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vazorg [7]
3 years ago
11

Consider two different flows over geometrically similar airfoil shapes, one airfoil being twice the size of the other. The flow

over the smaller airfoil has freestream properties given by T[infinity] = 200 K, rho[infinity] = 1.23 kg/m3 , and V[infinity] = 100 m/s. The flow over the larger airfoil is described by T[infinity] = 800 K, rho[infinity] = 1.739 kg/m3, and V[infinity] = 200 m/s. Assume that both μ and a are proportional to T1/2. Are the two flows dynamically similar?
Engineering
1 answer:
yulyashka [42]3 years ago
5 0

Answer

No;

The two flows are not dynamically similar

 Explanation:  Given

T∞,1 = 800k

V∞,1 = 200m/s

p∞,1 = 1.739kg/m³

T∞,2 = 200k

V∞,2 = 100m/s

p∞,2 = 1.23kg/m³

Size1 = 2 * Size2 (L1 = 2L2)  Assumptions Made

α ∝√T

μ∝√T  Two (2) conditions must be met if the two flows are to be considered similar.

Condition 1: Similar Parameters must be the same for both flows

Condition 2: The bodies and boundaries must be genetically true.  Condition 2 is true

Checking for the first condition...

Well need to calculate Reynold's Number for both flows

And Check if they have the same Reynold's Number  Using the following formula

Re = pVl/μ

Re1 = p1V1l1/μ1  Re2 = p2V2l2/μ2  Re1/Re2 = p1V1l1/μ1 ÷ p2V2l2/μ2

Re1/Re2 = p1V1l1/μ1 * μ2/p2V2l2

Re1/Re2 = p1V1l1μ2/p2V2l2μ1

Re1/Re2 = p1V1l1√T2 / p1V1l1√T1

Re1/Re2 = (1.739 * 200 * 2L2 * √200) / (1.23 * 100 * L2 * √800)

Re1/Re2 = 9837.2/3479

Re1/Re2 = 2,828/1

Re1:Re2 = 2.828:1

Re1 ≠ Re2,

So condition 1 is not satisfied  Since one of tbe conditions is not true, the two flows are not dynamically similar

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A 4KJ of energy is supplied to a machine used for lifting a mass. The force required is 800N. If the machine has an efficiency o
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A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg
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Answer:

L=107.6m

Explanation:

Cold water in: m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C

Hot water in: m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C

D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m

Step 1: Determine the rate of heat transfer in the heat exchanger

Q=m_{c}C_{c}(T_{c,out}-T_{c,in})

Q=1.2*4.18*(80-20)

Q=1.2*4.18*(80-20)

Q=300.96kW

Step 2: Determine outlet temperature of hot water

Q=m_{h}C_{h}(T_{h,in}-T_{h,out})

300.96=2*4.18*(160-T_{h,out})

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Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

dT_{1}=T_{h,in}-T_{c,out}

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Step 5: Determine length of heat exchanger

A_{s}=piDL

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L=107.57m

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