.88 %
An ideal Otto cycle has a compression ratio of 7. At the beginning of the compression process, P1 = 90 kPa, T1 = 27°C and V1 = 0.004 m³. The maximum cycle temperature is 1137°C. For each repetition of the cycle, calculate the heat rejection and the net work production. Also, calculate the thermal efficiency and mean effective pressure for this cycle. Use constant specific heats at room temperature. The properties of air at room temperature are Cp = 1.005kJ / kgK Cv = 0.718kJ / k gK R = 0.287kJ / k gK and k = 1.4 The heat rejection is kJ. kJ. The net work production is The thermal efficiency is The mean effective pressure kPa.
For air, R = 0.287 KJ/kgK CV = 0.718 KJ/kgK K = 1.4 P1 = 90 kPa T1 = 27°C = 300 K V1 = 0.004 m^3 Compression ratio r =7 r = V1/V2 = 7 m = P1V1/RT1 m = (90 kPa × 0.004m^3)/(0.287 KJ/kgK × 300 K) m = 0.00418 kg Process 1-2 isentropic, (P2/P1) = (V1/V2)^ɣ = (r)^ɣ (P2/P1) = (7)^1.4 P2 = 90 × (7)^1.4 = 1Read more
For air,
R = 0.287 KJ/kgK CV = 0.718 KJ/kgK K = 1.4 P1 = 90 kPa T1 = 27°C = 300 K V1 = 0.004 m^3
Compression ratio
r =7 r = V1/V2 = 7
m = P1V1/RT1
m = (90 kPa × 0.004m^3)/(0.287 KJ/kgK × 300 K)
m = 0.00418 kg
Process 1-2 isentropic,
(P2/P1) = (V1/V2)^ɣ = (r)^ɣ
(P2/P1) = (7)^1.4
P2 = 90 × (7)^1.4 = 1372.08 kPa
T2 = T1 (V1/V2)^ɣ-1
T2 = 300 × (7)^1.4-1
T2 = 653.37 K
Consider state 3,
P2V2/T2 = P3V3/T3
P3 = (T3/T3) × P2
P3 = (1400/653.37) × 1372.08
P3 = 2940 kPa
Process 3-4,
T4 = (T3)/(V4/V3)^ɣ-1
T4 = 1400/(7)^1.4-1
T4 = 642.8 K
Calculate the heat input to the cycle,
Qin = m×Cv(T3-T1)
Qin = 0.00418 × 0.718 (1400-653.37)
Qin = 2.24 KJ
Calculate the heat rejected from the cycle,
Qout = m×Cv(T4-T1)
Qout = 0.00418×0.718(642.8-300)
Qout = 1.028 KJ
Net Work = Qin -Qout
Net Work = (2.24 – 1.028) KJ
Net Work = 1.211 KJ
ηthermal = (Net Work/Qin) × 100
ηthermal = (1.24/2.24)×100
ηthermal = 54.08 %
Pmean = (Net Work)/V1(1-1/ɣ)
Pmean= (1.211)/0.04(1-1/1.4)
Pmean = 353 KPa
The following data refers to a test on an axial flow compressor. Atmospheric temperature an pressure at inlet are 18°C and I bar. Total head temperature in delivery pipe is 165°C. Tota head pressure in delivery pipe is 3.5 bar. Static pressure in delivery pipe is 3 bar. Adiabat index (y) or ratio of specific heats of flowing gas = 1.4 Calculate (a) total head isentropic efficiency. (b) Polytropic efficiency, and (c) air Velocity delivery pipe.
Given data: Inlet temperature, T, = 18° C = 291 K Inlet pressure, P1= 1 bar Outlet temperature, T₂ =165° C = 438 K Outlet pressure, P₂ = 3.5 bar Solution: (a) Total head isentropic efficiency Calculated the enthalpy of air at T1, and P1 using online calculator h1=291.544 kJ/kg Calculated the enthalpRead more
Given data:
Inlet temperature, T, = 18° C = 291 K
Inlet pressure, P1= 1 bar
Outlet temperature, T₂ =165° C = 438 K
Outlet pressure, P₂ = 3.5 bar
Solution:
(a) Total head isentropic efficiency
Calculated the enthalpy of air at T1, and P1 using online calculator
h1=291.544 kJ/kg
Calculated the enthalpy of air at T1 and P2, from using calculator
h₂ = 290.9465 kJ/kg
Calculated the enthalpy of air at T₂ and P2 using online calculator
h3 =443.36 kJ/kg
Calculated the enthalpy of air at T2, and P1 using online calculator
h4 = 440.0875 kJ/kg
Calculating the isentropic efficiency of the compressor,
ηIsentropic = (h1-h2/h4-1)x100_______________(1)
Substituting the values in the above equation (1).
ηIsentropic= (291.544-290.9456 /443.36-440.0875)× 100
ηIsentropic = 18.285%
Hence, the isentropic efficiency of the compressor is 18.285%
(b) Polytropic efficiency
Polytrophic path coefficient,
n = [1-(438/291)/(3.5/1)^-1
n = 1.754
ηpoly = (1.4-1/1.4)/(1.754-1/1.754) × 100
ηpoly = 66.464%
Here, 1.4 is the ratio of specific heat of the air.
Hence, the polytrophic efficiency of the compressor is 66.464%
See lessThe following data give reading for ten sample of size 8 each in the production of the certain component draw a control chart for mean and range and point out which sample if any are out of limit.
See lessThe demand for a product in the month of march turned out to be 20 units against an earlier made forecast of 20 units. The actual demand for April and May turned to be 25 and 26 units respectively. What will be the forecast for the month of June, using exponential smoothing method and taking smoothing constant as 0.2?
See less20 sample of size 100 are taken. The total number of defective items is 75. What is the LCL of the 3𝜎 (z=3) p-chart?
See lessIf the probability of acceptance of 1% defective lot is 0.95. then find the AOQ
Pac = 0.95 P = fraction defective = 1% = 0.01 AOQ = Pac × P AOQ = 0.95 × 0.01 = 0.0095
Pac = 0.95
P = fraction defective = 1% = 0.01
AOQ = Pac × P
AOQ = 0.95 × 0.01 = 0.0095
See less