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Sloan [31]
3 years ago
6

A supply fan is operating at 30000 cfm and 4 inch of water with an efficiency of 50%. (a) Calculate the fan power at the current

operating condition. (b) Calculate the pressure and power if the supply fan reduces its speed to deliver 20000 cfm.
Engineering
1 answer:
serg [7]3 years ago
5 0

Answer:

The pressure and power of fan is 1.77 and 11.18 Hp respectively.

Explanation:

Given:

Discharge Q_{1}  = 30000 cfm

Pressure difference \Delta P = 4 inch

Efficiency \eta = 50\%

(A)

From the formula of fan power,

     P _{1}  = \frac{Q \Delta P}{6356 \eta}

     P_{1}  = \frac{30000 \times 4}{6356 \times 0.5}

     P_{1}  = 37.76 Hp

(B)

Fan power and pressure is given by,

We know that pressure difference is proportional to the square of discharge.

    \frac{\Delta p_{2} }{\Delta P_{1} } = (\frac{Q_{2} }{Q_{1} } ) ^{2}

   \Delta P_{2}  = (\frac{20000}{30000} ) \times 4

   \Delta P_{2} = 1.77

Fan power proportional to the cube of discharge.

       \frac{P_{2} }{P^{1} }  = (\frac{Q_{2} }{Q_{1} } )^{3}

       P_{2} =   \  (\frac{20000}{30000} ) ^{3} \times 37.76

       P_{2} = 11.18 Hp

Therefore, the pressure and power of fan is 1.77 and 11.18 Hp respectively.

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Answer:The Urban heat island temperature will be REDUCED.

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1) provision of shade against Sunlight.

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3 years ago
Steam at 40 bar and 500o C enters the first-stage turbine with a volumetric flow rate of 90 m3 /min. Steam exits the turbine at
a_sh-v [17]

Answer:

(a) 62460 kg/hr

(b) 17,572.95 kW

(c) 3,814.57 kW

Explanation:

Volumetric flow rate, G = 30 m³ / 1 min => 90 / 60 => 1.5

Calculate for h₁ , h₂ , h₃

h₁ is h at P = 40 bar, 500°C => 3445.84 KJ/Kg

Specific volume steam, ц = 0.086441 m³kg⁻¹

h₂ is h at P = 20 bar, 400°C => 3248.23 KJ/Kg

h₃ is h at P = 20 bar, 500°C => 3468.09 KJ/Kg

h₄ is hg at P = 0.6 bar from saturated water table => 2652.85 KJ/Kg

a)

Mass flow rate of the steam, m = G / ц

m = 1.5 / 0.086441

m = 17.35 kg/s

mass per hour is m = 62460 kg/hr

b)

Total Power produced by two stages

= m (h₁ - h₂) + m (h₃ - h₁)

= m [(3445.84 - 3248.23) + (3468.09 - 2652.85)]

= m [ 197.61 + 815.24 ]

= 17.35 [1012.85]

= 17,572.95 kW

c)

Rate of heat transfer to the steam through reheater

= m (h₃ - h₂)

= 17.35 x (3468.09 - 3248.23)

= 17.35 x 219.86

= 3,814.57 kW

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