Question 1 Question 2 Question 3 Question 4 Question 5 Thanks A fresh-water heat

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Question 5

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A fresh-water heat-exchanger is used to cool the oil in a marine engine Water temperature into the exchanger is 15 degree C and water temperature out is 30 degree C Oil temperature in is 150 degree C and the oil flow ts 80L/mm The oil has a relative density of 0 9 and specific heat capacity of 1 80kj/kgoK The specific heat capacity of the water may be taken as A 19 kJ/kgoK If the outlet oil temperature a 100oC determine the flow rate of water A horizontal venturi tube in a 75mm diameter pipe has a throat diameter of 50mm. If water flows in the pipe such that the upstream pressure and velocity are 45kpa and 4 m/s respectively, determine the ideal velocity and pressure at the throat A water tank as shown in figure below discharges water through an orifice in the side of the tank Determine the velocity of the water at point 2 when h=2m if the air space pressure is; Atmospheric and 50 kPa (gauge) Neglect the losses Water flows in a pipe as shown in figure below At point 1 the diameter is 50mm. elevation 3 m, pressure 45 kPa and velocity 5 6 m/s At point 2 the diameter is 400mm and the elevation is 5m Determine the ideal pressure at 2. A pump system as shown in below is used to pump water at a rate of 30L/S. the following information is given: Determine the power input to the pump if the efficiency is 60% Neglecting losses between point 1 and 2; If the head loss between 1 and 2 is 8m

Explanation / Answer

question 1

let m be flow rate of water

oil density=900kg/m^3

water density=1000kg/m^3

so heat lost by oil=mCpdT

m=80L/min=900*80*10^-3Kg/min=72Kg/min

heat lost=72*1.8*50=6480KJ/min

heat gain by water=m*4.19*15

so mass flow rate of water=6480/(4.19*15)=103Kg/min=103L/min

question 2

here conservation of mass gives

A1V1=A2V2

so d1^2*V1=d2^2*V2

so pipe velocity=4m/s so throat velocity V2=(d1/d2)^2*V1=4*(75/50)^2=9m/s

and by benoli equation we get

P1/(rho*g)+V1^2/2g=P2/(rho*g)+V2^2/(2g)

so

P1/density+V1^2/2=P2/density+V2^2/2

so P2=12.5Kpa

question 3

a) when it is atmospheric we get

V=sqrt(2g*h)=6.26m/s (when g=9.8 ,h=2)

b) using bernioli equation we get

Pguage/(density*g)+h=V^2/2g

so V=sqrt(2*Pguage/density+2g*h)=9.44 m/s

question 4

we can solve by bernolis equation

first using mass conservation we find velocity

V2=A1V1/A2=(d1/d2)^2*V1=(50/400)^2*5.6=0.0875m/s

so P1/(rho*g)+h1+V1^2/(2g)=P2/(rho*g)+h2+V2^2/(2g)

so P2=41.076Kpa

question 5

a) neglecting head loss

mass flow=30L/s=30*10^-3m^3/s=A*V

so V1=3.81m/s

V2=(d1/d2)^2*V1=(100/75)^2*3.81=6.77m/s

so we get

25/(rhp*g)+3+3.81^2/(2g)+pumphead=80/(rho*g)+9+6.77^2/(2g)

pump head=13.21m

if efficiency=60% so pump head=22.01m

so power input=30*10^-3*22.01*1000*9.8=6470.94W

b) is head loss =6m

so pump head=13.21+6=19.21

if efficiency=60% so head=32.01m

so power input=30*10^-3*32.01*1000*9.8=9412.9W

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