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Where am I going wrong? Appreciate if someone can take a look at this and show me where I am going wrong.

A conical pipe with inlet diameter 80mm and inlet pressure 0.5 bar, is inclined so that the outlet is 2m higher than the inlet. The outlet diameter is 140mm, the fluid density is 1000 kg/m3, and the mass flow rate is 40 kg/s.

Calculate:

a) The inlet and outlet velocities

b) The outlet pressure

My workings are giving the outlet pressure higher than the inlet pressure,which must be wrong due to the head and increased diameter. Here is how I arrived at my answers:

First calculate the inlet velocity V1

Calculate CSA of inlet (A1)

A1 = (PI × 0.08^2)/4

= 5.02 × 10^-3 m2

Calculate CSA of outlet (A2)

A2 = (PI × 0.15^2)/4

= 0.0154 m2

V1 = Q/(A1 × ρ)

V1 = 40/(0.00502 × 1000)

= 7.968 m/s

V2 = Q/(A2 × ρ)

V2 = 40/(0.015 × 1000)

= 2.67 m/s

Inlet Pressure P1 = 0.5 bar = 50000 pa

Inlet head h1 = 0 m

Outlet head h2 = 2m

Gravity = 9.81

gh1+ 1/2 V1^2+ (P1/ρ) = gh2+ 1/2 V2^2+ (P2/ρ)

= (9.81 × 0) + 1/2 x 7.968^2 + (50000/1000) = (9.81 × 2) + 1/2 x 2.67^2 + (P2/1000)

81.74 = 23.18 + (P2/1000)

P2 = 1000(81.74-23.18)

P2 = 58560 pa

P2 = 0.58 bar

Attached is a diagram of the pipe.

A conical pipe with inlet diameter 80mm and inlet pressure 0.5 bar, is inclined so that the outlet is 2m higher than the inlet. The outlet diameter is 140mm, the fluid density is 1000 kg/m3, and the mass flow rate is 40 kg/s.

Calculate:

a) The inlet and outlet velocities

b) The outlet pressure

My workings are giving the outlet pressure higher than the inlet pressure,which must be wrong due to the head and increased diameter. Here is how I arrived at my answers:

First calculate the inlet velocity V1

Calculate CSA of inlet (A1)

A1 = (PI × 0.08^2)/4

= 5.02 × 10^-3 m2

Calculate CSA of outlet (A2)

A2 = (PI × 0.15^2)/4

= 0.0154 m2

**Inlet Velocity V1**V1 = Q/(A1 × ρ)

V1 = 40/(0.00502 × 1000)

= 7.968 m/s

**Outlet Velocity V2**V2 = Q/(A2 × ρ)

V2 = 40/(0.015 × 1000)

= 2.67 m/s

**Outlet Pressure Calc**Inlet Pressure P1 = 0.5 bar = 50000 pa

Inlet head h1 = 0 m

Outlet head h2 = 2m

Gravity = 9.81

gh1+ 1/2 V1^2+ (P1/ρ) = gh2+ 1/2 V2^2+ (P2/ρ)

= (9.81 × 0) + 1/2 x 7.968^2 + (50000/1000) = (9.81 × 2) + 1/2 x 2.67^2 + (P2/1000)

81.74 = 23.18 + (P2/1000)

P2 = 1000(81.74-23.18)

P2 = 58560 pa

P2 = 0.58 bar

Attached is a diagram of the pipe.