It is desired to filter a cell broth at a rate of 3000 liters/h on a rotary vacuum filter at a vacuum pressure of 70kPa. The cycle time for the drum will be 60 s and the cake formation time (filtering time) will be 20 s. The broth to be filtered has a viscosity of 2.0 cP and a cake solids (dry basis) per volume of filtrate of 10 g/liter. From laboratory test, the specific cake resistance has been determined to be 9 x 1010 cm/g. Determine the area of the filter that is required.

Answer:

sorry i don't know

Explanation:

Answer:

The thrust of the engine calculated using the cold air is 34227.35 N

Explanation:

For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as

[tex]\dot{m}=\rho AV_a[/tex]

Here

So the equation becomes

[tex]\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}[/tex]

Now in order to find the flow from the fan, the Bypass ratio is used.

[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}[/tex]

Here BPR is given as 8 so the equation becomes

[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}[/tex]

Now the exit velocity is calculated using the total energy balance which is given as below:

[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2[/tex]

Here

So the equation becomes

[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2[/tex]

Rearranging the equation gives

[tex]\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s[/tex]

Now using  the cold air approach, the thrust is given as follows

[tex]T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N[/tex]

So the thrust of the engine calculated using the cold air is 34227.35 N