Question

please request I need urgent solution ​

Answer 1

Answer:

please mark it as brainlist (ig:iamxhay)

Step-by-step explanation:

Centrifugal compressor Mass flow rate mf = 0.8Kg/sec

Condition at inlet:

P1 = 1 bar = 1 × 105 N/m2

V1 = 10m/sec; u1 = 30KJ/kg ν1 = 0.95m3/kg

Condition at exit:

P2 = 8bar = 8 × 105 N/m2

V2 = 6m/sec;

 u2 = 124KJ/kg

ν2 = 0.2m3/kg

The change in enthalpy is given by

h2 – h1 = (u2 + P2U2) – (u1 + P1U1)

= (124 × 103 + 8 × 105 × 0.2 – (30 × 103 + 1 × 105 × 0.95)

= 159000 J/Kg = 159KJ/kg ...

(i) Heat loss to cooling water  Q = – (dU + dW) = – (U2 – U1) – Ws KJ/sec

 Q = – (30 – 124) – Ws = – 96 – Ws ...

(ii)  From Sfee  Q – Ws = mf [(h2 – h1) + 1/2 (v22  – V1 2) + g (Z2 – Z1)]  

– 96 – Ws – Ws = 0.8 [159 + 1/2 (62 – 102)]  – 96 2Ws = 0.8 [159 + 1/2 (62 – 102)]  – 96 – 2Ws = 101.6  Ws = – 98.8KJ/sec .

(–ive sign indicate that work done on the system) Thus the power of motor required to drive the compressor is 54.60KW Mass flow rate at inlet = Mass flow rate at outlet

 = 0.8 A1 × 10/0.95

A1 = 0.076m2

Π/4.dintel 2 = 0.076

dinlet = 0.096 m = 96.77 mm .

Now; Mass flow rate of outlet

= mf2 = A2.V2/u2 0

.8 = A2 × 6/0.2 A2 = 0.0266m2

Π/4.doutlet 2 = 0.0266

doutlet = 0.03395 m = 33.95 mm

Answer 2

Answer:

The answer is 33.95mm.

Device: Centrifugal compressor Mass flow rate mf = 0.8Kg/sec Condition at inlet:

P1 = 1 bar = 1 × 10^5 N/m2 V1 = 10m/sec;

u1 = 30KJ/kg ν1 = 0.95m3/kg

Condition at exit:

P2 = 8bar = 8 × 10^5 N/m2 V2 = 6m/sec;

u2 = 124KJ/kg ν2 = 0.2m3/kg

The change in enthalpy is given by

h2 – h1 = (u2 + P2U2) – (u1 + P1U1) =

(124 × 10^3 + 8 × 10^5 × 0.2 – (30 × 10^3 + 1 × 10^5 × 0.95)

= 159000 J/Kg = 159KJ/kg ...(i)

Heat loss to cooling water

Q = – (dU + dW) = – (U2 – U1) – Ws KJ/sec

Q = – (30 – 124) – Ws = – 96 – Ws ...(ii)

From Sfee

Q – Ws = mf [(h2 – h1) + 1/2 (v2^2 – V1 ^2) + g (Z2 – Z1)]

– 96 – Ws – Ws = 0.8 [159 + 1/2 (6^2 – 10^2)]

– 96 2Ws = 0.8 [159 + 1/2 (6^2 – 10^2)]

– 96 – 2Ws = 101.6

Ws = – 98.8KJ/sec .

(–ive sign indicate that work done on the system) Thus the power of motor required to drive the compressor is 54.60KW Mass flow rate at inlet = Mass flow rate at outlet

= 0.8 A1 × 10/0.95

A1 = 0.076m2

Π/4.dintel 2 = 0.076

dinlet = 0.096 m = 96.77 mm .

Now; Mass flow rate of outlet = mf2 = A2.V2/u2

0.8 = A2 × 6/0.2

A2 = 0.0266m2

Π/4.doutlet 2 = 0.0266

doutlet = 0.03395 m = 33.95

Hope it helps....