gauge in position B? Solution: 1)The efficiency of the pump take whichever section as 1-l and 2-2. it means liquid flow starts from sectionl-I and reaches section 2-2 after a circle. For sections 1-1 and 2-2, the Bernoulli equation will be PI ∑h since PI=p The equation can be simplified to W=∑h=∑hAB+∑hB=98.1+49=1471J/kg The mass flow rate is =Vp=36×110/3600=11kg/s Then, the power of the pump N=Ww,/n=1471×11/0.7=2312W≈231kW 2)The reading of the gauge in position B Set up Bernoulli equation between section A and section B, and take the center of e section ++∑ In the equation Z,=0 ZR=7m p4=2.5×9807×104=245×103N/m( gage pressure) ∑b,A=98Jkg nput the data into the bernoulli equation PB=245×103-(981×7+98.1)×1100=62×104N/m2( gage pressure The reading of the gauge in position B 6.2×10 =0.63kg/c 987×10gauge in position B? Solution：1）The efficiency of the pump take whichever section as 1-1 and 2-2. it means liquid flow starts from section1-1 and reaches section 2-2 after a circle. For sections 1-1 and 2-2, the Bernoulli equation will be + + + e = + + +hf u p W gZ u p gZ   2 2 2 2 1 2 1 1 2 2 since 1 2 1 2 1 Z Z p p u u = = = The equation can be simplified to We =hf =hf ,AB +hf ,BA = 98.1+ 49 =147.1J / kg The mass flow rate is w V kg s s s =  = 361100/ 3600 =11 / Then, the power of the pump is N = Wews / =147.111/ 0.7 = 2312W  2.31k W 2）The reading of the gauge in position B Set up Bernoulli equation between section A and section B，and take the center of the section A as the horizon. + + = + + + f AB B B B A A A h u p gZ u p gZ , 2 2 2  2  In the equation Z A = 0 ZB = 7m 4 5 2 , 2.5 9.807 10 2.45 10 / (gage pressure 98.1 / A B A f AB u u p N m h J kg = =   =   = ） input the data into the Bernoulli equation 5 4 2 pB = 2.4510 − (9.817 + 98.1)1100 = 6.210 N / m （gage pressure） The reading of the gauge in position B 2 4 4 0.63 / 9.87 10 6.2 10 pB = kg cm   =