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vodka [1.7K]
3 years ago
15

A car radiator is a cross-flow heat exchanger with both fluids unmixed. Water, which has a flow rate of 0.05 kg/s, enters the ra

diator at 400 K and is to leave at 330 K. The water is cooled by air that enters at 0.75 kg/s and 300 K. If the overall heat transfer coefficient is 200 W/m2-K, what is the required heat transfer surface area?
Engineering
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

Explanation:

Known: flow rate and inlet temperature for automobile radiator.

Overall heat transfer coefficient.

Find: Area required to achieve a prescribed outlet temperature.

Assumptions: (1) Negligible heat loss to surroundings and kinetic and

potential energy changes, (2) Constant properties.

Analysis: The required heat transfer rate is

q = (m c)h (T h,i - T h,o) = 0.05 kg/s (4209J / kg.K) 70K = 14,732 W

Using the ε-NTU method,

Cmin = Ch = 210.45 W / K

Cmax = Cc = 755.25W / K

Hence, Cmin/Cmx(Th,i - Th,o) = 210.45W / K(100K) = 21,045W

and

ε=q/qmax = 14,732W / 21,045W = 0.700

NTU≅1.5, hence

A=NTU(cmin / U) = 1.5 x 210.45W / K(200W) / m² .K) = 1.58m²

1. the air outlet is..

Tc,o = Tc,i + q / Cc = 300K + (14,732W / 755.25W / K) = 319.5K

2. using the LMTD approach ΔTlm = 51.2 K,, R=0.279 and P=0.7

hence F≅0.95 and

A = q/FUΔTlm = (14,732W) / [0.95(200W / m².K) 51.2K] = 1.51m²

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Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)  

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What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its
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<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil (N_{p}) to the number of turns in its secondary coil (N_{s}) is equal to the ratio of the input voltage (V_{p}) to the output voltage (V_{s}) of the transformer. i.e

\frac{N_p}{N_s} = \frac{V_p}{V_s}            ----------------(i)

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N_{p} = number of turns in the primary coil = 8 turns

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3 years ago
Consider the mixing of 0.8 kg/s of hot water at 348 K and 1 kg/s of cool water at 298 K that is generating warm water. Assume no
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h_i = c_p. T_i   ... Eq 2

m_1 + m_2 = m_3   ... steady flow system (Eq 3)

Substitute Eq 2 and Eq3 in Eq1

m_3 = 0.8 + 1 = 1.8 kg/s

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T_3 = 355.3 / (1.8*4.18) = 47.22 C

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