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Assignment for calculating NPSH available and NPSH required
- 1. 1.(a)(i) NPSH available(NPSHa) is the net positive suction head resulting from system configuration and operation, it is the difference between the net inlet head and the head corresponding to the vapour pressure of the liquid required to make the liquid flow through the suction from the sump to the impeller .NPSH required(NPSHR) is an empirical quantity, obtained from the pump manufacturer’s tests,it is the total suction head ,varies with the pump design ,speed and capacity of the pump Calculations to find NPSHa NPSHa = (absolute pressure headat inlet of pump)-(vapor pressure head) + (velocity head at inlet ofpump) = 𝑃 𝑎𝑡𝑚 𝜌𝑔 − 𝑃𝑣 𝜌𝑔 + 𝑣 𝑠 2 2𝑔 = 𝑃 𝑎𝑡𝑚 𝜌𝑔 − ( 𝑣 𝑠 2 2𝑔 + ℎ 𝑠 + ℎ 𝑓𝑠) − 𝑃𝑣 𝜌𝑔 + 𝑣 𝑠 2 2𝑔 = 𝑃 𝑎𝑡𝑚 𝜌𝑔 − ℎ 𝑠 − ℎ 𝑓𝑠 − 𝑃𝑣 𝜌𝑔 = 𝐻 𝑎𝑡𝑚 − ℎ 𝑠 − ℎ 𝑓𝑠 − 𝐻𝑣 Where 𝑃𝑎𝑡𝑚 = 𝑎𝑡𝑚𝑜𝑠𝑝ℎ𝑒𝑟𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑃𝑣 = 𝑣𝑎𝑝𝑜𝑢𝑟 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ℎ 𝑠 = 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 ℎ 𝑓𝑠 = ℎ𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑠𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒 𝑃𝑎𝑡𝑚 =13.6× 749 1000 =10.1864m of water. =13.6× 26.2 1000 =0.35632m of water. ℎ 𝑠 =5.2m ℎ 𝑓𝑠 = 0.457𝑚 NPSHa= 10.1864-5.2-0.457 = 4.5294m 𝐻 𝑚 3 4 = 1450√0.0637 60×0.683 Hm =18.5277m 𝛿 𝑐 = 0.2 NPSHR = 𝛿 𝑐×Ham = 0.2×18.5277
- 2. =3.70554 Conclusion Since NPSHa=4.5294m> NPSHR= 3.66m, cavitation will not occur. Heat engines . They receive heat from high temperature source. They convert part of this heat to work. They reject the remaining waste heat to a low temperature sink. They operate on a cycle. Measure of performance is called thermal efficiency is=W net, out /Qin For heat engine the desiredresultisnetworkdone and the inputis the heatsuppliedtomake the cycle operate. Heat pumps
- 3. A device that transfers heat from low temperature medium to a high temperature one. Index of performance of a heat pump, is called coefficient of performance COP HP, Carnot = T H / (TH - T L) Pressure ( bars) Saturated temperature °C Specific enthalpy kJ/ kg hf Specific enthalpy kJ/kg hg Specific entropy kJ/kgk Sf 1.6 -19 19.11 179.37 0.0781 0.7077 9.0 37 71.88 202.26 0.2633 0.6833 (b)(i)s1 = sg@1.6 bar =0.7077kJ/kgk. [T3 = 37+273=310K] s1 = s2 ,s3 =sg@9.0 bar =0.6833kJ/kgk. [T2 = 80+273=353K] s2 = s3 +cpv ln (T2 /T3 )
- 4. 0.7077 = 0.6833+Cpvln( 353 310 ) C pv = 0.0244 0.1299
- 5. = 0.1878kJ/kgk h4 = hf@9.0bar =71.88kJ/kg h3 = hg@9.0bar =202.26kJ/kg h2 = h3 + cpv (T2 -T3 ). = 202.26 + 0.1878(353-310) = 210.3775kJ/kg Heat transfer from condenser = h2 - h4 = 210.3775-71.88 = 138.4975kJ/kg Refrigerant flow rate = (200kJ/min)/ (138.4975kJ/kg) = 1.4441kg/min = 0.02407kg/s (ii) Power input to the compressor,Wnet =m( h2 - h1) = 0.02407(210.3775-179.37) = 0.7464kW (iii) COPHP,Carnot =(h2 - h4 )/(h2 - h1 ). =(210.3775-71.88)/(210.3775-179.37) =4.4665