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zysi [14]
3 years ago
6

How are you today First to awnser is brainlyest.

Engineering
1 answer:
Leviafan [203]3 years ago
3 0

Answer:

Happy

Explanation:

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The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be
lakkis [162]

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = \frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}       .....................1

Q = \frac{T \infty - Tso}{\frac{1}{hA}}                         .....................2

now we compare both equation 1 and 2 and put here value

\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

3 0
3 years ago
Me ayudas plis noce ​
alekssr [168]

Answer:

Explanation:

7 0
3 years ago
11. Wet-cell batteries are commonly referred to as lead-acid batteries.<br> A) O True<br> B) O False
Alja [10]
Is true i think bye ansía
5 0
4 years ago
Air is compressed from 100 kPa, 300 K to 1000 kPa in a two-stage compressor with intercooling between stages. The intercooler pr
Brut [27]

Answer:

The total compressor work is 234.8 kJ/kg for a isentropic compression

Explanation:

Please look at the solution in the attached Word file

Download docx
6 0
4 years ago
Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 85 vol% carbide
velikii [3]

Answer:

Explanation:

k_max = 26.9 w/mk

  k_min  =  22.33 w/mk

Explanation:

a) the maximum thermal conductivity is given as

K_MAX = k_m v_m + k_p v_p

where k_m is thermal conductvitiy of metal

k_p is thermal conductvitiy of carbide

v_m = proportion of metal in the cement = 0.15

v_p = proportion of carbide in the cement = 0.85

K_MAX = k_m v_m + k_p v_p

            = 66*0.15 + 20*0.85

           k_max = 26.9 w/mk

b) the minimum thermal conductivity is given as

k_min = \frac{ k_{carbide} *k_{metal}}{k_{metal} v_{carbide} +k_{carbide} v_{metal}}

          = \frac{20*66}{20*0.15 +66*0.85}

        k_min  = 22.33 w/mk

5 0
3 years ago
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