[ values can be easily fixed. 100 0 obj <> endobj A 3 phase electric motor drawing 20.76 kW for 173.3 hours per month will use 3771.7 kWh (kilowatt hours) of electricity per month. There is no obligation to choose the transformer’s ratings as the base values, but it is an option. Unless maybe the meter is owing some debts which Ikeja deducts each time he vends. 0000071295 00000 n 0000003003 00000 n 0000002504 00000 n However, because they are interrelated, selecting base values for any two of them determines the base values of the remaining two. Some basic relationships follow and are different for single-phase systems and three-phase systems. Now the current per phase is also down to a third of what it would be with a single phase supply (about 30 amps per phase, rather than 100). 0000008328 00000 n Often, nameplate values are taken as base values but this need not be the case. Therefore, the power delivered by three phase is much more consistent and supply more power than a 3 single-phase power … > approximate cost per unit of 3 phase electricity. 0000004222 00000 n 2 Likes 1 Share },{ "name": "Home" The determination of per unit values is not the end objective of a study. While the actual values of the impedances of power system equipment, expressed in ohms, may vary over a wide range, depending on their rating, their per-unit values are restricted to a narrow range whatever the rating may be. xref 2. For the same amount of time, three-phase power lines can carry more power than that of single-phase power lines at a reduced cost. startxref For either a single-phase system or a three-phase system, the per units become, $\begin{align}  & {{V}_{pu}}=\frac{V}{{{V}_{base}}} \\ & {{I}_{pu}}=\frac{I}{{{I}_{base}}} \\ & {{S}_{pu}}=\frac{S}{{{S}_{base}}} \\ & {{Z}_{pu}}=\frac{Z}{{{Z}_{base}}} \\\end{align}$. 0000026395 00000 n I imagine the pit would have three phase power running through it, so cant see any issues there. A common power base of  ${{S}_{b}}=50MVA$  is chosen for the system, but the voltage bases are selected separately for each voltage level: Voltage base for buses 1, 2, and 3:${{V}_{b}}\left( 1 \right)={{V}_{b}}\left( 2 \right)={{V}_{b}}\left( 3 \right)=110kV$   Voltage base for bus 4: ${{V}_{b}}\left( 4 \right)=11kV$  Per-unit reactances of the generators and transformers are, \[\begin{align}  & \begin{matrix}   {{G}_{1}}: & {{X}_{G1}}=0.2\times \frac{50}{48}=0.0208pu  \\\end{matrix} \\ & \begin{matrix}   {{G}_{2}}: & {{X}_{G2}}=0.15\times \frac{50}{25}=0.3pu  \\\end{matrix} \\ & \begin{matrix}   {{T}_{1}}: & {{X}_{T1}}=0.08\times \frac{50}{50}=0.08pu  \\\end{matrix} \\ & \begin{matrix}   {{T}_{2}}: & {{X}_{T2}}=0.6\times \frac{50}{30}=0.1pu  \\\end{matrix} \\ & \begin{matrix}   {{T}_{3}}: & {{X}_{T3}}=0.1\times \frac{50}{50}=0.1pu  \\\end{matrix} \\\end{align}\], The impedance bases for lines 1, 2, and 3 are, \[{{Z}_{b\left( L1 \right)}}={{Z}_{b\left( L2 \right)}}={{Z}_{b\left( L3 \right)}}=\frac{{{\left( 110 \right)}^{2}}}{50}=242\Omega \], The per-unit reactances of lines 1, 2, and 3 are, \[\begin{align}  & \begin{matrix}   {{L}_{1}}: & {{X}_{L1}}=\frac{40}{242}=0.165pu  \\\end{matrix} \\ & \begin{matrix}   {{L}_{2}}: & {{X}_{L2}}=\frac{32}{242}=0.132pu  \\\end{matrix} \\ & \begin{matrix}   {{L}_{3}}: & {{X}_{L3}}=\frac{30}{242}=0.124pu  \\\end{matrix} \\\end{align}\], The power factor of the load is cos $\cos \varphi =0.75$ lagging; thus, the complex power of the load is ${{S}_{L}}=60\angle {{\cos }^{-1}}0.75=60\angle {{41.41}^{o}}MVA$ . "url": "https://electricalacademia.com/electric-power/per-unit-calculation-per-unit-system-examples/", 0000003039 00000 n Using the common base Sb = 50 MVA, draw the impedance diagram in per unit including the load impedance. "position": 3, Let's consider a sample consumption of 260 units per month. Go back to Three phase power measurements ↑ Wye and Delta comparison. Nevertheless, a person involved with three-phase electrical power would be well advised to understand the PU method, its mechanics, and how it is used. FIGURE 2 Per-phase equivalent impedance diagram for the system shown in Fig. "@type": "ListItem", 0000007693 00000 n This way, there is less chance of confusing line and phase voltages as well as between single- and three-phase power systems. Using the common base, The three-phase load at bus 4 absorbs 60 M, Voltage base for buses 1, 2, and 3:${{V}_{b}}\left( 1 \right)={{V}_{b}}\left( 2 \right)={{V}_{b}}\left( 3 \right)=110kV$, Voltage base for bus 4: ${{V}_{b}}\left( 4 \right)=11kV$, Per-unit reactances of the generators and transformers are, In all of these expressions, the numerator is a complex vector, whereas the denominator is a real number. I told them that I have a set of high tension power lines on part of my property and all I have to do is put a pole up with a transformer and I can have free electric for life for that price. 1. I currently pay $0.14 / KW and Run the unit 6 hours a day. 0000024875 00000 n } ] Calculate the total cost. \[{{I}_{base}}=\frac{{{S}_{base}}}{{{V}_{base}}\times \sqrt{3}}=\frac{700MVA}{145kV\times \sqrt{3}}=2.78kA\], \[{{Z}_{base}}=\frac{{{V}_{base}}}{{{I}_{base}}\times \sqrt{3}}=\frac{145kV}{2.78kA\times \sqrt{3}}=30.1\Omega \], If, say, a secondary voltage of 130 kV is under consideration, then the per unit value of that voltage is, \[V=\frac{V}{{{V}_{base}}}=\frac{130kV}{145kV}=0.89pu\]. "@type": "ListItem", 0000021615 00000 n Electricity unit price of domestic connections is Rs-12.09 per unit for 1 st 100 units and after 1 st 100 units it will increase into Rs-13.54 per unit. Rather determined per unit values serve to assist in the calculation of other system values. "@type": "BreadcrumbList", "name": "Per Unit Calculation | Per Unit System Examples" controlled locally by inbuilt electronics and can correct for +/-10% changes in the voltage will cost approximately $18k per single phase unit. 1. The fuel cost, or cost of imported natural gas, is tied to oil prices by commercial contracts, which change depending on global market conditions. 0000015543 00000 n Most domestic houses with gas central heating need single phase and have this as standard. 0000008168 00000 n "position": 2, Base Value for 3-ppyhase systems 0000015963 00000 n } Pt = total power of the circuit in watt (W) P1, P2, P3 = power of phase 1, phase 2 and phase 3 in watt (W) V = voltage in Volt (V) exemple : V1=230 V in 400 V three-phases circuit I1, I2, I3 = current of phase 1, phase 2 and phase 3 in ampere (A) Cos φ1, Cos φ2 , Cos φ3 = cosinus phi of each phase (without unit) The state that saw the great increase in prices for electricity is Rhode Island. 100 41 Three phase costs me (5.5 x .14 x 6) per day to operate or ($4.62) a difference of $2.03/day. 0000026747 00000 n \$\begingroup\$ If you need high power you should take the 3 phase system, like you said 2kW Motor, for low power devices you should take the single phase because you have lower costs for converting it into dc for example. The per-unit representation of physical values consists in expressing these values as fractions or percentages of some representative bases of the same physical nature. Now #3800 divided by 74.8 is about 50.8kwh per unit. A per unit quantity then becomes the ratio of a selected parameter to a selected base value. 0000005113 00000 n The manufacturer’s nominal ratings are given as follows: FIGURE 1 Single-line diagram of Example 1. I also need the slab rates like how much cost per unit for 0-100 units,100-200 units,200-300 units etc.I heard that we will get electric bill bi-monthly.But my landlord takes electric bill every month.He is insisting me to pay Rs 6/unit for every month (even if consumption is < 100 units) He told 3-phase power will come under commercial purpose, so the charge will be more.Is it true? 0000004489 00000 n To overcome this problem, all the system quantities are converted into a uniform normalized platform called, The per-unit representation of physical values consists in expressing these values as fractions or percentages of some representative, Electricity Tariffs Explained | Residential and Industrial Tariff, Grounding Electrical System | Purpose of Grounding Electrical Systems. }. Multiply the total kilowatt hours used by the rate per kilowatt hour charged by the power company to find the cost. For base values, a value of Sbase = 50 or 100 MVA is often selected. 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